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Page No 144:
Question 1:
If three angles of a quadrilateral are each equal to 75°, the fourth angle is
(a) 150°
(b) 135°
(c) 45°
(d) 75°
Answer:
Sum of angles of quadrilateral = 360º
Let fourth angle be xº
∴ 75º + 75º + 75º + x = 360º (âµ three angles = 75º each)
⇒ 225º + x = 360º
⇒ x = 135º
Hence, the correct answer is option B.
Page No 144:
Question 2:
For which of the following, diagonals bisect each other?
(a) Square
(b) Kite
(c) Trapezium
(d) Quadrilateral
Answer:
We know, Diagonals of squares bisect each other. But diagonals of kite, trapezium and quadrilateral do not bisect each other.
Hence, the correct answer is option A.
Page No 144:
Question 3:
For which of the following figures, all angles are equal?
(a) Rectangle
(b) Kite
(c) Trapezium
(d) Rhombus
Answer:
In rectangle, all angles are equal, i.e., each 90º.
Hence, the correct answer is option A.
Page No 144:
Question 4:
For which of the following figures, diagonals are perpendicular to each other?
(a) Parallelogram
(b) Kite
(c) Trapezium
(d) Rectangle
Answer:
Diagonals of kite are perpendicular to each other.
Hence, the correct answer is option B.
Page No 144:
Question 5:
For which of the following figures, diagonals are equal?
(a) Trapezium
(b) Rhombus
(c) Parallelogram
(d) Rectangle
Answer:
For rectangle, the diagonals are equal.
Hence, the correct answer is option D.
Page No 144:
Question 6:
Which of the following figures satisfy the following properties?
- All sides are congruent.
- All angles are right angles.
- Opposite sides are parallel.
(a) P
(b) Q
(c) R
(d) S
Answer:
All the properties are satisfy by figure R.
As, figure R has
(i) all its side congruent.
(ii) all angles are right angles.
(iii) opposite sides are parallel.
Hence, the correct answer is option C.
Page No 145:
Question 7:
Which of the following figures satisfy the following property?
- Has two pairs of congruent adjacent sides.
(a) P
(b) Q
(c) R
(d) S
Answer:
We know, Rhombus has two pair of congruent adjacent sides.
Hence, the correct answer is option B.
Page No 145:
Question 8:
Which of the following figures satisfy the following property?
- Only one pair of sides are parallel.
(a) P
(b) Q
(c) R
(d) S
Answer:
Given: Only one pair of sides are parallel. This property is satisfy by figure P.
Hence, the correct answer is option A.
Page No 145:
Question 9:
Which of the following figures do not satisfy any of the following properties?
- All sides are equal.
- All angles are right angles.
- Opposite sides are parallel.
(a) P
(b) Q
(c) R
(d) S
Answer:
Given properties are not satisfy by figure P only.
As figure P does not have
(i) all angles right angles.
(ii) parallel opposite sides.
Hence, the correct answer is option A.
Page No 145:
Question 10:
Which of the following properties describe a trapezium?
(a) A pair of opposite sides is parallel.
(b) The diagonals bisect each other.
(c) The diagonals are perpendicular to each other.
(d) The diagonals are equal.
Answer:
We know, In trapezium, a pair of opposite sides are parallel.
Hence, the correct answer is option A.
Page No 146:
Question 11:
Which of the following is a property of a parallelogram?
(a) Opposite sides are parallel.
(b) The diagonals bisect each other at right angles.
(c) The diagonals are perpendicular to each other.
(d) All angles are equal.
Answer:
In a parallelogram, opposite sides are parallel.
Hence, the correct answer is option A.
Page No 146:
Question 12:
What is the maximum number of obtuse angles that a quadrilateral can have ?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
We know, sum of angles of quadrilateral is 360º
Also, an obtuse angle is more than 90º
but less than 180º
Thus, almost 3 angles of quadrilateral can be obtuse.
Hence, the correct answer is option C.
Page No 146:
Question 13:
How many non-overlapping triangles can we make in a n-gon (polygon having n sides), by joining the vertices?
(a) n − 1
(b) n − 2
(c) n − 3
(d) n − 4
Answer:
We can make (n – 2) non-overlapping triangles in a n-gon by joining the vertices.
Hence, the correct answer is option B.
Page No 146:
Question 14:
What is the sum of all the angles of a pentagon?
(a) 180°
(b) 360°
(c) 540°
(d) 720°
Answer:
We know that,
Sum of interior angles of a polygon = (n − 2) × 180º
∴ Sum of interior angles of a pentagon = (5 − 2) × 180º = 540º
Hence, the correct answer is option C.
Page No 146:
Question 15:
What is the sum of all angles of a hexagon?
(a) 180°
(b) 360°
(c) 540°
(d) 720°
Answer:
We know that,
Sum of angles of a polygon = (n – 2) × 180º
Here n = 6 (for hexagon)
Sum of angles of a hexagon = (6 – 2) × 180º = 720º
Hence, the correct answer is option D.
Page No 146:
Question 16:
If two adjacent angles of a parallelogram are (5x − 5)° and (10x + 35)°, then the ratio of these angles is
(a) 1 : 3
(b) 2 : 3
(c) 1 : 4
(d) 1 : 2
Answer:
In parallelogram , sum of adjacent angles = 180º
∴ Required angles are
Now required ratio
Hence, the correct answer is option A.
Page No 146:
Question 17:
A quadrilateral whose all sides are equal, opposite angles are equal and the diagonals bisect each other at right angles is a __________.
(a) rhombus
(b) parallelogram
(c) square
(d) rectangle
Answer:
A quadrilateral, where all sides are equal, opposite angles are equal and diagonals bisect each other at right angle is rhombus.
Hence, the correct answer is option A.
Page No 146:
Question 18:
A quadrialateral whose opposite sides and all the angles are equal is a
(a) rectangle
(b) parallelogram
(c) square
(d) rhombus
Answer:
In rectangle, opposite sides are equal and angles are equal.
Hence, the correct answer is option A.
Page No 146:
Question 19:
A quadrilateral whose all sides, diagonals and angles are equal is a
(a) square
(b) trapezium
(c) rectangle
(d) rhombus
Answer:
In square, all sides, angles and diagonals are equal.
Hence, the correct answer is option A.
Page No 147:
Question 20:
How many diagonals does a hexagon have?
(a) 9
(b) 8
(c) 2
(d) 6
Answer:
Number of diagonal in a polygon having n sides =
Now, for hexagon (n = 6)
∴ Number of diagonal
âHence, the correct answer is option A.
Page No 147:
Question 21:
If the adjacent sides of a parallelogram are equal then parallelogram is a
(a) rectangle
(b) trapezium
(c) rhombus
(d) square
Answer:
Parallelogram, in which adjacent sides are equal is rhombus.
Hence, the correct answer is option C.
Page No 147:
Question 22:
If the diagonals of a quadrilateral are equal and bisect each other, then the quadrilateral is a
(a) rhombus
(b) rectangle
(c) square
(d) parallelogram
Answer:
In rectangle, diagonals are equal and bisect each other.
Hence, the correct answer is option B.
Page No 147:
Question 23:
The sum of all exterior angles of a triangle is
(a) 180°
(b) 360°
(c) 540°
(d) 720°
Answer:
Sum of exterior angle of a triangle = 360º.
Hence, the correct answer is option B.
Page No 147:
Question 24:
Which of the following is an equiangular and equilateral polygon?
(a) Square
(b) Rectangle
(c) Rhombus
(d) Right triangle
Answer:
In square, all sides and all angle are equal.
Thus, square is a equiangular and equilateral polygon.
Hence, the correct answer is option A.
Page No 147:
Question 25:
Which one has all the properties of a kite and a parallelogram?
(a) Trapezium
(b) Rhombus
(c) Rectangle
(d) Parallelogram
Answer:
Rhombus has all properties of parallelogram and kite.
Hence, the correct answer is option B.
Page No 147:
Question 26:
The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4. The smallest angle is
(a) 72°
(b) 144°
(c) 36°
(d) 18°
Answer:
Let angles be x, 2n, 3x, and 4x
Sum of angles of quadrilateral = 360º
Thus, smallest angle is 36º
Hence, the correct answer is option C.
Page No 147:
Question 27:
In the trapezium ABCD, the measure of ∠D is
(a) 55°
(b) 115°
(c) 135°
(d) 125°
Answer:
We know, sum of supplementary angles = 180º
Then,
Hence, the correct answer is option D.
Page No 147:
Question 28:
A quadrilateral has three acute angles. If each measures 80°, then the measure of the fourth angle is
(a) 150°
(b) 120°
(c) 105°
(d) 140°
Answer:
We know, sum of angles of quadrilateral = 360º
Let fourth angle be xº
Given: 3 angles are 80º each
3 × 80º + xº = 360º
x + 240º = 360º
x = 120º
Hence, the correct answer is option B.
Page No 147:
Question 29:
The number of sides of a regular polygon where each exterior angle has a measure of 45° is
(a) 8
(b) 10
(c) 4
(d) 6
Answer:
Sum of exterior angle of polygon = 360º
Each angle = 45º (given)
∴ Number of sides
Hence, the correct answer is option is A.
Page No 148:
Question 30:
In a parallelogram PQRS, if ∠P = 60°, then other three angles are
(a) 45°, 135°, 120°
(b) 60°, 120°, 120°
(c) 60°, 135°, 135°
(d) 45°, 135°, 135°
Answer:
In parallelogram,
sum of supplementary angles = 180º
Here, ∠P and ∠Q are supplementary
Now,
Hence, the correct answer is option B.
