Question

find the area of the shaded region AD = 12cm , BD=16cm, B= 48 And C=52cm​

Answer 1

${\underline{\underline{\huge{Answer ~:}}}}$

For get answer we have to find area of ∆ABD and ∆ABC after substract .

Area of ∆ABD :-

In ∆ABD ,

${{AB}^{2}={AD}^{2}+{AB}^{2}}$

$= \sqrt{ {(16)}^{2} + {(12)}^{2} } \\ \\ = \sqrt{400} \\ \\ = 20 \: cm$

angle D = 90°

area :)

$= \frac{1}{2} \times base \times height \\ \\ = \frac{1}{2} \times 12 \times 16 \\ \\ = 96 { \: cm}^{2}$

Area of ∆ABC :-

semi perimeter :)

$= \frac{48 + 52 + 20}{2} \\ \\ = 60 \: cm$

$= \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{60 \times 12 \times 8 \times 40} \\ \\ = 480 { \: cm}^{2}$

Now subtract :-

$= (480 - 96) { \: cm}^{2} \\ \\ = 384 { \: cm}^{2}$

${\color{Red}{Answer~is~{384cm}^{2}}$

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Answer 2

{\underline{\underline{\huge{Answer ~:}}}}

Answer :

For get answer we have to find area of ∆ABD and ∆ABC after substract .

Area of ∆ABD :-

In ∆ABD ,

{{AB}^{2}={AD}^{2}+{AB}^{2}}AB

2

=AD

2

+AB

2

\begin{lgathered}= \sqrt{ {(16)}^{2} + {(12)}^{2} } \\ \\ = \sqrt{400} \\ \\ = 20 \: cm\end{lgathered}

=

(16)

2

+(12)

2

=

400

=20cm

angle D = 90°

area :)

\begin{lgathered}= \frac{1}{2} \times base \times height \\ \\ = \frac{1}{2} \times 12 \times 16 \\ \\ = 96 { \: cm}^{2}\end{lgathered}

=

2

1

×base×height

=

2

1

×12×16

=96cm

2

Area of ∆ABC :-

semi perimeter :)

\begin{lgathered}= \frac{48 + 52 + 20}{2} \\ \\ = 60 \: cm\end{lgathered}

=

2

48+52+20

=60cm

\begin{lgathered}= \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{60 \times 12 \times 8 \times 40} \\ \\ = 480 { \: cm}^{2}\end{lgathered}

=

s(s−a)(s−b)(s−c)

=

60×12×8×40

=480cm

2

Now subtract :-

\begin{lgathered}= (480 - 96) { \: cm}^{2} \\ \\ = 384 { \: cm}^{2}\end{lgathered}

=(480−96)cm

2

=384cm

2

{\color{Red}{Answer~is~{384cm}^{2}}

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