In quadrilateral ABCD, angle A = 90^{circ}, A | vedclass

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(A) In $	riangle ABD$, $\angle A = 90^{\circ}$, $AB = 15 	ext{ cm}$, and $AD = 8 	ext{ cm}$.Using the Pythagorean theorem, $BD = \sqrt{AB^2 + AD^2}

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(A) In $	riangle ABD$, $\angle A = 90^{\circ}$, $AB = 15 	ext{ cm}$, and $AD = 8 	ext{ cm}$.Using the Pythagorean theorem, $BD = \sqrt{AB^2 + AD^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 	ext{ cm}$.Area of right-angled $	riangle ABD = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 15 	imes 8 = 60 	ext{ cm}^2$.In $	riangle BCD$, the sides are $a = 17 	ext{ cm}$, $b = 25 	ext{ cm}$, and $c = 12 	ext{ cm}$.Semi-perimeter $s = \frac{17 + 25 + 12}{2} = \frac{54}{2} = 27 	ext{ cm}$.Using Heron's formula, Area of $	riangle BCD = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{27(27-17)(27-25)(27-12)} = \sqrt{27 	imes 10 	imes 2 	imes 15} = \sqrt{8100} = 90 	ext{ cm}^2$.Area of quadrilateral $ABCD = 	ext{Area}(	riangle ABD) + 	ext{Area}(	riangle BCD) = 60 + 90 = 150 	ext{ cm}^2$.

In quadrilateral $ABCD$, $\angle A = 90^{\circ}$, $AB = 15 \text{ cm}$, $BC = 25 \text{ cm}$, $CD = 12 \text{ cm}$, and $DA = 8 \text{ cm}$. Find the area of quadrilateral $ABCD$. (in $\text{ cm}^2$)

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