Question

draw a triangle ABC ,right angle at B,such that AB-3cm,BC-4cm.now construct a triangle PBQ similar to triangle ABC,each of whose sides is 7/4 times the corresponding side of triangle ABC​

Answer 1

Steps shown below.

Step-by-step explanation:

Given:

We have to draw a triangle PBQ similar to triangle ABC right angle at B, and AB = 3 cm, BC = 4 cm.

Each of triangle PBQ, sides is $\frac{7}{4}$ times the corresponding side of triangle ABC​.

Steps of Construction:

1. First we have to draw a triangle of given dimensions.

2. Draw a line segment of length BC of length 4 cm.

3. Make an angle of 90° at B.

4. Cut an arc of radius 3 cm taking B as center on the ray BY. The arc cut point is A.

5. Join AC. ABC is the right angled triangle with the given dimensions.

6. Now we have to make a triangle which $\frac{7}{4}$ times of this triangle, that is bigger than this triangle.

7. So we extend the line BC along C to X axis and BA along A to Y axis.  

8. Draw a ray BZ making an acute angle with BC.  

9. Make 7 equal arcs along BZ starting from B then $B_1$ and so on till $B_7$.

10. Join $B _4C$.

11. From $B_7$ draw a ray parallel to $B_4C$ cutting the BX at Q.

12. From Q, draw another ray parallel to CA cutting BY at P.

13. Then PBQ is our required triangle.

Answer 2

Answer:

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Step-by-step explanation:

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