Question

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Answer 1

Answer:

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Answer 2

SOLUTION :

Thus, it is given that in a trapezium ABCD ,

The sides are :

AB = 25 cm

DC = 13 cm

BC = 15 cm

And, AD = 15 cm

Thus,here taking C ,draw CF || AD

And, also draw CE ⊥ AB.

Now, Finding FB .

Thus,

FB = AB - AF

FB = 25 - 13

FB = 12 cm

${\text{\sf{\pink{In the ∆FBC ,FC = BC = 15 cm.}}}}$

Thus,here we can understand that ∆ ABC is an isosceles triangle.

And, CE ⊥ FB.

So, E is the mid-point of FB.

Here,

$:\leadsto\sf{FE=}\dfrac{1}{2}FB$

$:\leadsto\sf\dfrac{1}{2}\times{12cm=6cm.}$

Now, In the ∆CEF (right-angled)

$\sf{CF^2=FE^2+CE^2}$

$\sf{15^2=6^2+CE^2}$

$\sf{CE^2=225-36}$

$\sf{CE^2=189}$

$\sf{CE=\sqrt{189}}$

$\sf\purple{CE=3\sqrt{21}}$

_________________________

Now, Find the Area of trapezium.

$\underline{\boxed{\sf{Area_{(trapezium)}=}\dfrac{1}{2}(AB+DC)\times{CE}}}$

$:\mapsto\dfrac{1}{2}{(25+13)}\times3\sqrt{21}$

$:\mapsto\dfrac{1}{\cancel{2}^1}\times{\cancel{38}^{19}}\times3\sqrt{21}$

$:\mapsto\sf{19\times{3}\sqrt{21}}$

$:\mapsto\sf{{57}\sqrt{21}}$

$\red\bullet{\text{\sf{\blue{Area of trapezium}}}}$$\sf\blue{=57\sqrt{21}}$.