RD Sharma Solutions Class 7 Chapter 17 Constructions

Exercise 17.2 

Question :1. Draw △ABC in which AB = 5.5 cm. BC = 6 cm and CA = 7 cm. Also, draw perpendicular bisector of side BC. 

Solution 1:

Question :2. Draw ∆PQR in which PQ = 3 cm, QR = 4 cm and RP = 5 cm. Also, draw the bisector of ∠Q 

Solution 2:

Question :3. Draw an equilateral triangle one of whose sides is of length 7 cm. 

Solution  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. 

Solution  4:

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. 

Solution  5:

Question :6. Draw △DEF such that DE= DF= 4 cm and EF = 6 cm. Measure ∠E and ∠F. 

Solution  6:

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. 

Solution  7:

Exercise 17.3 

Question :1. Draw △ABC in which AB = 3 cm, BC = 5 cm and ∠B = 70^o. 

Solution  1:

Question :2. Draw △ABC in which ∠A=70^o. AB = 4 cm and AC= 6 cm. Measure BC. 

Solution  2:

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^o. 

Solution  3:

Question :4. Draw △ABC in which ∠A = 120^o, AB = AC = 3 cm. Measure ∠B and ∠C. 

Solution  4:

Question :5. Draw △ABC in which ∠C = 90^o and AC = BC = 4 cm. 

Solution  5:

Construction:

Step 1. Make a 4 cm long line segment BC.

Step 2. Make ∠BCY=90° at C.

Step 3. At A, Make an arc on CY at a distance of 4 cm.

Step 4. connect AB, ABC is the required triangle.

 

 

Question :6. Draw a triangle ABC in which BC = 4 cm, AB = 3 cm and ∠B = 45^o. Also, draw a perpendicular from A on BC. 

Solution  6:

Construction:

Step 1. Make a 3 cm long line segment AB.

Step 2. At C, Make a 45^o angle and cut an arc with a radius of 4 cm at this angle.

Step 3. Connect AC to complete the triangle.

Step 4. Make intersecting arcs at M and N with A as the middle.

Step 5. Cut an arc on the opposite side of A with a middle M and a radius greater than half of MN.

Step 6. Cut an arc intersecting the previous arc at E with N as the center and the same radius as in the previous stage.

Step 7. Connect AE at D, where it meets BC, and AE becomes the appropriate perpendicular.

 

Question :7. Draw a triangle ABC with AB = 3 cm, BC = 4 cm and ∠B = 60^o. Also, draw the bisector of angles C and A of the triangle, meeting in a point O. Measure ∠COA. 

Solution  7:

Construction:

Step 1. Make a BC = 4 cm line segment.

Step 2. Make an angle ∠CBX = 60^o.

Step 3. Make a 3 cm radius arc on BX, cutting BX at A.

Step 4. Connect AC to complete the triangle. 

Angle bisector for angle A:

Step 5. Cut arcs of the same radius with A as the middle, cutting AB and AC at P and Q, respectively.

Step 6. Cut arcs of the same radius from P and Q, intersecting at T.

Step 7. Combine AT and A to get the angle bisector of A. 

Angle bisector for angle C:

Step 8.Cut arcs of the same radius with A as the middle, cutting CB and CA at M and N, respectively.

Step 9. Cut arcs of the same radius intersecting at R from M and N.

Step 10. To get the angle bisector of angle C, connect CR.

Step 11. Make an O at the intersection of CR and AT.

Step 12. Angle ∠COA = 120^o. 

Exercise 17.4 

 

Question :1. Construct ∆ABC in which BC = 4 cm, ∠B = 50^o and ∠C = 70^o.

Solution  1:

Construction:

Step 1. Make a 4 cm long line segment BC.

Step 2. Make ∠CBX, so that it equals ∠CBX=50^o.

Step 3. Make ∠BCY =70^o by placing Y on the same side of BC as X.

Step 4. Assume that CY and BX cross at A.

Step 5. The requisite triangle is ABC.

 

 

Question :2. Draw ∆ABC in which BC = 8 cm, ∠B = 50^o and ∠A = 50^o.

Solution  2:

Construction:

Step 1. Make an 8-cm-long line segment BC.

Step 2. Make ∠CBX so that it equals ∠CBX = 50^o.

Step 3. Make ∠BCY so that ∠BCY = 80^o, with Y on the same side of BC as X.

Step 4. Assume that CY and BX cross at A.