Page No 148:
Question 31:
If two adjacent angles of a parallelogram are in the ratio 2 : 3, then the measure of angles are
(a) 72°, 108°
(b) 36°, 54°
(c) 80°, 120°
(d) 96°, 144°
Answer:
Let the angles be 2x and 3x
In parallelogram,
sum of supplementary angles = 180º
2x + 3x = 180º
⇒ 5x = 180º
⇒ x = 36º
Thus, angles are 72º, and 108º.
Hence, the correct answer is option A.
Page No 148:
Question 32:
If PQRS is a parallelogram, then ∠P − ∠R is equal to
(a) 60°
(b) 90°
(c) 80°
(d) 0°
Answer:
In parallelogram, opposite angles are equal.
Hence, the correct answer is option D.
Page No 148:
Question 33:
The sum of adjacent angles of a parallelogram is
(a) 180°
(b) 120°
(c) 360°
(d) 90°
Answer:
In parallelogram
Sum of adjacent angles = 180º
Hence, the correct answer is option A.
Page No 148:
Question 34:
The angle between the two altitudes of a parallelogram through the same vertex of an obtuse angle of the parallelogram is 30°. The measure of the obtuse angle is
(a) 100°
(b) 150°
(c) 105°
(d) 120°
Answer:
![](/img/site_content/ask-answer/loader.gif)
∠EBF = 30º (given)
Sum of angles of quadrilateral = 360º
∴ ∠EBF + ∠BED + ∠EDF + ∠DFB = 360º
⇒ ∠EDF = 360º – (90 + 90 + 30º)
⇒ ∠EDF = 150º
Hence, the correct answer is option B.
Page No 148:
Question 35:
In the given figure, ABCD and BDCE are parallelograms with common base DC. If BC ⊥ BD, then ∠BEC =
(a) 60°
(b) 30°
(c) 150°
(d) 120°
Answer:
In parallelograms ABCD,
∠BCD = ∠A = 30º (âµ Opposite angles)
In â³DBC, using angle sum property, we get
∠DBC + ∠BCD + ∠BDC = 180º
⇒ ∠BDC = 180º – 120º = 60º
In parallelograms BDCE
∠BEC = ∠BDC = 60º (âµ Opposite angles)
Hence, the correct answer is option A.
Page No 148:
Question 36:
Length of one of the diagonals of a rectangle whose sides are 10 cm and 24 cm is
(a) 25 cm
(b) 20 cm
(c) 26 cm
(d) 3.5 cm
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: Ab = 24 cm, Bc = 10 cm
From phythagoras theorem
Hence, the correct answer is option C.
Page No 148:
Question 37:
If the adjacent angles of a parallelogram are equal, then the parallelogram is a
(a) rectangle
(b) trapezium
(c) rhombus
(d) any of the three
Answer:
If the adjacent angles of a parallelogram are equal, then since, adjacent angles are supplementary. Therefore, each angle will be 90º.
Thus, parallelogram will be rectangle.
Hence, the correct answer is option A.
Page No 148:
Question 38:
Which of the following can be four interior angles of a quadrilateral?
(a) 140°, 40°, 20°, 160°
(b) 270°, 150°, 30°, 20°
(c) 40°, 70°, 90°, 60°
(d) 110°, 40°, 30°, 180°
Answer:
Sum of interior angle of quadrilateral = 360º
So, 140º + 40º + 20º + 160º = 360º
Hence, the correct answer is option A.
Page No 149:
Question 39:
The sum of angles of a concave quadrilateral is
(a) more than 360°
(b) less than 360°
(c) equal to 360°
(d) twice of 360°
Answer:
Sum of interior angles of any polygon (concave or convex) = (n – 2) × 360º
Thus, sum of concave quadrilateral = 360º
Hence, the correct answer is option C.
Page No 149:
Question 40:
Which of the following can never be the measure of exterior angle of a regular polygon?
(a) 22°
(b) 36°
(c) 45°
(d) 30°
Answer:
Sum of exterior angle of polygon = 360º
∴ Measure of each angle =
Hence, n = number of sides.
Thus, measure of each exterior angle will always divide 360º connectively.
Thus, 22º can never be the measure of as exterior angle.
Hence, the correct answer is option A.
Page No 149:
Question 41:
In the figure, BEST is a rhombus, Then the value of y − x is
(a) 40°
(b) 50°
(c) 20°
(d) 10°
Answer:
Here, TS || BE
∠SBE = ∠BST = 40º
⇒ y = 90º (Diameters of a rhombus are intersect each other)
In âTSD
∠STO + ∠OST = ∠EOS (exterior angle property)
⇒ x + 40º = 90º
⇒ x = 50º
So, y – x = 90º – 50º = 40º
Hence, the correct answer is option A.
Page No 149:
Question 42:
The closed curve which is also a polygon is
Answer:
A polygon is a simple closed curve made up of line segments.
Hence, the correct answer is option A.
Page No 149:
Question 43:
Which of the following is not true for an exterior angle of a regular polygon with n sides?
(a)
(b) Exterior angle = 180° − interior angle
(c)
(d)
Answer:
Formula in option D is not correct.
Hence, the correct answer is option D.
Page No 149:
Question 44:
PQRS is a square. PR and SQ intersect at O. Then ∠POQ is a
(a) Right angle
(b) Straight angle
(c) Reflex angle
(d) Complete angle
Answer:
![](/img/site_content/ask-answer/loader.gif)
Diagonal in square intersect each other at 90º.
∴ ∠POQ = 90º
Hence, the correct answer is option A.
Page No 150:
Question 45:
Two adjacent angles of a parallelogram are in the ratio 1:5. Then all the angles of the parallelogram are
(a) 30°, 150°, 30°, 150°
(b) 85°, 95°, 85°, 95°
(c) 45°, 135°, 45°, 135°
(d) 30°, 180°, 30°, 180°
Answer:
Let the angles be x and 5x
x + 5x = 180º (Supplementary angles)
⇒ 6x = 180º
⇒ x = 30º
⇒ 5x = 150º
Thus, angles are 30º, 150º, 30º, 150º
Hence, the correct answer is option A.
Page No 150:
Question 46:
A parallelogram PQRS is constructed with sides QR = 6 cm, PQ = 4 cm and ∠PQR = 90°. Then PQRS is a
(a) square
(b) rectangle
(c) rhombus
(d) trapezium
Answer:
In a parallelogram, if one angle is 90º then all angles will be 90º.
Thus, parallelogram is rectangle.
Hence, the correct answer is option B.
Page No 150:
Question 47:
The angles P, Q, R and S of a quadrilateral are in the ratio 1:3:7:9. Then PQRS is a
(a) parallelogram
(b) trapezium with PQ || RS
(c) trapezium with QR||PS
(d) kite
Answer:
Let the angle be x, 3x, 7x and 9x, then sum of angles of quadrilateral = 360º
⇒ x + 3x + 7x + 9x = 360º
⇒ 20x = 360º
⇒ x = 18º
Thus, angles are 18º, 54º, 126º, and 162º
Since ∠P + ∠S = 180º , ∠Q + ∠R = 180º
Thus, PQ || RS.
Hence, the correct answer is option B.
Page No 150:
Question 48:
PQRS is a trapezium in which PQ||SR and ∠P=130°, ∠Q=110°. Then ∠R is equal to:
(a) 70°
(b) 50°
(c) 65°
(d) 55°
Answer:
Given PQ || SR
∠R + ∠Q = 180º (Supplementary angles)
⇒ ∠R + 110º = 180º
⇒ ∠R = 70º
Hence, the correct answer is option A.
Page No 150:
Question 49:
The number of sides of a regular polygon whose each interior angle is of 135° is
(a) 6
(b) 7
(c) 8
(d) 9
Answer:
Sum of interior angles of polygon = (n – 2) × 180º
Since, polygon is interior,
Hence, the correct answer is option C.
Page No 150:
Question 50:
If a diagonal of a quadrilateral bisects both the angles, then it is a
(a) kite
(b) parallelogram
(c) rhombus
(d) rectangle
Answer:
In rhombus, diagonals bisect both the angles.
Hence, the correct answer is option C.
Page No 150:
Question 51:
To construct a unique parallelogram, the minimum number of measurements required is
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
To construct a parallelogram, we require the measurement of any two non-parallel sides and measure of an angle.
Thus, a minimum of 3 measurement are required.
Hence, the correct answer is option B.
Page No 150:
Question 52:
To construct a unique rectangle, the minimum number of measurements required is
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
Since, in rectangle opposite sides are equal and parallel.
So, we need the measurement of only two adjacent sides.
Thus, we required only two measurement.
Hence, the correct answer is option C.
Page No 150:
Question 53:
Fill in the blanks to make the statements true.
In quadrilateral HOPE, the pairs of opposite sides are __________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Pair of opposite side = HO and PE, PO and HE.
Page No 150:
Question 54:
Fill in the blanks to make the statements true.
In quadrilateral ROPE, the pairs of adjacent angles are __________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In figure Rope, ∠R and ∠O, ∠O and ∠P, ∠P and ∠E, ∠E and ∠R are adjacent angles.
Page No 150:
Question 55:
Fill in the blanks to make the statements true.
In quadrilateral WXYZ, the pairs of opposite angles are __________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In figure WXYZ, ∠W and ∠Y, ∠X and ∠Z are opposite angles.
Page No 151:
Question 56:
Fill in the blanks to make the statements true.
The diagonals of the quadrilateral DEFG are __________ and __________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Hence, the diagonals of the quadrilateral DEFG are DF and EG.
Page No 151:
Question 57:
Fill in the blanks to make the statements true.
The sum of all __________ of a quadrilateral is 360°.
Answer:
Sum of all angles of quadrilateral is 360º.
Page No 151:
Question 58:
Fill in the blanks to make the statements true.
The measure of each exterior angle of a regular pentagon is __________.
Answer:
Measure of each exterior angle of a polygon of n side is .
∴ Measure of each exterior angle
Hence, the measure of each exterior angle of a regular pentagon is 72°.
