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

Solution:

Steps of Construction

1. Construct a line segment BC = 5 cm

2. Draw OQ which is the perpendicular bisector of line segment BC meeting BC at P’.

3. Consider B and C as centres construct two arcs of equal radius 6 cm which intersects each other at the point A

4. Now join BA and CA. ∆ABC is the required triangle.

5. From the point B, construct any ray BX which makes an acute angle ∠CBX

6. Let us locate four points B₁, B₂, B₃ and B₄ where BB₁ = B₁B₂ = B₂B₃ = B₃B₄

7. Join B₃C and from the point B₄ construct a line B₄R||B₃C which intersects the extended line

8. From the point R, construct RP||CA which meets BA produced at P

∆ PBR is the required triangle.

Justification:

Consider BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = x

From the construction

B₄R||B₃C

BC/CR = BB₃/B₃B₄ = 3x/x = 3

BC/CR = 3/1

Here

BR/BC = (BC + CR)/BC = BC/BC + CR/BC

= 1 + 1/3

= 4/3

From construction, RP||CA

So rABC is congruent to rPBR [by AAA criterion]

PB/AB = RP/CA = 4/3

The new triangle PBR is similar to isosceles triangle ABC and its sides are 4/3 times of the corresponding sides of ABC.

Therefore, a triangle PQR similar to triangle ABC is drawn.

✦ Try This: Draw an isosceles triangle ABC in which AB = AC = 3 cm and BC = 4 cm. Construct a triangle PQR similar to ∆ ABC in which PQ = 6cm. Also justify the construction.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 11

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NCERT Exemplar Class 10 Maths Exercise 10.4 Problem 4

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

Summary:

An isosceles triangle ABC with AB = AC = 6 cm and BC = 5 cm is drawn. A triangle PQR similar to ∆ ABC in which PQ = 8 cm is constructed

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