Question

the length of the parallel sides of an isosceles trapezium are 20 cm and 32 cm the length of each non Parallel side is 10 cm find the area of trapezium

Answer 1

Let parallel sides be a and b
Non parallel sides be c and d

Now, k = |a-b| = 32-20 = 12 cm

s = (c+d+k)/2
s = (10+10+12)/2
s = 16

Now, Area of trap.
= (a+b)/k × √(s(s-c)(s-d)(s-k))
= 52/12 × √((16)(6)(6)(4))
= 13/3 × (4×6×2)
= 13 × 4 × 2 × 2
= 208 cm^2

This formula has been made by me.

$\frac{(a + b)}{k} \sqrt{s(s - c)(s - d)(s - k)}$

Where a and b are parallel sides and c and d are non parallel sides.
k is difference of Parallel sides,
s is half the sum of non parallel sides and k.

Answer 2

​Clearly,AECD is a parallelogram
EC=AD=10cm
also,AE=EC=20cm
EB=AB-AE
= (32-20)cm =12cm

triangle CEB is isosceles
therefore, EL=1/2 (EB)=6cm
CL=√​CE^2-EL^2
=√​20^2-6^2
=√​436

Area of trapezium,
ABCD=1/2 (AB+CD)×distance between parallel sides
=1/2 (32+20)×√​436
=1/2×52×√​436