Question

find the area of a trapezium abcd in which ab// cd,ab=18 cm angle b=anglec=90° cd=12cm and ad=10cm.

Answer 1

Given: ABCD is a trapezium with AB // DC. AB = 18 cm, CD = 12 cm, AD = 10 cm,∠B = ∠C = 90° 
To Find: Area of trapezium. 

Construction: Draw altitude DE from D on AB.

Solution:
Area of Trapezium = 1/2 ( sum of parallel sides ) * h 
                             =  1/2 (AB +CD ) *DE

We need to find DE for this.
DE is altitude. Thus, ∆ AED is a right angled triangle. 
Also, AB = 18 cm and CD = EB = 12 cm 
Therefore,
AB = AE + EB
⇒ 18 = AE + 12
⇒ AE = 6 cm

Applying pythagoras theorem,
we get
 H² = P²+ B²
⇒ AD² =  DE² AE²
⇒ 10 ² = P ² + 6²
⇒ P ²= 100 − 36 = 64
 ⇒ P = √64 cm 
P= 8cm

DE = 8 cm

NOW 
Area of Δ = 1/2 (18 + 12  ) *8
= 30 * 4
120 cm²