Question

D
2. The area of an isosceles trapezium is 80 cm2. If the
parallel sides are 4 cm and 16 cm, find the (i) height
and (ii) length of non-parallel sides.
CSF Mathematics 9​

Answer 1

Appropriate Question:-

The area of an isosceles trapezium is 80 cm². If the parallel sides are 4 cm and 16 cm, find the

(i) height and

(ii) length of non-parallel sides.

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Given:-

• Area of an isosceles trapezium = 80 cm².

• Parallel Sides = 4cm and 16cm

To Find:-

• It's Height

• Length of Non-Parallel Sides

Formula Used:-

• ${\boxed{\bf{Area\:of\: Isoceleles\: Trapezium=\dfrac{1}{2}[Sum\:of\: Parallel\:Sides]\times Height}}}$

Solution:-

i) For it's height:-

$\sf :\implies\:Area\:=\dfrac{1}{2}\times [Sum\:of\: Parallel\:Sides]\times Height$

$\sf :\implies\:Area\:=\dfrac{1}{2}\times [AB+CD]\times h$

$\sf :\implies\:80\:=\dfrac{1}{2}\times [4+16]\times h$

$\sf :\implies\:80\times2\:=20\times h$

$\sf :\implies\:h\:=\dfrac{160}{20}$

$\sf :\implies\:h\:=\dfrac{16}{2}$

$\bf :\implies\:h\:=8\:cm$

Hence, The Height of the given Isoceleles trapezium is 8cm.

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ii) For length of it's non-parallel sides:-

Now,

$\sf :\implies\: AD=\sqrt{8^2+6^2}$

$\sf :\implies\: AD=\sqrt{64+36}$

$\sf :\implies\: AD=\sqrt{100}$

$\bf :\implies\: AD=10\:cm$

As we know, The opposite sides of the isoceleles trapezium are equal. And hence, BC = 10cm.

Hence, The Length of its Non-Parallel sides are 10cm each.

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