Question

answer me the question in the attachment provided..
@ARMIES anyone of my friends there.. (don't delete my question..please..)​

Answer 1

Answer:

refer the attachment hope its help u

Answer 2

If a line is drawn parallel to one side of a triangle to intersect the

other two sides in distinct points, the other two sides are divided in the same ratio.

As PQ∥BC

So

PB

AP

=

QC

AQ

∠AQP=∠ACB

∠APQ=∠ABC

So by AAA △AQP∼△ACB

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Hence

Area(ABC)

Area(APQ)

=

(AB)

2

(AP)

2

Area(ABC)

Area(APQ)

=

(AP+PB)

2

(AP)

2

Area(ABC)

Area(APQ)

=

(3x)

2

(x)

2

Area(ABC)

Area(APQ)

=1/9

Let Area(APQ)=k

Area(ABC)=9k

Area(BPQC)=Area(ABC)−Area(APQ)=9k−k=8k

Area(BPQC)

Area(APQ)

= 1/8

∴ the ratio of the △APQ and trapezium BPQC =

1/8

Hi army friend ʘ‿ʘ

will you be my Friend

i am Astha