Question

ABCD is a square of side 24 cm EF is parallel to BC and AE= 15 cm By how much does the perimeter of AEFD exceed the perimeter EBCF? ​

Answer 1

Answer:

i) 12 cm

ii)144cm^2

Step-by-step explanation:

Just refer the above diagram...

i)

\begin{gathered}P (AEFD) = 2×(24+15) < /p > < p > \\ = 2× 39 < /p > < p > \\ = 78 cm < /p > < p > \end{gathered}

P(AEFD)=2×(24+15)</p><p>

=2×39</p><p>

=78cm</p><p>

\begin{gathered} P(EBFC) = 2×(24+9) < /p > < p > \\ = 2×33 \\ 66cm\end{gathered}

P(EBFC)=2×(24+9)</p><p>

=2×33

66cm

Then, according to question their difference will be equal to :

= 78-66 = 12 cm^2.

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ii)

Ar (AEFD) = 24 \times 15 = 360cm {}^{2}Ar(AEFD)=24×15=360cm

2

Ar(EBFC) = 24 \times 9 = 216cm {}^{2}Ar(EBFC)=24×9=216cm

2

Then, according to question their difference will be equal to :

= 360-216 = 144 cm^2.

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Here, remember to apply the formulas of rectangle and not the square.

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Hope it helps...