Draw a triangle ABC, right angled at B such that, AB=3cm and BC=4cm. Now construct a triangle similar to AABC, each of whose sides 7/5 times the corresponding side of AABC.
Answers
Answer:
Step-by-step explanation:
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 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 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 and so on till .
10. Join .
11. From draw a ray parallel to 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.
Step-by-step explanation:
BQ=57BC=528=5.6
PB=57AB=521=4.2
On BA extended mark P at 4.2 cm from B on same side of A
On BC extended mark Q at 5.6 cm from B on same side of C
Join P,Q
∴△PBQ is formed.
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