Page No 151:
Question 59:
Fill in the blanks to make the statements true.
Sum of the angles of a hexagon is __________.
Answer:
We know, sum of interior angle of regular polygon = (n – 2) × 180º.
Here, n = 6
∴ Sum of angles = (6 – 2) × 180º = 720º
Hence, the sum of the angles of a hexagon is 720º.
Page No 151:
Question 60:
Fill in the blanks to make the statements true.
The measure of each exterior angle of a regular polygon of 18 sides is __________.
Answer:
Measure of each exterior angle of a polygon of n sides =
Here n = 18
∴ Measure of exterior angle
Hence, the measure of each exterior angle of a regular polygon of 18 sides is 20°.
Page No 151:
Question 61:
Fill in the blanks to make the statements true.
The number of sides of a regular polygon, where each exterior angle has a measure of 36°, is __________.
Answer:
We know, measure of each exterior of regular polygon =
Each exterior angle = 36º (given)
Hence, the number of sides of a regular polygon, where each exterior angle has a measure of 36°, is 10.
Page No 151:
Question 62:
Fill in the blanks to make the statements true.
![](/img/site_content/ask-answer/loader.gif)
Answer:
The closed curve entirely made up of line segments is called polygon.
Hence, the another name for this shape is polygon.
Page No 151:
Question 63:
Fill in the blanks to make the statements true.
A quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure is __________.
Answer:
Quadrilateral that is not a parallelogram but has exactly two opposite angles is kite.
Page No 151:
Question 64:
Fill in the blanks to make the statements true.
The measure of each angle of a regular pentagon is __________.
Answer:
Sum of angles of regular pentagon = (n – 2) × 180º
= (5 – 2) × 180º
= 540º
Measure of each interior angle = = 108º
Hence, the measure of each angle of a regular pentagon is 108º.
Page No 151:
Question 65:
Fill in the blanks to make the statements true.
The name of three-sided regular polygon is __________.
Answer:
Three sided regular polygon is known as triangle.
Page No 151:
Question 66:
Fill in the blanks to make the statements true.
The number of diagonals in a hexagon is __________.
Answer:
Number of diagonal is regular polygon =
Here, n = 6
Number of diagonals = = 9
Hence, the number of diagonals in a hexagon is 9.
Page No 151:
Question 67:
Fill in the blanks to make the statements true.
A polygon is a simple closed curve made up of only __________.
Answer:
Polygon is closed curve made up of only line segments.
Page No 151:
Question 68:
Fill in the blanks to make the statements true.
A regular polygon is a polygon whose all sides are equal and all __________ are equal.
Answer:
In regular polygon, all sides are equal and all angles are equal.
Page No 151:
Question 69:
Fill in the blanks to make the statements true.
The sum of interior angles of a polygon of n sides is __________right angles.
Answer:
Sum of interior angles of regular polygon = (2n – 4) × 90º
Hence, he sum of interior angles of a polygon of n sides is (2n – 4) right angles.
Page No 151:
Question 70:
Fill in the blanks to make the statements true.
The sum of all exterior angles of a polygon is __________.
Answer:
The sum of all exterior angles of a polygon is 360º.
Page No 151:
Question 71:
Fill in the blanks to make the statements true.
__________ is a regular quadrilateral.
Answer:
Square is a regular quadrilateral.
Page No 151:
Question 72:
Fill in the blanks to make the statements true.
A quadrilateral in which a pair of opposite sides is parallel is __________.
Answer:
In parallelogram, pair of opposite sides are parallel.
Hence, a quadrilateral in which a pair of opposite sides is parallel is parallelogram.
Page No 151:
Question 73:
Fill in the blanks to make the statements true.
If all sides of a quadrilateral are equal, it is a __________.
Answer:
In square and rhombus, all sides are equal.
Hence, if all sides of a quadrilateral are equal, it is a rhombus/square.
Page No 151:
Question 74:
Fill in the blanks to make the statements true.
In a rhombus diagonals intersect at __________ angles.
Answer:
In a rhombus, diagonal intersect at right angles.
Page No 151:
Question 75:
Fill in the blanks to make the statements true.
__________ measurements can determine a quadrilateral uniquely.
Answer:
We know to construct a unique quadrilateral, we require 5 measurement, i.e., four sides and one angle or three sides and two included angles or two adjacent sides and three angles are given.
Hence, 5 measurements can determine a quadrilateral uniquely.
Page No 152:
Question 76:
Fill in the blanks to make the statements true.
A quadrilateral can be constructed uniquely if its three sides and __________ angles are given.
Answer:
Two
A quadrilateral can be construct uniquely if its three sides and two angles are given.
Page No 152:
Question 77:
Fill in the blanks to make the statements true.
A rhombus is a parallelogram in which __________ sides are equal.
Answer:
All
In rhombus, all sides are equal.
Page No 152:
Question 78:
Fill in the blanks to make the statements true.
The measure of __________ angle of concave quadrilateral is more than 180°.
Answer:
One
In concave polygon, atleast one interior angle is more than 180º.
Page No 152:
Question 79:
Fill in the blanks to make the statements true.
A diagonal of a quadrilateral is a line segment that joins two __________ vertices of the quadrilateral.
Answer:
Opposite
Diagonal of a quadrilateral is a line segment that joins two opposite vertices.
Page No 152:
Question 80:
Fill in the blanks to make the statements true.
The number of sides in a regular polygon having measure of an exterior angle as 72° is __________.
Answer:
Sum of exterior angle of regular polygon =
Hence, number of sides is 5.
Page No 152:
Question 81:
Fill in the blanks to make the statements true.
If the diagonals of a quadrilateral bisect each other, it is a __________.
Answer:
Parallelogram.
If diagonal of quadrilateral bisect each other, than it is parallelogram.
Page No 152:
Question 82:
Fill in the blanks to make the statements true.
The adjacent sides of a parallelogram are 5 cm and 9 cm. Its perimeter is __________.
Answer:
Perimeter of parallelogram = 2 (length + breadth)
= 2 (5 + 9)
= 28 cm.
Hence, the perimeter is 28 cm.
Page No 152:
Question 83:
Fill in the blanks to make the statements true.
A nonagon has __________ sides.
Answer:
9
A nonagon has 9 sides.
Page No 152:
Question 84:
Fill in the blanks to make the statements true.
Diagonals of a rectangle are __________.
Answer:
Equal
Diagonal of a rectangle are equal.
Page No 152:
Question 85:
Fill in the blanks to make the statements true.
A polygon having 10 sides is known as __________.
Answer:
Decagon
Polygon having 10 sides is called Decagon.
Page No 152:
Question 86:
Fill in the blanks to make the statements true.
A rectangle whose adjacent sides are equal becomes a __________.
Answer:
Square
Rectangle whose adjacent sides are equal is called square.
Page No 152:
Question 87:
Fill in the blanks to make the statements true.
If one diagonal of a rectangle is 6 cm long, length of the other diagonal is __________.
Answer:
Since diagonals of rectangle are equal.
Hence, if one diagonal of a rectangle is 6 cm long, length of the other diagonal is also 6 cm.
Page No 152:
Question 88:
Fill in the blanks to make the statements true.
Adjacent angles of a parallelogram are __________.
Answer:
Sum of adjacent angles of parallelogram is 180º.
Hence, adjacent angles of a parallelogram are supplementary.
.
Page No 152:
Question 89:
Fill in the blanks to make the statements true.
If only one diagonal of a quadrilateral bisects the other, then the quadrilateral is known as __________.
Answer:
In Kite one diagonal bisects the other.
Hence, if only one diagonal of a quadrilateral bisects the other, then the quadrilateral is known as kite.
Page No 152:
Question 90:
Fill in the blanks to make the statements true.
In trapezium ABCD with AB||CD, if ∠A = 100°, then ∠D = __________.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Since, AB || Cd, ∠A and ∠D are supplementary
∴ ∠A + ∠D = 180º
⇒ 100º + ∠D = 180º
⇒ ∠D = 80º
Hence, in trapezium ABCD with AB||CD, if ∠A = 100°, then ∠D = 80º.
Page No 152:
Question 91:
Fill in the blanks to make the statements true.
The polygon in which sum of all exterior angles is equal to the sum of interior angles is called __________.
Answer:
A quadrilateral is a polygon in which the sum of both interior and exterior angles are equal.
Hence, the polygon in which sum of all exterior angles is equal to the sum of interior angles is called a quadrilateral.
Page No 152:
Question 92:
State whether the statements are true (T) or (F) false.
All angles of a trapezium are equal.
Answer:
False.
In a trapezium, all angles are not equal.
Page No 152:
Question 93:
State whether the statements are true (T) or (F) false.
All squares are rectangles.
Answer:
True.
Every square is a rectangle because it is quadiladeral with all 4 right angles.
Page No 152:
Question 94:
State whether the statements are true (T) or (F) false.
All kites are squares.
Answer:
False.
Because a kite is a quadilateral with two distinct pairs of adjacent congruent sides.
Page No 152:
Question 95:
State whether the statements are true (T) or (F) false.
All rectangles are parallelograms.
Answer:
True
A rectangle is a parallelogram with four right angles.
Page No 153:
Question 96:
State whether the statements are true (T) or (F) false.
All rhombuses are squares.
Answer:
False.
All squares are rhombuses, but not all rhombuses are squares. The opposite interior angles of rhombuses are congruent. Diagonals of a rhombus always bisect each other at right angles.
Page No 153:
Question 97:
State whether the statements are true (T) or (F) false.
Sum of all the angles of a quadrilateral is 180°.
Answer:
False
Sum of all the angles of quadrilateral is 360º.
Page No 153:
Question 98:
State whether the statements are true (T) or (F) false.
A quadrilateral has two diagonals.
