RD Sharma 2020, 2021 Solutions for Class 7 Maths Chapter 17 Constructions are provided here with simple step-by-step explanations. These solutions for Constructions are extremely popular among class 7 students for Maths Constructions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2020, 2021 Book of class 7 Maths Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2020, 2021 Solutions. All RD Sharma 2020, 2021 Solutions for class 7 Maths are prepared by experts and are 100% accurate.
Page No 17.1:
Question 1:
Draw an ∠BAC of measure 50° such that AB = 5 cm and AC = 7 cm. Through C draw a line parallel to AB and through B draw a line parallel to AC, intersecting each other at D. Measure BD and CD.
Answer:
Steps of construction:
- Draw angle BAC = 50 such that AB = 5 cm and AC = 7 cm.
- Cut an arc through C at an angle of 50.
- Draw a straight line passing through C and the arc. This line will be parallel to AB since .
- Alternate angles are equal; therefore the line is parallel to AB.
- Again through B, cut an arc at an angle of 50 and draw a line passing through B and this arc and say this intersects the line drawn parallel to AB at D.
- , since they are alternate angles. Therefore BDAC.
- Also we can measure BD = 7 cm and CD = 5 cm.
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Page No 17.1:
Question 2:
Draw a line PQ. Draw another line parallel to PQ at a distance of 3 cm from it.
Answer:
1. Draw a line PQ.
2. Take any two points A and B on the line.
3. Construct .
4. With A as centre and radius 3 cm cut AE at C.
5. With B as centre and radius 3 cm cut BF at D.
6. Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.
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Page No 17.1:
Question 3:
Take any three non-collinear points A, B, C and draw âABC. Through each vertex of the triangle, draw a line parallel to the opposite side.
Answer:
Steps of construction:
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Page No 17.2:
Question 4:
Draw two parallel lines at a distance 5 cm apart.
Answer:
Steps of construction:
3. Construct .
4. With A as centre and radius 5 cm cut AE at C.
5. With B as centre and radius 5 cm cut BF at D.
6. Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.
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Page No 17.3:
Question 1:
Draw â ABC in which AB = 5.5 cm, BC = 6 cm and CA = 7 cm. Also, draw perpendicular bisector of side BC.
Answer:
Steps of construction:
- Draw a line segment AB of length 5.5 cm.
- From B, cut an arc of radius 6 cm.
- With centre A, draw an arc of radius 7 cm intersecting the previously drawn arc at say, C.
- Join AC and BC to obtain the desired triangle.
- With centre B and radius more than , draw two arcs on both sides of BC.
- With centre C and the same radius as in the previous step, draw two arcs intersecting the arcs drawn in the previous step at X and Y.
- Join XY to get the perpendicular bisector of BC.
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Page No 17.3:
Question 2:
Draw â PQR in which PQ = 3 cm, QR = 4 cm and RP = 5 cm. Also, draw the bisector of ∠Q.
Answer:
Steps of construction:
- Draw a line segment PQ of length 3 cm.
- With Q as centre and radius 4 cm, draw an arc.
- With P as centre and radius 5 cm, draw an arc intersecting the previously drawn arc at R.
- Join PR and QR to obtain the required triangle.
- From Q, cut arcs of equal radius intersecting PQ and QR at M and N, respectively.
- From M and N, cut arcs of equal radius intersecting at point S.
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Page No 17.3:
Question 3:
Draw an equilateral triangle one of whose sides is of length 7 cm.
Answer:
Steps of construction:
- Draw a line segment AB of length 7 cm.
- With centre A, draw an arc of radius 7 cm.
- With centre B, draw an arc of radius 7 cm intersecting the previously drawn arc at C.
- Join AC and BC to get the required triangle.
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Page No 17.3:
Question 4:
Draw a triangle whose sides are of lengths 4 cm, 5 cm and 7 cm. Draw the perpendicular bisector of the largest side.
Answer:
Steps of construction:
- Draw a line segment PR of length 7 cm.
- With centre P, draw an arc of radius 5 cm.
- With centre R, draw an arc of radius 4 cm intersecting the previously drawn arc at Q.
- Join PQ and QR to obtain the required triangle.
- From P, draw arcs with radius more than half of PR on either sides.
- With the same radius as in the previous step, draw arcs from R on either sides of PR intersecting the arcs drawn in the previous step at M and N.
