Question

In the given figure, ABC is a triangle, a circle with diameter BC is drawn, intersecting AB and AC at D and E respectively. If the lengths of AB, AC and CD are 30 cm, 25 cm and 20 cm respectively, then the sum of lengths of AT and BE is

(AT is tangent to the circle)

• A.
  $3\sqrt{2}(5+4\sqrt{2})cm$
• B.
  $3\sqrt{2}(4+5\sqrt{2})cm$
• C.
  $3(5+4\sqrt{2})cm$
• D.
  $6(5+4\sqrt{2})cm$

Answer 1

The correct option is A

$3\sqrt{2}(5+4\sqrt{2})cm$
Given that AB = 30 cm, AC = 25 cm, CD = 20 cm and AT is a tangent to the circle.

Here, $\angle BDC={90}^{\circ }and\angle BEC={90}^{\circ }$
(Angle formed in a semi-circle is a right angle)

Also,$\angle BDC+\angle ADC={180}^{\circ }\left(LinearPair\right)\Rightarrow {90}^{\circ }+\angle ADC={180}^{\circ }\Rightarrow angleADC={90}^{\circ }Also,\angle BEC={90}^{\circ }\Rightarrow \angle AEB={90}^{\circ }\left(Linearpair\right)$

In $\Delta ADC,$ by Pythagoras theorem,

$A{D}^2+D{C}^2=A{C}^2\Rightarrow A{D}^2+(20{)}^2=(25{)}^2\Rightarrow A{D}^2=(25{)}^2–(20{)}^2\Rightarrow A{D}^2=625–400=225\Rightarrow AD=15cm$

Now, Apply tangent-secant theorem,

$\therefore A{T}^2=AD\times AB\Rightarrow A{T}^2=15\times 30\Rightarrow AT=\sqrt{450}\Rightarrow AT=15\sqrt{2}cm.....\left(i\right)Similarly,AE\times AC=A{T}^2\Rightarrow AE\times 25=450\Rightarrow AE=18cmIn\Delta AEB,\Delta AEB={90}^{\circ }.\therefore A{E}^2+E{B}^2=A{B}^2\left(ByPythagorastheorem\right)\Rightarrow \left(18\right)2+\left(BE\right)2=\left(30\right)2\Rightarrow B{E}^2=900–324\Rightarrow B{E}^2=576\Rightarrow BE=24cm\dots ..\left(ii\right)Thus,AT+BE=15\sqrt{2}+24\left[From\right(i\left)and\right(ii\left)\right]=3\sqrt{2}(5+4\sqrt{2})cm$

Hence, the correct answer is option a.