Question

8. In the following figure, angle ACB = angle AQP and AP is x cm.

c

i. Find x.

ii. Find the length of QP.

iii. Find the ratio between area of quadrilateral QPCB and area of triangle AQP.

p​

Answer 1

Answer:

i. 5cm

ii. 5cm

iii. 1:8

Step-by-step explanation:

In tri ABC and tri AQP

ang PAQ=ang CAB[Common]

ang ACB=ang AQP[Given]

So, ang APQ=ang ABC[3rd ang]

hence, tri AQP ~ tri ACB [By AAA axiom]

By Similarity,

QP/BC=AP/AB=AQ/AC

so, QP/15=x/15=3/x+4

so, x/15=3/x+4

so, x(x+4)=15×3=45

so, x^2+4x-45=0

so, x^2+(9-5)x-45=0

so, x(x+9)-5(x+9)=0

so, (x+9)(x-5)=0

so, either x=-9 or x=5

As length cannot be negative so x= 5cm [Ans..i]

so, QP/15=5/15

so, QP=5cm [Ans..ii]

ar tri AQP= 1/2×5×5=12.5cm^2

ar tri ABC=1/2×15×15=112.5cm^2

so, ar quad QPCB= 112.5-12.5=100cm^2

hence,

ar quad QPCB:ar tri AQP

=100/12.5

=1/8

so, ratio =1:8 [Ans..iii]