Find the area and the perimeter of quadrilateral $A B C D$ , given above; if, $A B = 8$ cm, $A D = 10$ cm, $B D = 12$ cm, $D C = 13$ cm and $∠ D B C = 90^{\circ}$ .

69.7 c m^{2}; 36

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80 c m^{2}; 43

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69.7 c m^{2}; 48

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70 c m^{2}; 36

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Solution

The correct option is A $69.7 c m^{2}; 36$ cm
Given,
$A B = 8$ cm and $A D = 10$ cm
$B D = 12$ cm and $C D = 13$ cm
$∠ D B C = 90^{\circ}$
$△ D B C$ is a right angle triangle.
$∴ D C^{2} = D B^{2} + B C^{2}$
$\Rightarrow B C^{2} = D C^{2} - D B^{2}$
$\Rightarrow B C^{2} = 13^{2} - 12^{2}$
$\Rightarrow B C^{2} = 169 - 144$
$\Rightarrow B C^{2} = 25$
$\Rightarrow B C = 5$ cm
$∴$ area of $△ A B C = \frac{1}{2} \times B D \times B C$
$= \frac{1}{2} \times 13 \times 5$
$= 30 c m^{2}$
In $△ A B D, S = \frac{A B + B D + A D}{2}$
$= \frac{8 + 12 + 10}{2}$
$= 15$ cm
Area of $△ A B D = \sqrt{S (S - A C) (S - B D) (S - A D)}$
$= \sqrt{15 (15 - 8) (15 - 12) (15 - 10)}$
$= \sqrt{15 (7) (3) (5)}$
$= 39.68 c m^{2}$
$∴$ area of Quadrilateral $A B C D =$ Area of $△ A B D +$ Area of $△ D B C$
$= (39.68 + 30)$
$= 69.68 c m^{2}$
$≃ 69.7 c m^{2}$
Perimeter of Quadrilateral $A B C D = A B + B C + C D + A D$
$= (8 + 5 + 13 + 10)$
$= 36$ cm

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