Question

AB=4, BC=5 CA=3 DRAW TRIANGLE 5/3​

Answer 1

Answer:

draw base AB=4cm

2. with A as center, with a scale of 5cm as radius draw an arc and with B as center, with a scale of 6cm as radius draw an arc.

3. let C be the point of intersection of the above 2 lines.

4. draw a ray AX making an acute angle with line AB on the opposite side if the vertex C.

5. mark 3 points A1,A2,A3 such tha A1A2=A2A3=A3A1

6. join A3B and draw a parallel line B

′

A2

7. draw a line parallel to BC to intersect AC at C

′

Thus AB

′

C

′

is a required triangle.

Justification;

By construction,

AB

AB

′

=

AA3

AA2

=

3

2

also B

′

C

′

∥BC

∴∠AB

′

C

′

=∠ABC

In △AB

′

C

′

,△ABC

∠A=∠A

∠AB

′

C

′

=∠ABC

△AB

′

C

′

∼△ABC

⇒ ABAB ′

= ACAC ′

= BCB C ′∴ ABAB ′ = ACAC , = BCB ′C ,= 32solution