Question

The given diagram shows two isosceles triangles which are similar also. In the given diagram, PQ and BC are not parallel and AP = PQ.

Given that PC =4 cm, AQ=3 cm, QB =12 cm, BC =15 cm.

Calculate the length of AP.

• A. 4 cm
• B. 8 cm
• C. 5 cm
• D. 6 cm

Answer 1

The correct option is C

5 cm

$\because \Delta APQ\sim \Delta ABC\left(given\right)$

$\therefore \frac{AQ}{AC}=\frac{AP}{BC}$

$\Rightarrow \frac{3}{AP+PC}=\frac{AP}{15}(\because AQ=3,BC=15)$

$\Rightarrow \frac{3}{AP+4}=\frac{AP}{15}(\because PC=4)$

AP (AP+4)=45

Let AP =x, then

$x(x+4)=45\Rightarrow x^2+4x-45=0$

$\Rightarrow x^2+9x-5x-45=0$

$\Rightarrow x(x+9)-5(x+9)=0$

$\Rightarrow (x+9)(x-5)=0$

Either x+9=0, then x =-9 which is not possible.

Or x-5=0, then x=5

$\therefore AP=5cm$