Question

In the given figure ABCD is pentagonal park, in which DE = DC, AB = BC = CE = EA = 25m and its total height is 41 m. Find the area of the park.

Answer 1

Step-by-step explanation:

HEIGHT OF THE TRIANGLE = 41-25

                                                =16cm

AREA OF PARK = AREA OF SQUARE ABCE+ AREA OF TRIANGLE DEC

                          =25*25+1/2*25*16

                          =625+200

                          =825 cm²

Answer 2

$\huge \mathfrak \red{answer}$

$\bf \underline \pink{area \: of \: park = 825 {cm}^{2}}$

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❁Question:

In the given figure ABCD is pentagonal park, in which DE = DC, AB = BC = CE = EA = 25m and its total height is 41 m. Find the area of the park.

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※Step by step explanation:

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✥According to question:

$\tt{height \: of \: triangle = 41 - 25 = 16cm}$

formula:

$\bf{ \boxed{ \underline{ \green{ \tt{area \: of \: park = area \: of \: sqaure \: ABCE + area \: of \: triangleDEC \: }}}}}$

from the question

• DE = DC
• AB = BC = CE = EA = 25m
• total height is 41 m.
• ABCD is pentagonal park

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• Area of the park = ?

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$\rm \red{ = 25 \times 25 + \frac{1}{2} \times 25 \times 16}$

$\rm \red{ = 625 + 200}$

$\rm \red{ = 825 {cm}^{2}}$

I hope it's help uh