Question

From the given figure, calculate :
(a) The length of AB
(b) area of trapezium ABCD
(c) area of ΔBCD​

Answer 1

Answer:

ADB is a triangle

thus p2 +b2=h2

√(26)2-(10)2=√36 x16[(a2-b2=(a+b)(a-b)]

=24

thus AB =24 cm

area of trapezium =1/2*h*( sum of parallel sides)

=1/2*10*(24+12)=1/2*10*36=180

area of ABCD= 180cm2

or you could subtract the area , as

area of ABCD-Area of ADB

=180-(1/2*24*10)

180-120=60cm2 =area of BCD

Answer 2

3 rd part do it by yourself