Answer:
True.
Quadrilateral is a polygon which has 4 side and two diagonals.
Page No 153:
Question 99:
State whether the statements are true (T) or (F) false.
Triangle is a polygon whose sum of exterior angles is double the sum of interior angles.
Answer:
True.
In triangle,
Sum of interior angle = 180º
Sum of exterior angle = 360º
Hence, triangle is a polygon whose sum of exterior angle is double the sum of interior angles.
Page No 153:
Question 100:
State whether the statements are true (T) or (F) false.
![](/img/site_content/ask-answer/loader.gif)
Answer:
False.
It is not a polygon, because it is not a simple closed curve as it intersects with itself more than once.
Page No 153:
Question 101:
State whether the statements are true (T) or (F) false.
A kite is not a convex quadrilateral.
Answer:
False.
A kite is a convex quadrilateral.
Page No 153:
Question 102:
State whether the statements are true (T) or (F) false.
The sum of interior angles and the sum of exterior angles taken in an order are equal in case of quadrilaterals only.
Answer:
True.
Sum of interior and exterior angles of quadilateral, both are equal to 360º.
Page No 153:
Question 103:
State whether the statements are true (T) or (F) false.
If the sum of interior angles is double the sum of exterior angles taken in an order of a polygon, then it is a hexagon.
Answer:
True.
Since, the sum of exterior angle of a hexagon is 360º and the sum of interior angles of a hexagon is 720º. i.e., double sum of exterior angles.
Page No 153:
Question 104:
State whether the statements are true (T) or (F) false.
A polygon is regular if all of its sides are equal.
Answer:
False.
Because, a polygon is regular if all sides and all angles are equal.
Page No 153:
Question 105:
State whether the statements are true (T) or (F) false.
Rectangle is a regular quadrilateral.
Answer:
False.
Because, in rectangle all angle are equal to 90º but all sides are not equal.
Page No 153:
Question 106:
State whether the statements are true (T) or (F) false.
If diagonals of a quadrilateral are equal, it must be a rectangle.
Answer:
True.
If diagonals of quadrilateral are equal, it must be a rectangle.
Page No 153:
Question 107:
State whether the statements are true (T) or (F) false.
If opposite angles of a quadrilateral are equal, it must be a parallelogram.
Answer:
True.
If opposite angles are equal, it must be a parellogram.
Page No 153:
Question 108:
State whether the statements are true (T) or (F) false.
The interior angles of a triangle are in the ratio 1 : 2 : 3, then the ratio of its exterior angles is 3 : 2 : 1.
Answer:
False.
Let the angles be x, 2x and 3x (âµ Ratio = 1 : 2 : 3)
∴ x + 2x + 3x = 180º
⇒ 6x = 180º
⇒ x = 30º
Thus, angles are 30º, 60º, 90º.
Let exterior angles be .
Now = 180º – 30º
= 150º
Also, = 180º – 60º
= 120º
And = 180º – 120º
= 60º
Now,
= 150º : 120º : 60º = 5 : 4 : 3
Page No 153:
Question 109:
State whether the statements are true (T) or (F) false.
![](/img/site_content/ask-answer/loader.gif)
Answer:
False.
Pentagon has five sides, but here we have 6 sides.
Hence, it is a concave hexagon.
Page No 153:
Question 110:
State whether the statements are true (T) or (F) false.
Diagonals of a rhombus are equal and perpendicular to each other.
Answer:
False.
Because diagonal of rhombus are perpendicular but are not equal to each other.
Page No 153:
Question 111:
State whether the statements are true (T) or (F) false.
Diagonals of a rectangle are equal.
Answer:
True.
Diagonal of a rectangle are equal.
Page No 153:
Question 112:
State whether the statements are true (T) or (F) false.
Diagonals of rectangle bisect each other at right angles.
Answer:
False.
Diagonals of a rectangle do not bisect each other at right angles.
Page No 153:
Question 113:
State whether the statements are true (T) or (F) false.
Every kite is a parallelogram.
Answer:
False.
Kite is not a parallelogram, as its opposite sides are not equal and parallel.
Page No 154:
Question 114:
State whether the statements are true (T) or (F) false.
Every trapezium is a parallelogram.
Answer:
False.
Because in trapezium only one pair of sides is parallel.
Page No 154:
Question 115:
State whether the statements are true (T) or (F) false.
Every parallelogram is a rectangle.
Answer:
False.
Every rectangle is a parellogram but the inverse is not true.
Page No 154:
Question 116:
State whether the statements are true (T) or (F) false.
Every trapezium is a rectangle.
Answer:
False.
Since in a rectangle, opposite sides are equal and parallel but in a trapezium it is not.
So, every rectangle is a trapezium but every trapezium is not a rectangle.
Page No 154:
Question 117:
State whether the statements are true (T) or (F) false.
Every rectangle is a trapezium.
Answer:
True.
Every rectangle is a trapezium as a rectangle satisfy all the properties of trapezium.
Page No 154:
Question 118:
State whether the statements are true (T) or (F) false.
Every square is a rhombus.
Answer:
True.
Because a square possess all the properties of a rhombus but vice-versa is not true.
Page No 154:
Question 119:
State whether the statements are true (T) or (F) false.
Every square is a parallelogram.
Answer:
True.
Because a square possess all the properties of a parallelogram.
Page No 154:
Question 120:
State whether the statements are true (T) or (F) false.
Every square is a trapezium.
Answer:
True.
Because a square possess all the properties of a trapezium.
Page No 154:
Question 121:
State whether the statements are true (T) or (F) false.
Every rhombus is a trapezium.
Answer:
True.
Because a rhombus possess all the properties of a trapezium but vice-versa is not true.
Page No 154:
Question 122:
State whether the statements are true (T) or (F) false.
A quadrilateral can be drawn if only measures of four sides are given.
Answer:
False.
Because we require atleast five measurement to determine a quadilateral uniquely.
Page No 154:
Question 123:
State whether the statements are true (T) or (F) false.
A quadrilateral can have all four angles as obtuse.
Answer:
False.
Because, if all four angles are obtuse, then the sum will exceed 360º.
Page No 154:
Question 124:
State whether the statements are true (T) or (F) false.
A quadrilateral can be drawn if all four sides and one diagonal is known.
Answer:
True
A quadrilateral can be drawn if four sides and one diagonal is known.
Page No 154:
Question 125:
State whether the statements are true (T) or (F) false.
A quadrilateral can be drawn when all the four angles and one side is given.
Answer:
False
We cannot draw a unique-quadrilateral, if four angles and one side is known.
Page No 154:
Question 126:
State whether the statements are true (T) or (F) false.
A quadrilateral can be drawn if all four sides and one angle is known.
Answer:
True
A quadrilateral can be drawn, if all four sides and one angle is known.
Page No 154:
Question 127:
State whether the statements are true (T) or (F) false.
A quadrilateral can be drawn if three sides and two diagonals are given.
Answer:
True
A quadrilateral can be drawn, if three sides and two diagonals are given.
Page No 154:
Question 128:
State whether the statements are true (T) or (F) false.
If diagonals of a quadrilateral bisect each other, it must be a parallelogram.
Answer:
True
If diagonals of a quadrilateral bisect each other, it must be a parallelogram. It is the property of a parallelogram.
Page No 154:
Question 129:
State whether the statements are true (T) or (F) false.
A quadrilateral can be constructed uniquely if three angles and any two sides are given.
Answer:
True
We can construct a unique quadrilateral can be constructed uniquely if three angles and any two sides are given.
Page No 154:
Question 130:
State whether the statements are true (T) or (F) false.
A parallelogram can be constructed uniquely if both diagonals and the angle between them is given.
Answer:
True
We can construct a unique parallelogram, if both diagonals and the angle between them is given.
Page No 154:
Question 131:
State whether the statements are true (T) or (F) false.
A rhombus can be constructed uniquely if both diagonals are given.
Answer:
True
A rhombus can be constructed uniquely, if both diagonals are given.
Page No 154:
Question 132:
Solve the following :
The diagonals of a rhombus are 8 cm and 15 cm. Find its side.
Answer:
Let ABCD be a rhombus.
Given: AC = 15 cm and BD = 8 cm
Since the diagonals of a rhombus bisect each other at 90º.
Therefore, in âAOB,
AB2 = OA2 + OB2
Since it is a rhombus all sides are equal.
Hence, length of each side is 8.5 cm.
Page No 154:
Question 133:
Solve the following :
Two adjacent angles of a parallelogram are in the ratio 1 : 3. Find its angles.
Answer:
Let the adjacent angles of a parallelogram be x and 3x.
Since adjacent angles of parallelogram are supplementary
∴ x + 3x = 180°
⇒ 4x = 180°
⇒ x = 45°
Thus, the angles are 45°, 135°.
Opposite angles in a parallelogram are equal.
Hence, the angles are 45°, 135°, 45°, 135°.
Page No 154:
Question 134:
Solve the following :
Of the four quadrilaterals— square, rectangle, rhombus and trapezium— one is somewhat different from the others because of its design. Find it and give justification.
Answer:
In square, rectangle and rhombus, opposite sides are parallel and equal.
Also, opposite angles are equal, i.e. they all are parallelograms.
But in trapezium, there is only one pair of parallel sides, i.e. it is not a parallelogram.
Hence, trapezium has different design.
Page No 155:
Question 135:
Solve the following :
In a rectangle ABCD, AB = 25 cm and BC = 15. In what ratio does the bisector of ∠C divide AB?
Answer:
We have, AB = 25 cm and BC = 15 cm
Let AO be x, then
OB = AB − x = 25 − x
In rectangle ABCD,
CO is the bisector of ∠C and divides AB. So,
∠OCB = ∠OCD = 45º
In âOCB,
∠CBO + ∠OCB + ∠COB = 180º
⇒ ∠COB = 180º − 90º − 45º
⇒ ∠COB = 45º
In âOCB,
∠OCB = ∠COB
As the angle opposite to equal sides are equal.
