$ \mathrm{ABCD} $ is a square of side $ 24 \mathrm{~cm} $. EF is parallel to $ B C $ and $ A E=15 \mathrm{~cm} . $ By how much does

(i) the perimeter of AEFD exceed the perimeter of EBCF?
(ii) the area of AEFD exceed the area of EBCF?

Given:

\( \mathrm{ABCD} \) is a square of side \( 24 \mathrm{~cm} \). EF is parallel to \( B C \) and \( A E=15 \mathrm{~cm} . \)

To do:

We have to find by how much does

(i) the perimeter of AEFD exceed the perimeter of EBCF
(ii) the area of AEFD exceed the area of EBCF

Solution:

$BE=AB-AE=(24-15)\ cm=9\ cm$
$CF=BE=9\ cm$

Perimeter of $AEFD=2(15+24)\ cm=2(39)\ cm=78\ cm$

Perimeter of $EBCF=2(9+24)\ cm=2(33)\ cm=66\ cm$

Area of rectangle $BCFE=9\times24\ cm^2=216\ cm^2$
Area of square $ABCD=(24\ cm)^2=576\ cm^2$

Area of rectangle $AEFD=(576-216)\ cm^2=360\ cm^2$

Therefore,

(i) Perimeter of AEFD exceed the perimeter of EBCF by $(78-66)\ cm=12\ cm$.

(ii) The area of AEFD exceed the area of EBCF by $(360-216)\ cm^2=144\ cm^2$.

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Updated on: 10-Oct-2022

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