Question

PQ is parallel to BC in a triangle find PQ when PB=3AP and QC=3AQ.​

Answer 1

Answer:

Step-by-step explanation:

R.E.F image

Given, AB=3AP...(1)

we can prove the two triangles

are similar using

1) AA (Angle-Angle)

2) SSS (side-side-side)

3) SAS (side-Angle-side)

In this case.

∠A is common to both the

ΔABC and ΔAPQ

Also, ∠B≅∠P (corresponding angles)

∴ using AA theorem it is proved

that ΔABC∼ΔAPQ

[Note :- Similarity sign is ∼]

Therm : The two similar Δs , the ratio of

their area is the square of the

ratio of their sides.

Hence.

AreaofΔAPQ

AreaofΔABC

​

=

(AP)

2

(AB)

2

​

=(

AP

AB

​

)

2

from (1)

AreaofΔABC

AreaofΔAPQ

​

=(

AB

AP

​

)

2

=(

3

1

​

)

2

=

9

1

​