Question

In an isosceles trapezium the lengths of the parallel sides are 26 cm. and 14 cm, and
the lengths of non parallel sides are 10 cm, each, then find the area of the trapezium.
a) 149 cm², b) 160 cm², c) 180 cm² d) 200 cm²​

Answer 1

Answer:

Through C, draw CF||AD

Also DRAW CE perpendicular to AB .

NOW, FB=AB-AF= 26 - 14 =12 cm.

In ∆FBC, FC =BC=15cm

Also, E is the mid point of FB.

FE =1/2×12= 6cm

In a right angled ∆CEF,

CF^2= FE^2 +CE^2

=> 10^2 = 6^2 + CE^2

=> CE = √98

Area = 1/2 ×40×√98

= 20√98

maybe ur options are wrong

Answer 2

1. 200 cm
2. ${?}^{2}$