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Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to AABC in which PQ = 8 cm. Also justify the construction.

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Steps of construction

1. Draw a line segment BC = 5 cm.

2. Construct OQ the perpendicular bisector of line segment BC meeting BC at P’.

3. Taking B and C as centres draw two arcs of equal radius 6 cm intersecting each other at A

4. Join BA and CA. So, ΔABC is the required isosceles triangle.

5. From B, draw any ray BX making an acute ∠CBX

6. Locate four points B1, B2, B3 and B4 on BX such that BB1 = B1B2 = B2B3 = B3B4

7. Join B3C and from B4 draw a line B4R || B3C intersecting the extended line segment BC at R.

8. From point R, draw RP||CA meeting BA produced at P

Then, ΔPBR is the required triangle.

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