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  • The area of a quadrilateral ABCD in which AC=17 m, BC = 15 m, CD = 12 m, DA = 9m and ∠ B = 90º is (in m2)<div class='qna-option'

The area of a quadrilateral ABCD in which AC=17 m, BC = 15 m, CD = 12 m, DA = 9m and ∠ B = 90º is (in m2)

Option: 1

60+4\sqrt{665}


Option: 2

64+2\sqrt{665}


Option: 3

60+2\sqrt{600}


Option: 4

60+2\sqrt{665}


Answers (1)

best_answer

Let ABCD be the given quadrilateral

AC=17 m, BC = 15 m, CD = 12 m, DA = 9m and ∠ B = 90º

In ? ABC, by Pythagoras’ theorem, we have

            AC2 = AB2 + BC2

           172 = AB2 + 152  

             AB = 8

\begin{aligned} \text { Area of } \Delta A B C &=\frac{1}{2} \times A B \times B C \\ &=\left(\frac{1}{2} \times 8 \times 15\right) \mathrm{m}^{2}=60 \mathrm{m}^{2} \end{aligned}

In ? ABD, Let a = AC = 17 m, b = CD = 12 m and  c = AD = 9 m.

\therefore \;\text{s}=\frac{1}{2}\left ( 17+12+9 \right )\text{m}=19\text{ m} 

(s - a) = (19 - 17) = 2 m

(s - b) = (19 - 12) = 7 m

(s - c) = (19 - 9) = 10 m

\\\therefore \quad \text { area }(\Delta A C D)=\sqrt{s(s-a)(s-b)(s-c)}\\\text{ }\quad\quad\quad\quad\quad\quad\quad\;\;\;\;=\sqrt{19\times 2\times 7\times 10}\;m^2 \\\text{ }\quad\quad\quad\quad\quad\quad\quad\;\;\;\;=2\sqrt{665}m^2

\\\therefore \quad \text { area of quad. } A B C D=\operatorname{area}(\triangle A B C)+\operatorname{area}(\triangle A C D)\\\text{}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\;\;=(60+2\sqrt{665})\text{ m}^2

 

Posted by

sudhir.kumar

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