∴ OB = BC = 15 cm
⇒ 25 − x = 15
⇒ x = 10 cm
Hence, AO : OB = x : 25 − x = 10 : 15 = 2 : 3.
Page No 155:
Question 136:
Solve the following :
PQRS is a rectangle. The perpendicular ST from S on PR divides ∠S in the ratio 2:3. Find ∠TPQ.
Answer:
Given, ST ⊥ PR and ST divides ∠S in the ratio 2 : 3.
So,
∠TSP =
∠TSR =
Also, by angle sum property of a triangle,
∠TPS = 180º − (∠STP + ∠TSP)
= 180º − (90º + 36º)
= 54º
Now, ∠SPQ = 90º
⇒ ∠TPS + ∠TPQ = 90º
⇒ ∠TPQ = 36º
Page No 155:
Question 137:
Solve the following :
A photo frame is in the shape of a quadrilateral. With one diagonal longer than the other. Is it a rectangle? Why or why not?
Answer:
No, the photo frame cannot be a rectangle as, in a rectangle both the diagonals are of equal lengths.
Page No 155:
Question 138:
Solve the following :
The adjacent angles of a parallelogram are (2x – 4)° and (3x – 1)°. Find the measures of all angles of the parallelogram.
Answer:
(2x – 4)° + (3x – 1)° = 180° (The adjacent angles of a parallelogram are supplementary)
Thus, the adjacent angles are:
Hence, all the angles are 70°, 110°, 70° and 110°.
Page No 155:
Question 139:
Solve the following :
The point of intersection of diagonals of a quadrilateral divides one diagonal in the ratio 1:2. Can it be a parallelogram? Why or why not?
Answer:
No, it can never be a parallelogram, as the diagonals of a parallelogram intersect each other in the ratio 1 : 1.
Page No 155:
Question 140:
Solve the following :
The ratio between exterior angle and interior angle of a regular polygon is 1 : 5. Find the number of sides of the polygon.
Answer:
Let the exterior and interior angle be x and 5x respectively.
Since exterior angle and corresponding interior angles are supplementary.
Then,
x + 5x = 180º
⇒ 6x = 180º
⇒ x = 30º
Hence, there are 12 sides in the polygon.
Page No 155:
Question 141:
Solve the following :
Two sticks each of length 5 cm are crossing each other such that they bisect each other. What shape is formed by joining their end points? Give reason.
Answer:
Sticks can be taken as the diagonals of a quadrilateral.
Now, since they are bisecting each other, therefore the shape formed by joining their end points will be a parallelogram.
Hence, it may be a rectangle or a square depending on the angle between the sticks.
Page No 155:
Question 142:
Solve the following :
Two sticks each of length 7 cm are crossing each other such that they bisect each other at right angles. What shape is formed by joining their end points? Give reason.
Answer:
Sticks can be treated as the diagonals of a quadrilateral.
Since the diagonals are equal and bisecting each other at right angles, therefore the shape formed by joining their end points will be a square.
Page No 155:
Question 143:
Solve the following :
A playground in the town is in the form of a kite. The perimeter is 106 metres. If one of its sides is 23 metres, what are the lengths of other three sides?
Answer:
Let the length of other non-consecutive sides be x metres.
Then, perimeter of the playground = 23 + 23 + x + x
⇒ 106 = 2(23 + x)
⇒ 46 + 2x = 106
⇒ 2x = 106 – 46
⇒ 2x = 60
⇒ x = 30 m
Since a kite has two pairs of equal consecutive sides.
Hence, the lengths of other three sides are 23 m, 30 m and 30 m.
Page No 155:
Question 144:
In rectangle READ, find ∠EAR, ∠RAD and ∠ROD.
Answer:
∠EOA = 180º − 60º = 120º (Linear pair angles)
Now, in âEOA,
∠OEA = ∠OAE
⇒ OE = OA (In triangle sides opposite to equal angles are equal)
So,
∠EOA + ∠OEA + ∠OAE = 180º (Angle sum property of a triangle)
⇒ ∠OAE = ∠EAR = 30º
∠RAD = 90º − 30º = 60º
∠ROD = ∠EOA = 120º (Vertically opposite angles)
Hence, ∠EAR = 30º, ∠RAD = 60º and ∠ROD = 120º.
Page No 156:
Question 145:
In rectangle PAIR, find ∠ARI, ∠RMI and ∠PMA.
Answer:
Given: ∠RAI = 35º
So, ∠PRA = 35º (Alternate interior angles)
In â³ARI, using Angle sum property, we get
∠ARI + ∠RAI + ∠RIA = 180º
⇒ ∠ARI + 90º + 35º = 180º
⇒ ∠ARI = 55º
Now, AM = MI (Diagonals of a rectangle bisect each other)
So, ∠RAI = ∠PIA = 35º (Angle opposite to equal sides of a triangle are equal)
In âAMI, using Angle sum property, we get
∠MAI + ∠MIA + ∠AMI = 180º
⇒ 35º + 35º + ∠AMI = 180º
⇒ ∠AMI = 110º
Now,
∠AMI + ∠RMI = 180º (Linear pair of angles)
⇒ 110º + ∠RMI = 180º
⇒ ∠RMI = 70º
And, ∠PMA = ∠RMI = 70º (Vertically opposite angles)
Hence, ∠ARI = 55º, ∠RMI = 70º and ∠PMA = 70º.
Page No 156:
Question 146:
In parallelogram ABCD, find ∠B, ∠C and ∠D.
Answer:
∠C = ∠A = 80º (Opposite angles in a parallelogram are equal)
Also,
∠A + ∠D = 180º (Adjacent angles in a parallelogram are supplementary)
⇒ 80º + ∠D = 180º
⇒ ∠D = 100º
And,
∠D = ∠B = 100º
Hence, ∠B = 100º, ∠C = 80º and ∠D = 100º.
Page No 156:
Question 147:
In parallelogram PQRS, O is the mid point of SQ. Find ∠S, ∠R, PQ, QR and diagonal PR.
Answer:
∠PQR + ∠RQY = 180º (Linear pair of angles)
⇒ ∠PQR + 60º = 180º
⇒ ∠PQR = 120º
Now,
∠PQR = ∠S = 120º (Opposite angles in a parallelogram are equal)
Also,
∠RQY = ∠R = 60º (SR || PQY, RQ is a transversal and the angles are alternate interior angles)
And,
∠P = ∠R = 60º (Opposite angles in a parallelogram are equal)
Now, PQ = SR = 15 cm (Opposite sides of a parallelogram are equal)
And, PS = QR = 11 cm (Opposite sides of a parallelogram are equal)
Diagonal PR = 2 × PO (Diagonals of a parallelogram bisect each other)
⇒ PR = 12 cm
Hence, ∠S = 120º, ∠R = 60º, PQ = 15 cm, QR = 11 cm and diagonal PR = 12 cm.
Page No 156:
Question 148:
In rhombus BEAM, find ∠AME and ∠AEM.
Answer:
Since, the diagonals of a rhombus bisect each other at 90º, ∠AOM = 90º.
So, in âAMO,
∠AOM + ∠OMA + ∠MAO = 180º
⇒ 90º + ∠OMA + 70º = 180º
⇒ ∠OMA = 20º
⇒ ∠AME = 20º (âµ ∠OMA = ∠AME)
Also, AM = AE so,
Since angles opposite to equal sides are equal.
∴ ∠OMA = ∠AEM = 20º
Page No 156:
Question 149:
In parallelogram FIST, find ∠SFT, ∠OST and ∠STO.
Answer:
In parallelogram FIST, FI || TS and TI is a transversal. So,
∠FIT = ∠STI = ∠STO = 35º
∠SIT = ∠OTF = 25º
And,
∠FOT + ∠SOT = 180º (Linear pair)
⇒ 110º + ∠SOT = 180º
⇒ ∠SOT = 70º
In âSOT,
∠SOT + ∠STO + ∠OST = 180º (Angle sum property)
⇒ 70º + 35º + ∠OST = 180º
⇒ ∠OST = 75º
In âFOT,
∠FOT + ∠OTF + ∠TFO = 180º (Angle sum property)
⇒ 110º + 25º + ∠TFO = 180º
⇒ ∠TFO = 45º = ∠SFT
Hence, ∠SFT = 45º, ∠OST = 75º and ∠STO = 35º.
Page No 157:
Question 150:
In the given parallelogram YOUR, ∠RUO = 120° and OY is extended to point S such that ∠SRY = 50°. Find ∠YSR.
Answer:
∠RUO = ∠RYO = 120° (Opposite angles of a quadrilateral are equal)
And,
∠RYO + ∠RYS = 180° (Linear pair)
⇒ 120° + ∠RYS = 180°
⇒ ∠RYS = 180° − 120° = 60°
In âRSY,
∠YSR + ∠SRY + ∠RYS = 180°
⇒ ∠YSR + 50° + 60° = 180°
⇒ ∠YSR = 70°
Page No 157:
Question 151:
In kite WEAR, ∠WEA = 70° and ∠ARW = 80°. Find the remaining two angles.
Answer:
In âWEA,
∠WEA + ∠EAW + ∠AWE = 180º
⇒ 70º + 2∠EAW = 180º (∠EAW = ∠AWE as angles opposite to equal sides are equal)
⇒ 2∠EAW = 110º
⇒ ∠EAW = 55º = ∠AWE
Similarly in âWRA,
∠WRA + ∠RAW + ∠AWR = 180º
⇒ 80º + 2∠RAW = 180º (∠RAW = ∠AWR as angles opposite to equal sides are equal)
⇒ 2∠RAW = 100º
⇒ ∠RAW = 50º = ∠AWR
So, ∠RWE = ∠AWR + ∠AWE = 50º + 55º = 105º
And, ∠RAE = ∠RAW + ∠EAW = 50º + 55º = 105º
Page No 157:
Question 152:
A rectangular MORE is shown below:
Answer the following questions by giving appropriate reason.
