Question

in a trapezium ABCD , AB//DC ,AB = 30cm BC =15cm ,DC =44cm and AD = 13cm find the area of trapezium​

Answer 1

In a trapezium ABCD , AB//DC ,AB = 30cm BC =15cm ,DC =44cm and AD = 13cm. To find the area of trapezium​ :

• To answer this question we first draw a line AE parallel to BC.
• Now we have a triangle ADE, where AE=15cm and DE=14cm.
• Using Heron's Formula we can find the area of this triangle as follows :

A = √(s(s-a)(s-b)(s-c)), where s = (a+b+c)/2

So, we find the area of this triangle using the above formula as ADE as 84 cm sq.

Now, using the formula of triangle of ADE as = 1/2(base x height) = 84 cm sq, we can find the height of the triangle taking base as DE=14cm.

Calculating the question we get, height = 12cm.

Now we know that the area of trapezium as = (a+b)h/2, where a and b are length of parallel sides of trapezium and h is the height.

Using the formula we get,

Area of trapezium = ((30+44) x 12) / 2 = 444 cm sq.

Hence the area of the trapezium is 444 cm sq.

Answer 2

Area of trapezium​ = 444  cm²

Step-by-step explanation:

AB = 30 cm

CD = 44

Draw AM ⊥ CD  & BN ⊥CD

AM = BN = y

MN = 30 cm

DM = x cm

=> CN = 44 - (30 + x) = 14 - x  cm

AD = 13 cm

AD² = AM² + DM²

=> 13² = y² + x²

BC = 15 cm

=> BC² = BN² + CN²

=> 15² = y²  + (14 - x)²

=> 15² - 13² =  (14 - x)² - x²

=> 28 * 2 = (14) (1 4 - 2x)

=> 4 = 14 - 2x

=> 2x = 10

=> x = 5

13² = y² + x²

=>   13² = y² + 5²

=> y² = 144

=> y = 12

AM = BN = 12 cm

Area of Trapezium = (1/2) (AB + CD) * AM

= (1/2)(30 + 44) * 12

= 6 * 74

= 444  cm²

area of trapezium​ = 444  cm²

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