Question

4. Find the perimeter and area of quadrilateral ABCD in which
AB = 9 cm, AD = 12 cm, BD
CD = 17 cm and <CBD = 90°.
15 cm,​

Answer 1

Question ⤵⤵⤵⤵⤵

Find the perimeter and area of quadrilateral ABCD in which

AB = 9 cm, AD = 12 cm, BD

CD = 17 cm and <CBD = 90°.

15 cm,

Answer ⤵⤵⤵⤵⤵

$<b> </p><p></p><p>In ΔBCD,</p><p></p><p>\angle BCD=90° ∠BCD=90° </p><p></p><p>{BC}^{2} ={BD}^{2}-{DC}^{2} BC </p><p>2</p><p> =BD </p><p>2</p><p> −DC </p><p>2</p><p> </p><p></p><p>\begin{gathered} = \sqrt{ {17}^{2} - {15}^{2} } \\ \\ = \sqrt{64} \\ \\ = 8 \: cm\end{gathered} </p><p>= </p><p>17 </p><p>2</p><p> −15 </p><p>2</p><p> </p><p> </p><p> </p><p>= </p><p>64</p><p> </p><p> </p><p>=8cm</p><p> </p><p> </p><p></p><p>Area of quadrilateral = ar(ΔBCD+ΔABD)</p><p></p><p>Area of ΔBCD :</p><p></p><p>\begin{gathered} = \frac{1}{2} \times base \times height \\ \\ = \frac{1}{2} \times 15 \times 8 \\ \\ = 60 { \: cm}^{2} \end{gathered} </p><p>= </p><p>2</p><p>1</p><p> </p><p> ×base×height</p><p>= </p><p>2</p><p>1</p><p> </p><p> ×15×8</p><p>=60cm </p><p>2</p><p> </p><p> </p><p> </p><p></p><p>Area of ΔABD : </p><p></p><p>semi perimeter :</p><p></p><p>\begin{gathered} = \frac{a + b + c}{2} \\ \\ = \frac{15 + 12 + 9}{2} \\ \\ = 18 \: cm\end{gathered} </p><p>= </p><p>2</p><p>a+b+c</p><p> </p><p> </p><p>= </p><p>2</p><p>15+12+9</p><p> </p><p> </p><p>=18cm</p><p> </p><p> </p><p></p><p>Area :</p><p></p><p>\begin{gathered} = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{18 \times 3 \times 6 \times 9} \\ \\ = 54 \: {cm}^{2} \end{gathered} </p><p>= </p><p>s(s−a)(s−b)(s−c)</p><p> </p><p> </p><p>= </p><p>18×3×6×9</p><p> </p><p> </p><p>=54cm </p><p>2</p><p> </p><p> </p><p> </p><p></p><p>Area of quadrilateral :</p><p></p><p>\begin{gathered} = (54 + 60) { \: cm}^{2} \\ \\ = 114 { \: cm}^{2} \end{gathered} </p><p>=(54+60)cm </p><p>2</p><p> </p><p>=114cm </p><p>2</p><p> </p><p>$

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I hope it is helpful for you

@Aman jha