$A B C D$ is a square of side $24$ cm. $E F$ is parallel to $B C$ and $A E = 15$ cm.By how much does the area of $A E F D$ exceed the area of $E B C F$ ?

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Solution

$A B C D$ - square of side $24$ cm

$A E = 15$ cm $∴ A B = B C = C D = A D = 24 c m$

$A B = A E + E B$

$24 = 15 + E B$ $∴ E B = 9 c m$

Area of rectangle $E B C F = E B \times B C = l \times b$

$= 9 \times 24$

$= 216 c m^{2}$

Area of rectangle $A E F D = A E \times A D = l \times b$

$= 15 \times 24 = 360$

$∴ A (□ A E F D) - A (□ E B F C) = 360 - 216$

$= 144.$

Area AEFD exceeds Area of EBCF by $144 c m^{2}$

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

ABCD - square of side 24 cmAE=15 cm ∴AB=BC=CD=AD=24cmAB=AE+EB24=15+EB ∴EB=9cmArea of rectangle EBCF=EB×BC=l×b=9×24=216cm2Area of rectangle AEFD=AE×AD=l×b=15×24=360∴A(□AEFD)−A(□EBFC)=360−216=144.Area AEFD exceeds Area of EBCF by 144cm2

$A B C D$ is a square of side $24$ cm. $E F$ is parallel to $B C$ and $A E = 15$ cm.By how much does the area of $A E F D$ exceed the area of $E B C F$ ?