Question

in the given figure calculate the area of quard. ABCD​

Answer 1

The area of quadrilateral ABCD is $114$ $cm^{2}$.

Explanation:

∆ABD and ∆BDC are right-angled triangles.

So,  In ∆BDC

      $BC^{2}$ + $BD^{2}$ = $CD^{2}$

                 $BD^{2}$ = $CD^{2}$ - $BC^{2}$

                 $BD^{2}$ = $(17)^{2}$ - $(8)^{2}$

                 $BD^{2}$ = $289$ - $64$

                 $BD^{2}$ = $225$

                  = 15 cm

and, In ∆ABD

      $AD^{2}$ + $AB^{2}$ = $BD^{2}$

                 $AB^{2}$ = $BD^{2}$ - $AD^{2}$

                 $AB^{2}$ = $(15)^{2}$ - $(9)^{2}$

                 $AB^{2}$ = $225$ - $81$

                 $AB^{2}$ = 144

                 $AB$ = 12 cm

Now,  

Area of ∆ ABD

ar(∆ ABD) = 1/2(base X height)

ar(∆ ABD) = 1/2(AB X AD)

ar(∆ ABD) = 1/2(12 X 9)

ar(∆ ABD) = 1/2(108)

ar(∆ ABD) = 54 $cm^{2}$

Area of ∆ BDC

ar(∆ BDC) = 1/2(base X height)

ar(∆ BDC) = 1/2(BC X BD)

ar(∆ BDC) = 1/2(8 X 15)

ar(∆ BDC) = 1/2(120)

ar(∆ BDC) = 60 $cm^{2}$

ar(quadrilateral ABCD) = ar(∆ABD) + ar(∆BDC)

ar(quadrilateral ABCD) = 54 $cm^{2}$ + 60

ar(quadrilateral ABCD) = 114 $cm^{2}$

∴ The area of quadrilateral ABCD is $114$ $cm^{2}$

(Hope it helps you!)