 

Question :3. Draw ∆ABC in which ∠Q = 80^o, ∠R = 55^o and QR = 4.5 cm. Draw the perpendicular bisector of side QR. 

Solution  3:

Construction:

Step 1. Make a QR = 4.5 cm line section.

Step 2. Make ∠RQX = 80° and ∠QRY = 55^o.

Step 3. Assume that QX and RY intersect at P, forming the appropriate triangle PQR.

Step 4. Make arcs on both sides of QR with Q as the middle and a radius greater than 4.5 cm.

Step 5. Make arcs intersecting the previous arcs at M and N with R as the middle and a radius greater than 4.5 cm.

Step 6. Connect MN.

Step 7. MN is the perpendicular bisector of QR that is necessary.

 

Question :4. Construct ∆ABC in which AB = 6.4 cm, ∠A = 45^o and ∠B = 60^o 

Solution  4:

 

Construction: 

Step 1. Make a line segment with the length AB = 6.4 cm. 

Step 2. Make ∠BAX = 45^o angle. 

Step 3. ∠ABY should be drawn with Y on the same side of AB as X, so that ∠ABY = 60^o. 

Step 4. Assume the AX and BY cross at C. 

Step 5. The requisite triangle is ABC. 

 

 

Question :5. Draw ∆ABC in which AC = 6 cm, ∠A = 90^o and ∠B = 60^o 

Solution 5:

Construction:

Step 1. Make a line segment with the length AC = 6 cm.

Step 2. Make ∠ACX = 30^o angle.

Step 3. Make ∠CAY so that Y is on the same side of AC as X and ∠CAY = 90^o.

Step 4. Team up with CX and AY. Allow these to cross at B.

Step 5. The required triangle is ABC, with an angle of ∠ABC = 60^o. 

Exercise 17.5 

Question :1. Draw a right triangle with hypotenuse of length 5 cm and one side of length 4 cm. 

Solution  1:

 

Construction:

Step 1. Make a QR = 4 cm line section.

Step 2. Make an angle 90o ∠QRX.

Step 3. Make an arc of the triangle with a middle Q and a radius PQ = 5 cm to intersect ray RX at P.

Step 4. Combine PQ and PQR to form the ideal triangle PQR.

Step 5. The required triangle is PQR.

 

 

Question :2. Draw a right triangle whose hypotenuse is of length 4 cm and one side is of length 2.5 cm. 

Solution 2:

 

Construction:

Step 1. Make a 2.5-cm-long line segment QR.

Step 2. Make a 90o ∠QRX.

Step 3. Make an arc of the triangle with a middle Q and a radius PQ = 4 cm to intersect ray RX at P.

Step 4. Combine PQ and PQR to form the ideal triangle PQR.

Step 5. The required triangle is PQR.

 

 

Question :3. Draw a right triangle having hypotenuse of length 5.4 cm, and one of the acute angles of measure 30^o 

Solution  3:

 

Let ABC be the right triangle at A such that hypotenuse BC = 5.4 cm. Let C = 30^o.

Therefore ∠A + ∠B + ∠C = 180^o

∠B = 180^o − 30^o− 90^o = 60^o

Construction:

Step 1. Make a BC = 5.4 cm line section.

Step 2. Make a 60^o angle CBY.

Step 3. Make a 30^o angle BCX on the same side of BC as Y, with X on the same side of BC as Y.

Step 4. Assume that BY and CX cross at A.

Step 5. The requisite triangle is then ABC.

 

Question :4. Construct a right triangle ABC in which AB = 5.8 cm, BC = 4.5 cm and ∠C = 90^o. 

Solution  4:

Construction:

Step 1. Make a BC = 4.5 cm line section.

Step 2. Make a BCX with a 90 o angle.

Step 3. Make an arc of the triangle to intersect ray CX at A, with center B and radius AB = 5.8 cm

Step 4. Connect AB to make the triangle ABC you want.

Step 5. The requisite triangle is ABC.

 

Question :5. Construct a right triangle, right angled at C in which AB = 5.2 cm and BC= 4.6 cm. 

Solution  5:

Construction:

Step 1. Make a BC = 4.6 cm line section.

Step 2. Make ∠BCX of measure 90^o

Step 3. Make an arc of the triangle with a center B and a radius AB = 5.2 cm that intersects ray CX at A.

Step 4. To get the desired triangle ABC, connect AB.

Step 5. The requisite triangle is ABC.