In the given figure, ABC is a triangle and GHED is a rectangle. If BC=12 cm, HE=6 cm, FC=BF and altitude AF is 24 cm, then find the area of the rectangle.

It is given that ABC is a triangle and GHED is a rectangle.

BC = 12 cm

HE = 6 cm

FC = BF

and altitude AF is 24 cm.

ΔAFC

and

ΔEHC

∠ AFC = ∠ EHC

(Each

90^{\circ}

)

∠ FCA = ∠ ECH

(Common)

\Rightarrow ΔAFC \sim ΔEHC

(By AA similarity criterion)
We know that the corresponding parts of similar triangles are propositional.
Thus,

\frac{AF}{EH} = \frac{FC}{HC}

So,

\frac{AF}{FC} = \frac{EH}{HC}

\Rightarrow \frac{24}{6} = \frac{6}{HC}

\Rightarrow HC = 6 \times \frac{6}{24}

\Rightarrow HC = \frac{36}{24}

\Rightarrow HC = 1.50 cm

From the figure,

FH = FC - HC

\Rightarrow FH = 6 - 1.50

\Rightarrow FH = 4.50 cm

Now,

GH = 2 FH

\Rightarrow GH = 2 (4.50)

\Rightarrow 9 cm

Thus, the area of rectangle GHED is,

A = lengt h \times breadt h

\Rightarrow A = 9 \times 6

\Rightarrow A = 54 {cm}^{2}

Therefore, the area of rectangle GHED is

54 {cm}^{2}

.
Hence, option 1 is correct.

In the given figure, ABC is a triangle and GHED is a rectangle. If $BC = 12 cm$ , $HE = 6 cm$ , $FC = BF$ and altitude AF is 24 cm, then find the area of the rectangle.