Question

In a parallelogram abcd of area 72 sq cm, the sides cd and ad have lengths 9 cm and 16 cm, respectively. Let p be a point on cd such that ap is perpendicular to cd. Then the area, in sq cm, of triangle apd is

Answer 1

$\mathfrak{The\:Answer\:is}$

Area( parallelogram ABCD) = a side * corresponding altitude = 72 sq cm

=> 9 x AP = 72

=> AP = 72/9 = 8 cm

In right triangle APD

DP = √(AD² - AP² ) = √( 16² - 8²)

= √ 24 * 8 = √( 8² * 3)

=> DP = 8√3 cm

Now, area ( tri APD) = 1/2 * DP * AP

= 1/2 * 8√3 * 8

= 32 √3 cm²



$\boxed{Hope\:This\:Helps}$