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A kite in the shape of a square with a diagonal $32\ cm$ and an isosceles triangle of base $8\ cm$ and sides $6\ cm$ each is to be made of three different shades as shown in Fig. $12 .17$ How much paper of each shade has been used in it? Fig. $ 12.16 $
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Given: A kite in the shape of a square with a diagonal $32\ cm$ and an isosceles triangle of base $8\ cm$ and sides $6\ cm$ each is to be made of three different shades as shown in the fig.

To do: To find how much paper of each shade has been used in it?

Solution:



Let the kite is made with square $ABCD$ & an isosceles $\vartriangle DEF$.

Given, sides of a $\vartriangle DEF$ are $DE=DF=6\ cm$ & $EF=8\ cm$ & Diagonal of a square $ABCD=32\ cm$.

We know that,

As the diagonals of a square bisect each other at right angle.

$OA=OB=OC=OD=\frac{32}{2}=16\ cm$

$AO$ perpendicular $BC$ & $DO$ perpendicular $BC$.

Area of region I$=$Area of $\vartriangle ABC=$Area of right Triangle$=\frac{1}{2}\times base\times height=\frac{1}{2}\times BC\times OA$

Area of region I$=\frac{1}{2}\times 32\times 16=256\ cm^2$ 

Similarly area of region II $=256\ cm^2$ 

For the III section,

Now, in $\vartriangle DEF$

Let the sides $a=6\ cm,\ b=6\ cm$ & $c=8\ cm$

Semi perimeter of triangle, $s=\frac{a+b+c}{2}=\frac{( 6+6+8)}{2}\ cm=10\ cm$

Using heron’s formula,

Area of the III triangular piece$=\sqrt{s( s-a)( s-b)( s-c)}$

$=\sqrt{10(10-6)(10-6)(10-8)}$

$=\sqrt{10\times 4\times 4\times 2}$

$=\sqrt{2\times 5\times 4\times 4\times 2}$

$=\sqrt{2\times 2\times 4\times 4\times 5}$

$=2\times 4\sqrt{5}$

$=8\times 2.24=17.92\ cm^2$ 

Hence, area of paper of I colour used in making kite $=256\ cm^2$

Area of paper of II colour used in making kite $=256\ cm^2$ 

And area of paper of III colour used in making kite $=17.92\ cm^2$

Updated on: 10-Oct-2022

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