(i) Is RE = OM?
(ii) Is ∠MYO = ∠RXE?
(iii) Is ∠MOY = ∠REX?
(iv) Is âMYO ≅ âRXE?
(v) Is MY = RX?
Answer:
(i) Yes, RE = OM as opposite sides of a rectangle are equal.
(ii) Yes, ∠MYO = ∠RXE as RX and MY are perpendicular to OE.
(iii) Yes, ∠MOY = ∠REX as RE || OM, OE is a transversal and the angles are alternate interior angles.
(iv) Yes, âMYO ≅ âRXE as RE = OM, ∠MYO = ∠RXE and ∠MOY = ∠REX so ASA criteria applies.
(v) Yes, MY = RX by CPCT as âMYO ≅ âRXE.
Page No 158:
Question 153:
In parallelogram LOST, SN⊥OL and SM⊥LT. Find ∠STM, ∠SON and ∠NSM.
Answer:
In âSTM,
∠STM + ∠SMT + ∠TSM = 180º (Angle sum property)
⇒ ∠STM + 90º + 40º = 180º
⇒ ∠STM = 50º
Now,
∠STM = ∠SON = 50º (Opposite angles in a parallelogram are equal)
And, in âSON,
∠SON + ∠SNO + ∠NSO = 180º (Angle sum property)
⇒ 50º + 90º + ∠NSO = 180º
⇒ ∠NSO = 40º
Also,
∠SON + ∠TSO = 180º
⇒ ∠SON + ∠TSM + ∠NSO + ∠NSM = 180º
⇒ 50º + 40º + 40º + ∠NSM = 180º
⇒ ∠NSM = 50º
Hence, ∠STM = 50º, ∠SON = 50º and ∠NSM = 50º.
Page No 158:
Question 154:
In trapezium HARE, EP and RP are bisectors of ∠E and ∠R respectively. Find ∠HAR and ∠EHA.
Answer:
Since, EP and RP are angle bisectors of ∠E and ∠R therefore,
∠E = 2 × 25º = 50º
∠R = 2 × 30º = 60âº
Also, ER || AH and EH is a transversal so,
∠E + ∠H = 180º (Interior angles on the same side of the transversal sum up to 180º)
⇒ 50º + ∠H = 180º
⇒ ∠H = 130º = ∠EHA
Similarly,
∠R + ∠A = 180º
⇒ 60º + ∠A = 180º
⇒ ∠A = 120º = ∠HAR
Hence, ∠HAR = 120º and ∠HAR = 120º.
Page No 158:
Question 155:
In parallelogram MODE, the bisector of ∠M and ∠O meet at Q, find the measure of ∠MQO.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Since, MODE is a parallelogram,
∠EMO + ∠DOM = 180º (Adjacent angles are supplementary)
⇒ ∠OMQ + ∠MOQ = 90º
Now, in âOMQ,
∠OMQ + ∠MOQ + ∠MQO = 180º
⇒ 90º + ∠MQO = 180º
⇒ ∠MQO = 90º
Page No 158:
Question 156:
A playground is in the form of a rectangle ATEF. Two players are standing at the points F and B where EF = EB. Find the values of x and y.
Answer:
In âEFB,
∠EFB + ∠FBE + ∠BEF = 180º (Angle sum property)
⇒ 2∠FBE + 90º = 180º (∠FBE = ∠BFE as angles opposite to equal sides are equal)
⇒ 2∠FBE = 90º
⇒ ∠FBE = 45º = ∠BFE
Since, every angle in a rectangle is 90º therefore,
∠BFE + y = 90º
⇒ 45º + y = 90º
⇒ y = 45º
And,
∠FBE + x = 180º (Linear pair)
⇒ 45º + x = 180º
⇒ x = 135º
Hence, x = 135º and y = 45º.
Page No 158:
Question 157:
In the following figure of a ship, ABDH and CEFG are two parallelograms. Find the value of x.
Answer:
![](/img/site_content/ask-answer/loader.gif)
FE || GC and FG is a transversal so,
∠EFG = ∠CGH = 30º (Corresponding angles)
In parallelogram ABDH,
∠DBA = ∠DHA = 130º (Opposite angles are equal)
Now, ∠DHA + ∠DHG = 180º
⇒ 130º + ∠DHG = 180º
⇒ ∠DHG = 50º
By Angle sum property, we have
∠DHG + ∠CGH + x = 180º
⇒ 50º + 30º + x = 180º
⇒ x = 100º
Page No 159:
Question 158:
A Rangoli has been drawn on a floor of a house. ABCD and PQRS both are in the shape of a rhombus. Find the radius of semicircle drawn on each side of rhombus ABCD.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In âAOB,
(AB)2 = (OA)2 + (OB)2
⇒ (AB)2 = 42 + 32
⇒ (AB)2 = 25
⇒ AB = 5 units
AB is the diameter of the semicircle.
So, radius = units.
Page No 159:
Question 159:
ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. Find ∠AMC.
Answer:
AM is perpendicular to DC.
∴ ∠AMC = 90º
Page No 159:
Question 160:
Quadrilateral EFGH is a rectangle in which J is the point of intersection of the diagonals. Find the value of x if JF = 8x + 4 and EG = 24x – 8.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Here, EG = 2 × JF (Diagonals of a rectangle are equal and bisect each other)
Page No 159:
Question 161:
Find the values of x and y in the following parallelogram.
Answer:
6y = 120º (Opposite angles of a parallelogram are equal)
⇒ y = 20º
Also,
5x + 10º + 6y = 180º (Adjacent angles of a parallelogram are supplementary)
⇒ 5x + 10º + 120º = 180º
⇒ 5x = 50º
⇒ x = 10º
Page No 160:
Question 162:
Find the values of x and y in the following kite.
Answer:
y = 110º (One pair of angles of a kite are equal)
Also,
x + y + 110º + 60º = 360º (Angle sum property of quadrilaterals)
⇒ x + 110º + 110º + 60º = 360º
⇒ x + 280º = 360º
⇒ x = 80º
Hence, x = 80º and y = 110º.
Page No 160:
Question 163:
Find the value of x in the trapezium ABCD given below.
Answer:
Since the adjacent angles of a trapezium are supplementary,
Page No 160:
Question 164:
Two angles of a quadrilateral are each of measure 75° and the other two angles are equal. What is the measure of these two angles? Name the possible figures so formed.
Answer:
Let the equal angles be x.
Then,
The possible figure is a parallelogram since opposite angles are equal.
Page No 160:
Question 165:
In a quadrilateral PQRS, ∠P = 50°, ∠Q = 50°, ∠R = 60°. Find ∠S. Is this quadrilateral convex or concave?
Answer:
∠P + ∠Q + ∠R + ∠S = 360º (Angle sum property)
⇒ 50° + 50° + 60° + ∠S = 360º
⇒ ∠S = 200°
Since, one angle is obtuse, the quadrilateral is concave.
Page No 160:
Question 166:
Both the pairs of opposite angles of a quadrilateral are equal and supplementary. Find the measure of each angle.
Answer:
Let the angles of the quadrilateral be ∠A, ∠B, ∠C, ∠D.
Then, ∠A = ∠C and ∠B = ∠D.
And,
∠A + ∠C = 180º
⇒ ∠A + ∠A = 180º
⇒ 2∠A = 180º
⇒ ∠A = 90º = ∠C
⇒ ∠A = ∠C = 90º
Similarly, ∠B = ∠D = 90º.
Page No 160:
Question 167:
Find the measure of each angle of a regular octagon.
Answer:
Sum of interior angles of a regular polygon = (n − 2) × 180º
So, sum of interior angles of a regular octagon = (8 − 2) × 180º (Number of sides of an octagon = 8)
= 6 × 180º
= 1080º
So, measure of each angle =
Page No 160:
Question 168:
Find the measure of an exterior angle of a regular pentagon and an exterior angle of a regular decagon. What is the ratio between these two angles?
Answer:
Exterior angle of a regular polygon = (Where n is the number of sides of the polygon)
So,
Exterior angle of a regular pentagon = (Number of sides of a pentagon = 5)
Exterior angle of a regular decagon = (Number of sides of a decagon = 10)
∴ Required ratio =
Page No 160:
Question 169:
In the figure, find the value of x.
Answer:
Sum of exterior angles of a regular polygon = 360º
Hence, the value of x is 74.
Page No 161:
Question 170:
Three angles of a quadrilateral are equal. Fourth angle is of measure 120°. What is the measure of equal angles?
Answer:
Let the three equal angles be x each.
Then,
Hence, the measure of each equal angle is 80°.
Page No 161:
Question 171:
In a quadrilateral HOPE, PS and ES are bisectors of ∠P and ∠E respectively. Give reason.
Answer:
**Disclaimer: The data given in this question is insufficient to solve the problem.
Page No 161:
Question 172:
ABCD is a parallelogram. Find the value of x, y and z.
Answer:
![](/img/site_content/ask-answer/loader.gif)
x + 100º = 180º (Linear pair of angles)
⇒ x = 80º
Also, AD || BC and BD is a transversal. So,
z = 30º (Alternate interior angles)
Now, in âBOC,
∠BOC + ∠COB + ∠BCO = 180º
⇒ 80º + 30º + y = 180º
⇒ y = 70º
Hence, x = 80º, y = 70º and z = 30º.
Page No 161:
Question 173:
Diagonals of a quadrilateral are perpendicular to each other. Is such a quadrilateral always a rhombus? Give a figure to justify your answer.
