Question

Two parallel sides of a trapezium are 120cm, 150cm and other sides are 50cm and 52 cm. Find the area of the trapezium.​

Answer 1

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Refer to the attachment.....

Formula used

1) Area of a trapezium of parallel sides a and b and of altitude h is ...

=[(1/2)×(a+b)×h]

2)Area of a triangle of sides a, b, c and of half perimeter s is ....

$\implies\sqrt{s(s-a)(s-b)(s-c)}$

3)Area of a triangle of base b and height h is..

$\implies\frac{1}{2}\times b \times h$

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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Answer 2

The area of trapezium is 6576 cm².

Two parallel sides of a trapezium are 120cm, 154cm [you made a mistake in typing its 154 instead of 150 cm ] and other sides are 50cm and 52cm.

We have to find the area of the trapezium.

We know the area of trapezium is given by,

$\text{Area of Trapezium}=\frac{1}{2}(\text{sum of two parallel sides})\times\text{height}$

In question, Two parallel sides of the trapezium are given but height is not given. So we need to find height of trapezium to find the area of it.

       First draw a rough diagram of given trapezium as shown in figure. Here ABCD is a trapezium where AB and CD are two parallel sides of it and AD and BC are two other sides. Let AE and BF are perpendicular upon CD from A and B respectively.

Let DE = x and FC = y.

∴ x + y = CD - AB = 154 - 120 = 34 ...(1)

In Δ ADE,

x² + h² = 50²   [ From Pythagoras theorem]

In BFC,

y² + h² = 52²  [ From pythagoras theorem ]

∴ (y² + h²) - (x² + h²)  = 52² - 50² = 204

⇒(y - x)(y + x) = 204

⇒y - x = 204/34 = 6

⇒ y - x = 6 ...(2)

From equations (1) and (2) we get,

y = 20 and x = 14

∴ h² = 50² - x² = 50² - 14² = (50 - 14)(50 + 14) = 36 × 64

⇒ h = 6 × 8 = 48 cm

$\text{Now, the area of trapezium}=\frac{1}{2}(120+154)\times48$

$=274\times24$

= 6576 cm²

Therefore the area of trapezium is 6576 cm².

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