The parallel sides of a trapezium are 20 cm and30 m and its non-parall

1.	The parallel sides of a trapezium are 20 cm and30 m and its non-parallel sieles avend the area of the trapezium
Answer» Let ABCD be the trapezium in which AB is parallel to DC. Let AB = 30cm, DC=20 cm,and the non parallel sides AD =6cm and BC = 8cm. Draw perpendiculars DE and CF from D and C to AB respectively. So DE = CF. DC=EF = 20cm. Let AE = x cm . Then BF = 10-x. In right triangle ADE DE^2 = AD^2 - AE^2 In right triangle BCF , CF^2 = BC^2 - BF^2 => AD^2-AE^2 = BC^2-BF^2 => 36-x^2 = 64 -(10-x)^2 => 20x = 72 => x=3.6 So DE^2 = 36 - 3.6^2 = 36 - 12.96 =23.04 So DE =√23.04 = 4.8 cm So area of trapezium = (1/2)h(a+b) = (1/2)×4.8×(30+20) = 2.4×50 =120cm² Like my answer if you find it useful!

The parallel sides of a trapezium are 20 cm and30 m and its non-parallel sieles avend the area of the trapezium

Answer»

Let ABCD be the trapezium in which AB is parallel to DC.

Let AB = 30cm, DC=20 cm,and the non parallel sides AD =6cm and BC = 8cm.

Draw perpendiculars DE and CF from D and C to AB respectively. So DE = CF. DC=EF = 20cm.

Let AE = x cm . Then BF = 10-x.

In right triangle ADE DE^2 = AD^2 - AE^2

In right triangle BCF , CF^2 = BC^2 - BF^2

=> AD^2-AE^2 = BC^2-BF^2

=> 36-x^2 = 64 -(10-x)^2 => 20x = 72 => x=3.6

So DE^2 = 36 - 3.6^2 = 36 - 12.96 =23.04

So DE =√23.04 = 4.8 cm

So area of trapezium = (1/2)h(a+b)

= (1/2)×4.8×(30+20) = 2.4×50 =120cm²

Like my answer if you find it useful!