Question

2. Find the area of a quadrilateral ABCD in which AB 3 cm, BC = 4 cm. CD = 4 cm,
DA 5cm and AC-5 cm.
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Answer 1

Step-by-step explanation:

$\bf \underline{Question} \:$

2. Find the area of a quadrilateral ABCD in which AB 3 cm, BC = 4 cm. CD = 4 cm,

DA 5cm and AC-5 cm.

______________________________

$\bf \underline{Given} \:$

• AB 3 cm,
• BC = 4 cm.
• CD = 4 cm,
• DA 5cm
• AC-5 cm.

______________________________

$\bf \underline{To..Find} \:$

• The area of a quadrilateral ABCD

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According To the ∆ ABC

$\tt \: AC^2 = AB^2 + BC^2 \\ </p><p> \tt \: (5)^2 = (3)^2 + (4)^2$

Therefore, ΔABC is a right-angled triangle, right-angled at point B.

$\bf \: we \: have \: the \: formula \to$

$\tt \: Area \: of \: ABC = \frac{1}{2} \times AB \times BC$

$\bf \: putting \: the \: all \: values$

$\tt \: = \frac{1}{2} \times 3 \times 4 = 6 {cm}^{2}$

For ΔADC,

$</p><p> \tt \: Perimeter = 2s = AC + CD + DA\\ \tt \:= (5 + 4 + 5) cm = 14 cm$

$\tt \: s = 7 cm$

By Heron’s formula,

$\bf area \: of \: triangle \: = \sqrt{s(s - a)(s - b)(s - c)}\\ \\ \tt \: putting \: the \: all \: values \\\\ \tt \: area \: of \: adc = \sqrt{7(7 - 5)(7 - 5)(7 - 4)}\\ \\ \tt \to \: \sqrt{7 \times 2 \times 2 \times 3}\\ \\ \tt \to \: 2\sqrt{21} = 9.166 {cm}^{2}$

$\tt \: Area of ABCD = Area of ABC + Area of ACD$

$\bf \: putting \: all \: values$

$\tt \: = (6 + 9.166) {cm}^{2} \\ \tt = 15.166 {cm}^{2} \\ \tt \: = 15.2 {cm}^{2} (approximately)</p><p></p><p>$