Answer:
Diagonals of a trapezium can also be perpendicular to each other. Hence, the quadrilateral does not necessarily need to be a rhombus.
Page No 161:
Question 174:
ABCD is a trapezium such that AB || CD, ∠A : ∠D = 2 : 1, ∠B : ∠C = 7 : 5. Find the angles of the trapezium.
Answer:
Let the angles be ∠A = 2x, ∠D = x, ∠B = 7y and ∠C = 5y.
Then,
∠A + ∠D = 180º (Adjacent angles are supplementary)
⇒ 2x + x = 180º
⇒ 3x = 180º
⇒ x = 60º
Also,
∠B + ∠C = 180º (Adjacent angles are supplementary)
⇒ 7y + 5y = 180º
⇒ 12y = 180º
⇒ y = 15º
So, angles of the trapezium are
∠A = 2 × 60º = 120º
∠B = 7 × 15º = 105º
∠C = 5 × 15º = 75º
∠D = 1 × 60º = 60º
Page No 161:
Question 175:
A line l is parallel to line m and a transversal p intersects them at X, Y respectively. Bisectors of interior angles at X and Y intersect at P and Q. Is PXQY a rectangle? Given reason.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Since, l || m and p is a transversal,
∠DXY = ∠AYX (Alternate interior angles)
⇒ ∠DXY = ∠AYX
⇒ ∠PXY = ∠QYX ( XP and YQ are angle bisectors)
So, XP || YQ as alternate interior angles are equal.
Similarly, XQ || YP.
Also,
∠CXY + ∠DXY = 180º (Linear pair)
⇒ ∠CXY + ∠DXY = 90º
⇒ ∠QYX + ∠PXY = 90º
⇒ ∠QYP = 90º
Since opposite sides are parallel and one angle is 90º, PXQY forms a rectangle.
Page No 161:
Question 176:
ABCD is a parallelogram. The bisector of angle A intersects CD at X and bisector of angle C intersects AB at Y. Is AXCY a parallelogram? Give reason.
Answer:
![](/img/site_content/ask-answer/loader.gif)
AY || XC (Since, AB || CD)
Also,
∠A = ∠C
⇒ ∠A = ∠C
⇒ ∠XAY = ∠YCX
Also, ∠YCX = ∠CYB (Alternate interior angles)
⇒ ∠CYB = ∠XAY
So, XA || CY (Corresponding angles are equal)
Hence, AXCY is a parallelogram.
Page No 161:
Question 177:
A diagonal of a parallelogram bisects an angle. Will it also bisect the other angle? Give reason.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: Parallelogram ABCD bisects ∠A
To prove: Parallelogram ABCD bisects ∠C
Proof: ∠1 = ∠4 and ∠2 = ∠3 (AB || CD and the angles are alternate interior angles)
But, ∠1 = ∠2 (Since, AC bisects ∠A)
So, ∠3 = ∠4
Hence, proved.
Page No 161:
Question 178:
The angle between the two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 45°. Find the angles of the parallelogram.
Answer:
![](/img/site_content/ask-answer/loader.gif)
∠BED + ∠BEC = 180º (Linear pair)
⇒ 90º + ∠BEC = 180º
⇒ ∠BEC = 90º
In âBEC,
∠BEC + ∠ECB + ∠CBE = 180º (Angle sum property)
⇒ 90º + x + ∠CBE = 180º
⇒ ∠CBE = 90º − x
Similarly in âBAF,
∠BAF + ∠AFB + ∠FBA = 180º (Angle sum property)
⇒ 90º + x + ∠FBA = 180º
⇒ ∠FBA = 90º − x
Now,
∠A + ∠B = 180º (Adjacent angles of a parallelogram are supplementary)
⇒ x + ∠FBA + ∠FBE + ∠CBE = 180º
⇒ x + 90º − x + 45º + 90º − x = 180º
⇒ x = 45º
So,
∠A = 45º
∠C = 45º
∠B = ∠FBA + ∠FBE + ∠CBE
= 90º − x + 45º + 90º − x
= 90º − 45º + 45º + 90º − 45º
= 135º
∠D = ∠B = 135º
Page No 161:
Question 179:
ABCD is a rhombus such that the perpendicular bisector of AB passes through D. Find the angles of the rhombus.
Hint: Join BD. Then â ABD is equilateral.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Let ABCD be a rhombus and let DE be the perpendicular bisector of AB.
Now in âADE and âBDE,
AE = BE (DE is the perpendicular bisector)
∠AED = ∠BED (Both 90º)
DE = DE (Common)
⇒ âADE â âBDE by SAS congruency rule.
⇒ AD = DB = AB (CPCT)
So, âABD is an equilateral triangle.
⇒ ∠DAB = ∠ABD = ∠BDA = 60º
⇒ ∠DCB = ∠DAB = 60º (Opposite angle of a rhombus are equal)
Now, ∠DAB + ∠ABC = 180º (Adjacent angles are supplementary)
⇒ 60º + ∠ABD + ∠DBC = 180º
⇒ 60º + 60º + ∠DBC = 180º
⇒ ∠DBC = 60º
So, ∠ABC = 120º = ∠ADC
Page No 161:
Question 180:
ABCD is a parallelogram. Points P and Q are taken on the sides AB and AD respectively and the parallelogram PRQA is formed. If ∠C = 45°, find ∠R.
Answer:
![](/img/site_content/ask-answer/loader.gif)
∠C = 45° = ∠A (Opposite angles in a parallelogram are equal)
Then,
∠R = ∠A = 45° (Opposite angles in a parallelogram are equal)
Page No 162:
Question 181:
In parallelogram ABCD, the angle bisector of ∠A bisects BC. Will angle bisector of B also bisect AD? Give reason.
Answer:
Let ABCD be a parallelogram and the angle bisector of ∠A meet BC at F.
Construction: Join FE || AB.
Then, ABFE is a parallelogram by construction.
So, ∠1 = ∠6 (Alternate interior angles)
But, ∠1 = ∠2 (AF bisect ∠A)
So, ∠2 = ∠6.
⇒ AB = BF (Sides opposite to equal angles are equal)
⇒ ABFE is a rhombus. (Parallelogram with equal sides is a rhombus)
Now, in âABO and âFBO,
OA = OF (Diagonals of a rhombus bisect each other)
AB = BF (Sides of a rhombus)
OB = OB (Common)
So, âABO â âFBO by SSS criteria.
⇒ ∠3 = ∠4
Also, BF = BC (Given)
⇒ BF = AD (BC = AD)
⇒ EA = AD (BF = EA)
Hence, the angle bisector of B also bisects AD.
Page No 162:
Question 182:
A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. Find ∠BCF.
Answer:
Construction: Join FC
Now,
Measure of each interior angle of a regular polygon =
Measure of each interior angle of a regular pentagon =
So, ∠CBA = 108º
Now,
∠CBA + ∠ABF + ∠FBC = 360º
⇒ 108º + 90º + ∠FBC = 360º
⇒ ∠FBC = 162º
âFBC is an isosceles triangle as FB = BC.
So, ∠BCF = ∠BFC
In âFBC,
∠BCF + ∠BFC + ∠FBC = 180º
⇒ ∠BCF + ∠BCF + 162º = 180º
⇒ 2∠BCF = 18º
⇒ ∠BCF = 9º
Page No 162:
Question 183:
Find maximum number of acute angles which a convex, a quadrilateral, a pentagon and a hexagon can have. Observe the pattern and generalize the result for any polygon.
Answer:
If an angle is acute, then the corresponding exterior angle is greater than 90°.
Now, suppose a convex polygon has four or more acute angles.
Since, the polygon is convex, all the exterior angles are positive. So, sum of the exterior angle is at least the sum of the interior angles.
Now, supplementary of the four acute angles, which is greater than 4 × 90° = 360°
However, this is impossible.
Since, the sum of exterior angle of a polygon must equal to 360° and cannot be greater than it.
It follows that the maximum number of acute angle in convex polygon is 3.
Page No 162:
Question 184:
In the following figure, FD||BC||AE and AC||ED. Find the value of x.
Answer:
Produce DF to meet AB at G.
In âABC,
∠ABC + ∠BCA + ∠CAB = 180º
⇒ 64º + ∠BCA + 52º = 180º
⇒ ∠BCA = 64º
Now BC || FG so,
∠BCA = ∠GFA = 64º (Corresponding angles)
And,
∠GFA + ∠AFD = 180º (Linear pair)
⇒ ∠AFD = 116º
Since opposite angles of a parallelogram are equal.
∴ x = 116º
Page No 162:
Question 185:
In the following figure, AB||DC and AD = BC. Find the value of x.
Answer:
Construction: Draw a line through D parallel to BC that intersects AB at E.
Then, DEBC is a parallelogram. So,
CD = EB = 20 cm
DE = BC = AD = 10 cm
In âADE, AD = DE so,
∠DAE = ∠DEA = 60º
So, âADE is an equilateral triangle by angle sum property.
⇒ AD = AE = 10 cm
Hence, x = AE + EB = 10 + 20 = 30 cm
Page No 162:
Question 186:
Construct a trapezium ABCD in which AB||DC, ∠A = 105°, AD = 3 cm, AB = 4 cm and CD = 8 cm.
Answer:
∠A + ∠D = 180° (Adjacent angles are supplementary)
⇒ ∠D = 75°
Steps of construction:
Step 1: Draw AB = 4 cm
Step 2: Draw AX such that ∠BAX = 105°
Step 3: Cut an arc on AX at 3 cm that cuts AX at D.
Step 4: Draw DY such that ∠ADY = 75°
Step 5: Cut an arc on DY at 8 cm that cuts DY at C.
Step 6: Join BC
ABCD is the required trapezium.
Page No 162:
Question 187:
Construct a parallelogram ABCD in which AB = 4 cm, BC = 5 cm and ∠B = 60°.