- MN is the required perpendicular bisector of the largest side.
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Page No 17.3:
Question 5:
Draw a triangle ABC with AB = 6 cm, BC = 7 cm and CA = 8 cm. Using ruler and compass alone, draw (i) the bisector AD of ∠A and (ii) perpendicular AL from A on BC. Measure LAD.
Answer:
Steps of construction:
- Draw a line segment BC of length 7 cm.
- With centre B, draw an arc of radius 6 cm.
- With centre C, draw an arc of radius 8 cm intersecting the previously drawn arc at A.
- Join AC and BC to get the required triangle.
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Page No 17.3:
Question 6:
Draw â DEF such that DE = DF = 4 cm and EF = 6 cm. Measure ∠E and ∠F.
Answer:
Steps of construction:
- Draw a line segment EF of length 6 cm.
- With E as centre, draw an arc of radius 4 cm.
- With F as centre, draw an arc of radius 4 cm intersecting the previous arc at D.
- Join DE and DF to get the desired triangle.DF, .
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Page No 17.3:
Question 7:
Draw any triangle ABC. Bisect side AB at D. Through D, draw a line parallel to BC, meeting AC in E. Measure AE and EC.
Answer:
We first draw a triangle ABC with each side = 6 cm.
5. DE is the required parallel line.
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Page No 17.5:
Question 1:
Draw â ABC in which AB = 3 cm, BC = 5 cm and ∠B = 70°.
Answer:
Steps of construction:
- Draw a line segment AB of length 3 cm.
- Draw .
- Cut an arc on BX at a distance of 5 cm at C.
- Join AC to get the required triangle.
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Page No 17.5:
Question 2:
Draw â ABC in which ∠A = 70°, AB = 4 cm and AC = 6 cm. Measure BC.
Answer:
Steps of construction:
- Draw a line segment AC of length 6 cm.
- Draw XAC = 70.
- Cut an arc on AX at a distance of 4 cm at B.
- Join BC to get the desired triangle.
- We see that BC = 6 cm.
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Page No 17.5:
Question 3:
Draw an isosceles triangle in which each of the equal sides is of length 3 cm and the angle between them is 45°.
Answer:
Steps of construction:
- Draw a line segment PQ of length 3 cm.
- Draw .
- Cut an arc on PX at a distance of 3 cm at R.
- Join QR to get the required triangle.
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Page No 17.5:
Question 4:
Draw â ABC in which ∠A = 120°, AB = AC = 3 cm. Measure ∠B and ∠C.
Answer:
Steps of construction:
- Draw a line segment AC of length 3 cm.
- Draw .
- Cut an arc on AX at a distance of 3 cm at B.
- Join BC to get the required triangle.
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By measuring,we get
.
Page No 17.5:
Question 5:
Draw â ABC in which ∠C = 90° and AC = BC = 4 cm.
Answer:
Steps of construction:
- Draw a line segment BC of length 4 cm.
- AT C, draw .
- Cut an arc on CY at a distance of 4 cm at A.
- Join AB.
- ABC is the required triangle.
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Page No 17.5:
Question 6:
Draw a triangle ABC in which BC = 4 cm, AB = 3 cm and ∠B = 45°. Also, draw a perpendicular from A on BC.
Answer:
Steps of construction:
- Draw a line segment AB of length 3 cm.
- Draw an angle of 45 and cut an arc at this angle at a radius of 4 cm at C.
- Join AC to get the required triangle.
- With A as centre, draw intersecting arcs at M and N.
- With centre M and radius more that , cut an arc on the opposite side of A.
- With N as centre and radius the same as in the previous step, cut an arc intersecting the previous arc at E.
- Join AE, it meets BC at D, then AE is the required perpendicular.
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Page No 17.5:
Question 7:
Draw a triangle ABC with AB = 3 cm, BC = 4 cm and ∠B = 60°. Also, draw the bisector of angles C and A of the triangle, meeting in a point O. Measure ∠COA.
Answer:
Steps of construction:
- Draw a line segment BC = 4 cm.
- Draw .
- Draw an arc on BX at a radius of 3 cm cutting BX at A.
- Join AC to get the required triangle.
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Page No 17.6:
Question 1:
Construct â ABC in which BC = 4 cm, ∠B = 50° and ∠C = 70°.
