Question

2. Find the area of a kite ABCD in which AB=BC=12 cm, AD=CD=16 cm and diagonal
BD=20 cm.
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Answer 1

Answer:

{Area of quadrilateral ABCD=336}\:cm^2

Step-by-step explanation:

\text{Area of quadrilateral= Area of triangle ABD+ Area of triangle BCD}

\textbf{Area of triangle ABD:}

a=16 cm, b=20 cm, c=12 cm

s=\frac{a+b+c}{2}=\frac{16+20+12}{2}

\implies\:s=\frac{48}{2}=24\:cm

\text{Area of triangle ABD}

=\sqrt{s(s-a)(s-b)(s-c)}

=\sqrt{24(24-16)(24-20)(24-12)}

=\sqrt{24(8)(4)(12)}

=\sqrt{3*8*8*4*3*4}

=3*8*4

=96\:cm^2

\textbf{Area of triangle ABD:}

a=26 cm, b=26 cm, c=20 cm

s=\frac{a+b+c}{2}=\frac{26+20+26}{2}

\implies\:s=\frac{72}{2}=36\:cm

\text{Area of triangle ABD}

=\sqrt{s(s-a)(s-b)(s-c)}

=\sqrt{36(36-26)(36-26)(36-20)}

=\sqrt{36(10)(10)(16)}

=6*10*4

=240\:cm^2

\text{Area of quadrilateral ABCD= Area of triangle ABD+ Area of triangle BCD}

\text{Area of quadrilateral ABCD=96+240}

\therefore\:\textbf{Area of quadrilateral ABCD=336}\:cm^2

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