Answer:
Since opposite sides of a parallelogram are equal,
AB = DC = 4 cm
BC = AD = 5 cm
Also, since adjacent angles of a parallelogram are supplementary,
∠B + ∠A = 180°
⇒ ∠A = 120°
Steps of construction:
Step 1: Draw AB = 4 cm.
Step 2: Draw BX such that ∠ABX = 60º.
Step 3: Cut an arc at 5 cm on BX and mark it C.
Step 4: Draw AY such that ∠BAY = 120º.
Step 5: Cut an arc at 5 cm on AY and mark it D.
Step 6: Join CD.
ABCD is the required parallelogram.
Page No 162:
Question 188:
Construct a rhombus whose side is 5 cm and one angle is of 60°.
Answer:
If one angle is 60°, the adjacent angle will be 120°. (Adjacent angles are supplementary)
Steps of construction:
Step 1: Draw AB = 5 cm
Step 2: Draw AY such that ∠BAY = 120º
Step 3: Cut an arc at 5 cm on AY and mark it D
Step 4: Draw BX such that ∠ABX = 60º
Step 5: Cut an arc at 5 cm on BX and mark it C
Step 6: Join CD.
ABCD is the required rhombus.
Page No 162:
Question 189:
Construct a rectangle whose one side is 3 cm and a diagonal equal to 5 cm.
Answer:
Steps of construction:
Step 1: Draw AB = 3 cm
Step 2: Draw BX such that ∠ABX = 90º
Step 3: Draw an arc on BX such that it is 5 cm away from point A. Mark the intersection as C.
Step 4: Draw an arc of radius 5 cm with B as centre and with C as centre draw another arc of radius 3 cm.
Step 5: Mark the intersection D.
Step 6: Join AD and CD.
ABCD is the required rectangle.
Page No 162:
Question 190:
Construct a square of side 4 cm.
Answer:
Steps of construction:
Step 1: Draw AB = 4 cm
Step 2: Draw BX such that ∠ABX = 90º
Step 3: With B as the centre and radius = 4 cm, cut an arc at BX and mark the intersection C.
Step 4: With C as the centre and radius = 4 cm, draw an arc.
Step 5: With A as the centre and radius = 4 cm, draw another arc cutting the previous arc.
Step 6: Mark the intersection as D. Join AD and CD.
ABCD is the required square.
Page No 162:
Question 191:
Construct a rhombus CLUE in which CL = 7.5 cm and LE = 6 cm.
Answer:
Steps of construction:
Step 1: Draw LC = 7.5 cm
Step 2: With C as the centre and radius = 7.5 cm, draw an arc.
Step 3: With L as the centre and radius = 6 cm, draw an arc intersecting the previous arc. Mark the point E.
Step 4: Join CE
Step 5: With L as the centre and radius = 7.5 cm, draw an arc.
Step 6: With E as the centre and radius = 7.5 cm, draw an arc intersecting the previous arc. Mark the point U.
Step 7: Join LU and UE.
CLUE is the required rhombus.
Page No 162:
Question 192:
Construct a quadrilateral BEAR in which BE = 6 cm, EA = 7 cm, RB = RE = 5 cm and BA = 9 cm. Measure its fourth side.
Answer:
Steps of construction:
Step 1: Draw BE = 6 cm
Step 2: With B as centre and radius = 5 cm, draw an arc.
Step 3: With E as centre and radius = 5 cm, draw another arc intersecting the previous arc at R.
Step 4: Join BR
Step 5: With E as centre and radius = 7 cm, draw an arc.
Step 3: With B as centre and radius = 9 cm, draw another arc intersecting the previous arc at A.
Step 4: Join EA and RA.
Now when we measure the fourth side i.e., EA by the scale, the length of the EA will be 4 cm.
Page No 163:
Question 193:
Construct a parallelogram POUR in which, PO=5.5 cm, OU = 7.2 cm and ∠O = 70°.
Answer:
Steps of construction:
Step 1: Draw PO = 5.5 cm
Step 2: Construct ∠POX = 70º
Step 3: With O as centre and radius = 7.2 cm, draw an arc cutting OX at U.
Step 4: With U as centre and radius = 5.5 cm, draw an arc.
Step 5: With P as centre and radius = 7.2 cm, draw another arc cutting the previous arc. Mark it R.
Step 6: Join PR and UR
POUR is the required parallelogram.
Page No 163:
Question 194:
Draw a circle of radius 3 cm and draw its diameter and label it as AC. Construct its perpendicular bisector and let it intersect the circle at B and D. What type of quadrilateral is ABCD? Justify your answer.
Answer:
Steps of construction:
Step 1: Draw a circle of radius = OC = 3 cm
Step 2: Draw the diameter AC.
Step 3: Draw the perpendicular bisector of AC and let it cut the circle at points B and D.
Then, the quadrilateral is a cyclic quadrilateral since all its vertices are lying on the circle.
Page No 163:
Question 195:
Construct a parallelogram HOME with HO = 6 cm, HE = 4 cm and OE = 3 cm.
Answer:
Steps of construction:
Step 1: Draw HO = 6 cm
Step 2: With H as centre and radius = 4 cm, draw an arc.
Step 3: With O as centre and radius = 3 cm, draw another arc intersecting the previous arc at E.
Step 4: Join HE
Step 5: With E as centre and radius = 6 cm, draw an arc.
Step 6: With O as centre and radius = 4 cm, draw another arc intersecting the previous arc at M.
Step 4: Join OM and EM.
HOME is the required parallelogram.
Page No 163:
Question 196:
Is it possible to construct a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 5.4 cm, DA = 5.9 cm and diagonal AC = 8 cm? If not, why?
Answer:
No, it is not possible to construct such a quadrilateral.
Given measures are AB = 3 cm, BC = 4 cm, CD = 5.4 cm, DA = 59 cm and AC = 8cm.
Now, AB + BC = 3 + 4 = 7 cm and AC = 8 cm
i.e. the sum of two sides of a triangle is less than the third side, which is absurd.
Page No 163:
Question 197:
Is it possible to construct a quadrilateral ROAM in which RO = 4 cm, OA = 5 cm, ∠O = 120°, ∠R = 105° and ∠A = 135°? If not, why?
Answer:
No, it is not possible to construct such a quadrilateral.
Given measures are RO = 4 cm, OA = 5 cm, ∠O = 120°, ∠R = 105° and ∠A = 135°.
Now, ∠O + ∠R + ∠A = 360°.
i.e. the sum of three angles of a quadrilateral is 360°, which is absurd.
Page No 163:
Question 198:
Construct a square in which each diagonal is 5cm long.
Answer:
Steps of construction:
Step 1: Draw AC = 5 cm
Step 2: With A and C as centres respectively, and radius more than their halves, cut two arcs each intersecting each other.
Step 3: Join the intersection points and let the line cut AC at O.
Step 4: With O as centre, cut off OB = OD = 2.5 cm along the perpendicular bisector.
Step 5: Join AB, BC, CD and DA.
ABCD is the required square.
Page No 163:
Question 199:
Construct a quadrilateral NEWS in which NE = 7cm, EW = 6 cm, ∠N = 60°, ∠E = 110° and ∠S = 85°.
Answer:
Steps of construction:
Step 1: Draw NE = 7 cm
Step 2: Make ∠NEX = 110º
Step 3: With E as a centre and radius 6 cm, draw an arc cutting EX at W.
Step 4: Make ∠EWY = 105º
Step 5: Make ∠ENZ = 60º such that NZ and WY intersect each other at S.
âThen, NEWS is the required quadrilateral.
Page No 163:
Question 200:
Construct a parallelogram when one of its side is 4 cm and its two diagonals are 5.6 cm and 7cm. Measure the other side.
Answer:
Steps of construction:
Step 1: Draw AB = 4 cm
Step 2: With A as centre and radius = 2.8 cm, draw an arc.
Step 3: With B as centre and radius = 3.5 cm, draw an arc. Mark the intersection O.
Step 4: Join OA and OB
Step 5: Produce AO to C such that AO = OC and produce BO to D such that OB = OD.
Step 6: Join AD, BC and DC.
ABCD is the required parallelogram.
Page No 163:
Question 201:
Find the measure of each angle of a regular polygon of 20 sides?
Answer:
Sum of interior angles of an n-polygon = (n − 2) × 180º
Since, n = 20, sum of the interior angles = (20 − 2) × 180º
= 3240º
So, measure of each angle =
Page No 163:
Question 202:
Construct a trapezium RISK in which RI||KS, RI = 7 cm, IS = 5 cm, RK=6.5 cm and ∠I = 60°.
Answer:
Steps of construction:
Step 1: Draw RI = 7 cm
Step 2: Make ∠RIX = 60º
Step 3: With I as centre and radius = 5 cm, draw an arc on IX and mark the intersection S.
Step 4: Make ∠ISY = 120º
Step 5: With R as centre and radius = 6.5 cm, draw an arc on SY and mark the intersection K.
Step 6: Join RK.
RISK is the required parallelogram.
Page No 163:
Question 203:
Construct a trapezium ABCD where AB||CD, AD = BC = 3.2cm, AB = 6.4 cm and CD = 9.6 cm. Measure ∠B and ∠A.
Answer:
Since, CD − AB = 3.2 cm, âBEC is an equilateral triangle with each angle = 60º.
Steps of construction:
Step 1: Draw DC = 9.6 cm
Step 2: With D as the centre, draw an angle of 60º. Draw an arc of radius 3.2 cm and mark the intersection A.
Step 3: Draw AB || CD and AB = 6.4 cm
Step 4: Join CB.
Step 5: Make ∠CBE = 60º.
ABCD is the required trapezium with ∠A = 120º and ∠B = 60º + 60º = 120º.
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