Question

ans q16


plzzs help test tmrr

Answer 1

In trap ABCD

ADllBC and AB = 28 , BC = 78, CD = 30, AD = 52

Draw DE ll AB and DF perpendicular to BC

NOW ABED is ll gm

EC = 78 - 52 = 26

THus in triangle DEF

a = 26, b = 28 , c = 30

26 + 28 + 30 / 2 = 13 + 14 + 15 42

Area of triangle DEC = 

T = $\sqrt{s(s-a) (s-b) (s-c)}$

= square root of [ (42(42 - 26)(42 - 28)(42 - 30) ]

= 336

Again area of triangle DEF = 1/2 * 26 * DF = 336

DF = 336 / 13

Area of trap ABCD = 1/2(AD + BC) *  DF

= 1/2 (52 + 78) * DF

1/2(52 + 78) * 336/13

= 65 * 336 / 13

= 336 * 5

= 1680cm2