Question

9. The parallel sides of a trapezium are 12 cm and 36 cm. Its non-parallel sides are equal,
each being 15 cm. find the area of the trapezium.​

Answer 1

Answer:

• 317.8 cm²

Given :-

• The parallel sides of a trapezium are 12 cm and 36 cm.
• non-parallel sides are equal,each being 15 cm.

To Find :-

• Area of trapezium

Diagram :-

• Refer to the attachment

Step-by-step explanation:

According to the diagram,

AD = BE = 12 cm

And EC = 36 - 12 cm = 24 cm

Let the height of ∆DEC be DF = h cm

Now, the height of ∆DEC

$\implies \sf \: h = \sqrt{ \dfrac{ {a}^{2} - {b}^{2} }{2} }$

• a and b are the two sides of the traingle

$\implies \sf \: h = \sqrt{ \dfrac{ {24}^{2} - {15}^{2} }{2} }$

$\implies \sf \: h = \sqrt{13.24}$

Area of the triangle,

$\implies \sf \triangle \: DEC = \dfrac{13.24 \times 24}{2}$

$\implies \sf \triangle \: DEC = 158.9$

Area of the parallelogram,ABED

$\implies \sf \: base \times height$

$\implies \sf \: 12 \times 13.24$

$\implies \sf \: 158.9$

Area of the trapezium,

$\boxed{ \red{ \bold{158.9 + 158.9 = 317.8 {cm}^{2} }}}$

Answer 2

QUESTION

The parallel sides of a trapezium are 12 cm and 36 cm. Its non-parallel sides are equal, each being 15 cm. find the area of the trapezium.

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Solution

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Given that

ABCD is a trapezium in which

• AB =12cm
• DC =36cm
• AD=15cm

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therefore BC= 15cm

→FC = DC- AB

→FC=36-12cm

→FC= 24cm

→FC=EC= 24/2= 12cm

now in triangle BEC

• BC= 15cm
• EC=12cm
• BE=?

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${by\ Pythagoras\ theorem{\boxed{\sf{\red{h²=b²+p²}}}}}$

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$\tt→{BC²= EC²+ BE²}$

$\tt→{15²= 12²+ BE²}$

$\tt→{225= 144+ BE²}$

$\tt→{225-144=BE²}$

$\tt→{81=BE²}$

$\tt→{\sqrt{81}= BE}$

$\tt→{\sqrt{9²}= BE}$

$\tt→{BE= 9cm}$

Now in Trapezium ABCD

• AB =12cm
• DC =36cm
• BE=9cm

$\sf{\red{AREA\ OF\ Trapezium\ ABCD= \frac{1}{2} × (sum\ of\ parallel\ sides )× height}}$

$\sf→{AREA\ OF\ Trapezium\ ABCD= 1/2 (12+36)×9}$

$\tt→{AREA\ OF\ Trapezium\ ABCD= 24×9}$

$\tt→{AREA\ OF\ Trapezium\ ABCD= 216m²}$

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