Question

9. Find the area of a trapezium if its parallel sides are 32 cm and 20 cm, and the non-parallel sides are equ
each being 10 cm.​

Answer 1

Answer:

$\huge\underline\mathfrak\pink{♡SoLuTiOn♡}$

To find:

Area of Trapezium ABCD

Construction:

We draw AE⊥DC and AF||BC(Attachment)

Prove:

Since,AB||DC=>AB||FC

From construction AF||BC

In quadrilateral ABCF, we have:

AB||FC and AF||BC

Therefore, quadrilateral ABCF is parallelogram.

It means, AB=FC (Opposite side of ||gm are equal)

=>AB=FC=20cm------(i)

Now, we can write DF as:

=>DF=DC-FC

=>DF=(32-20)cm

=>DF=12cm-----------(ii)

Since AD=BC=10cm

Therefore, AD=BC

And AF=BC (Opposite side of ||gm are equal)

So we can write,AD=AF

In ∆ADF, AD=AF

Therefore it is isosceles∆.

Perpendicular drawn from the vertex to its opposite side in an isosceles∆ bisects the sides.

DF=2EF

=>12cm=1/2DE (From ii)

=>EF=6cm

In right ∆AEF, we have:

AF²=EF²+AE² (From Pythagoras' Theorem)

=>AE²=(10²-6²)

=>AE²=64

=>AE=8cm

Now area of Trapezium ABCD:

$\frac{1}{2} (sum \: of \: parallel \: sides \times height)$

$= \frac{1}{2} (20 + 32) \times 8 \\ = \frac{1}{2} \times 52 \times 8 \\ = 208 {cm}^{2}$

Here's your required answer ❤️

Answer 2

Answer:

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Step-by-step explanation:

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