Answer:
Steps of construction:
- Draw a line segment BC of length 4 cm.
- Draw .
- Draw with Y on the same side of BC as X such that BCY = 70.
- Let CY and BX intersect at A.
- ABC is the required triangle.
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Page No 17.6:
Question 2:
Draw â ABC in which BC = 8 cm, ∠B = 50° and ∠A = 50°.
Answer:
Steps of construction:
- Draw a line segment BC of length 8 cm.
- Draw .
- Draw BCY with Y on the same side of BC as X such that BCY = 80.
- Let CY and BX intersect at A.
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Page No 17.6:
Question 3:
Draw â PQR in which ∠Q = 80°, ∠R = 55° and QR = 4.5 cm. Draw the perpendicular bisector of side QR.
Answer:
Steps of construction:
- Draw a line segment QR = 4.5 cm.
- Draw .
- Let QX and RY intersect at P so that PQR is the required triangle.
- With Q as centre and radius more that 2.25 cm, draw arcs on either sides of QR.
- With R as centre and radius more than 2.25 cm, draw arcs intersecting the previous arcs at M and N.
- Join MN; MN is the required perpendicular bisector of QR.
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Page No 17.6:
Question 4:
Construct â ABC in which AB = 6.4 cm, ∠A = 45° and ∠B = 60°.
Answer:
Steps of construction:
- Draw a line segment AB = 6.4 cm.
- Draw .
- Draw ABY with Y on the same side of AB as X such that ABY = 60.
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Page No 17.6:
Question 5:
Draw â ABC in which AC = 6 cm, ∠A = 90° and ∠B = 60°.
Answer:
We can see that ∠A+∠B+∠C = 180°. Therefore ∠C = 180 − 60 − 90 = 30°.
Steps of construction:
- Draw a line segment AC = 6 cm.
- Draw .
- Draw CAY with Y on the same side of AC as X such that CAY = 90.
- Join CX and AY. Let these intersect at B.
- ABC is the required triangle where angle ABC = 60.
Page No 17.8:
Question 1:
Draw a right triangle with hypotenuse of length 5 cm and one side of length 4 cm.
Answer:
Steps of construction:
- Draw a line segment QR = 4 cm.
- Draw QRX of measure 90.
- With centre Q and radius PQ = 5 cm, draw an arc of the circle to intersect ray RX at P.
- Join PQ to obtain the desired triangle PQR.
- PQR is the required triangle.
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Page No 17.8:
Question 2:
Draw a right triangle whose hypotenuse is of length 4 cm and one side is of length 2.5 cm.
Answer:
Steps of construction:
- Draw a line segment QR = 2.5 cm.
- Draw QRX of measure 90.
- With centre Q and radius PQ = 4 cm, draw an arc of the circle to intersect ray RX at P.
- Join PQ to obtain the desired triangle PQR.
- PQR is the required triangle.
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Page No 17.8:
Question 3:
Draw a right triangle having hypotenuse of length 5.4 cm, and one of the acute angles of measure 30°.
Answer:
Let ABC be the right triangle at A such that hypotenuse BC = 5.4 cm. Let .
Therefore
Steps of construction:
- Draw a line segment BC = 5.4 cm.
- Draw angle CBY = 60o.
- Draw angle BCX of measure 30o with X on the same side of BC as Y.
- Let BY and CX intersect at A.
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Page No 17.8:
Question 4:
Construct a right triangle ABC in which AB = 5.8 cm, BC = 4.5 cm and ∠C = 90°.
Answer:
Steps of construction:
- Draw a line segment BC = 4.5 cm.
- Draw BCX of measure 90.
- With centre B and radius AB =5.8 cm, draw an arc of the circle to intersect ray BX at A.
- Join AB to obtain the desired triangle ABC.
- ABC is the required triangle.
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Page No 17.8:
Question 5:
Construct a right triangle, right angled at C in which AB = 5.2 cm and BC = 4.6 cm.
Answer:
Steps of construction:
- Draw a line segment BC = 4.6 cm.
- Draw BCX of measure 90.
- With centre B and radius AB = 5.2 cm, draw an arc of the circle to intersect ray CX at A.
- Join AB to obtain the desired triangle ABC.
- ABC is the required triangle.
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