What is the perimeter of the figure ACDE?

8 + $4 \sqrt{3}$ cm

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8 + $4 \sqrt{2}$ cm

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16 + $4 \sqrt{2}$ cm

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4 + $4 \sqrt{2}$ cm

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Solution

The correct option is C
16 + $4 \sqrt{2}$ cm

Perimeter of ACDE = AC + CD + DE + AE

Using Pythagoras theorem,

AC = $\sqrt{A B^{2} + B C^{2}}$

AC = $\sqrt{4^{2} + 4^{2}}$ = $\sqrt{32}$ = $4 \sqrt{2}$ cm

AD = $\sqrt{A C^{2} + C D^{2}}$

AD = $\sqrt{(4 \sqrt{2})^{2} + 4^{2}}$ = $\sqrt{48}$ = $4 \sqrt{3}$ cm

AE = $\sqrt{A D^{2} + D E^{2}}$

AE = $\sqrt{(4 \sqrt{3})^{2} + 4^{2}}$ = $\sqrt{64}$ = 8 cm

$∴$ Perimeter of ACDE = AC + CD + DE + AE

= $4 \sqrt{2}$ + 4 + 4 + 8

= 16 + $4 \sqrt{2}$ cm

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