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Rs aggarwal solutions

3: Construct a ∆PQR, in which PQ = 6 cm, QR = 7 cm and PR = 8 cm. Then, construct another a triangle whose sides are 4/5 times the corresponding sides of ∆PQR.

Steps of construction:

- Draw a line segment PQ = 6 cm.
- Draw an arc, using P as a center and radius = 8 cm
- Draw another arc, using Q as a center and radius = 7 cm
- Now, join PR and QR to get △PQR
- Draw a ray PX by making an acute angle, angle QPX
- Divide PX into 4 equal parts 7.
- Join P5Q
- Draw a line P4Q' which is parallel to P5Q
- Similar to step 8, draw a line Q’R’ which is parallel to QR
- Therefore, △PQ’R’ is the required triangle.

Steps of construction:

- Draw a line segment PQ = 6 cm.
- Draw an arc, using P as a center and radius = 8 cm
- Draw another arc, using Q as a center and radius = 7 cm
- Now, join PR and QR to get △PQR
- Draw a ray PX by making an acute angle, angle QPX
- Divide PX into 4 equal parts 7.
- Join P5Q
- Draw a line P4Q' which is parallel to P5Q
- Similar to step 8, draw a line Q’R’ which is parallel to QR
- Therefore, △PQ’R’ is the required triangle.

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