Question

In the given figure if AB = 20 cm DC = 10 cm and BC = 24 cm, then AD = _________ cm.

• A. 30
• B. 13
• C. 25
• D. 26

Answer 1

The correct option is D

26

Drop a perpendicular from D to AB, meeting AB at E.
Then, BEDC is a rectangle.

BE = DC = 10 cm

AE = 20 - 10 = 10 cm

ED = BC = 24 cm

Using Pythagoras' theorem in $△AED$ we have,

$A{D}^2=A{E}^2+E{D}^2$

$A{D}^2={10}^2+{24}^2$

$A{D}^2=(2\times 5{)}^2+(2\times 12{)}^2$

$A{D}^2=2^2(5^2+{12}^2)$

$A{D}^2=2^2\left({13}^2\right)$ $.......(5,12,13)\text{Pythagorean Triplet}$

$AD=26cm$