Question

In the given figure, AE and BC intersect each other at point D. If $\angle CDE={90}^0$, AB = 5cm, BD = 4cm and CD = 9 cm, find AE.

• A. 20 cm
• B. 10 cm
• C. 13 cm
• D. 15 cm

Answer 1

The correct option is D

15 cm

In the given circle, Chords AE and BC interesect each other at D as right angle i.e., $\angle CDE={90}^0$, AB is joined AB = 5 cm, BD= 4cm, CD= 9 cm

Now we have to find AE.

Let DE= xm

Now in right $\Delta ABD$,

$A{B}^2=A{D}^2+B{D}^2$ (Pythagoras Theorem)

$\Rightarrow (5{)}^2=A{D}^2+(4{)}^2\Rightarrow 25=A{D}^2+16$

$\Rightarrow A{D}^2=25-16=9(3{)}^2\therefore AD$ = 3cm

$\because$ Chords AE and BC intersect each other at D inside the circle

$\therefore AD\times DE=BD\times DC\Rightarrow 3\times x=4\times 9$

$\Rightarrow x=\frac{4\times 9}{3}=12cm\Rightarrow DE$ = 12 cm

$\therefore$ AE =AD +DE = 3 + 12 = 15 cm