Question

In the figure AP = 3 cm , AR = 4.5 cm, AQ = 6 cm. AB = 5 cm and AC = 10 cm. Find the length of AD. PQ and AD intersect at R.

Answer 1

From the given information we get, (AB/AP) = (AC/AQ)

In triangle ABC, (AB/AP) = (AC/AQ)

By using converse of “Thales theorem” PQ is parallel to BC.

RD = x

In triangle ABD, PR is parallel to BD

AD = AR + RD

AD = 4.5 + x

(AB/AP) = (AD/AR)

(5/3) = (4.5 + x)/4.5

(5 x 4.5)/3 = 4.5 + x

7.5 = 4.5 + x

x = 7.5 – 4.5

x = 3

Here we need to find the length of AD = 4.5 + x

                                                       = 4.5 + 3

                                                       = 7.5 cm

Hope This Helps :)

Answer 2

Answer:

In tri. ABC and tri . APQ

<A=<A (COMMON)

AP/AB=AQ/AC =3/5

So, ABC~APQ

So, <aqr=<acd

(same part of similar tri.)

In AQR. and ADC

<DAC = <RAQ(Common)

<aqr=<acd

So,tri.aqr~adc

So,ar/ad=aq/ac

4.5/ad=3/5

22.5=3ad

AD=7.5 cm

Hope it helps