Question

From the given figure,find the area of trapezium ABCD​

Answer 1

Step-by-step explanation:

Given, trapezium ABCD such that AB∥CD.

Draw a perpendicular from A on CD produced to meet at E succh that ABCE is a rectangle.

In △AED

AE=4cm and AD=5cm

AD^2 =AE^2+ED^2

(Pythagoras Theorem)

5^2 =4^2+ED^2

25−16=ED^2

ED=3cm

Now,

A(□ABCD)=A(□ABCD)−A(△ADE)

A(□ABCD)=(AB×BC)−1/2 AE×ED

A(□ABCD)=5×4=1/2 (3)(4)

A(□ABCD)=20−6

A(□ABCD)=14cm

Answer 2

Answer:

$\huge\bold\purple\bigstar{AnswEr}$

Given, trapezium ABCD such that AB∥CD. Draw a perpendicular from A on CD produced to meet at E succh that ABCE is a rectangle.

In △AED

AED

AE=4cm and AD=5cm

AD² =AE² +ED²(Pythagoras theorem)

5²=4²+ED²

25-16=ED²

ED=3cm

Now,

A(□ABCD)=A(□ABCD)−A(△ADE)

A(□ABCD)=(AB×BC)−½AE × ED

A(□ABCD)=5×4=½(3)(4)

A(□ABCD)=20-6

A(□ABCD)=14cm²