Question

Find the arra of shaded region in figure. ​

Answer 1

Area of that (unshaded) right angle triangle is 30 cm by applying formula.

So, for area of shaded part.

Subtract to get area of shaded part:-

AREA OF FULL TRIANGLE - AREA OF NONSHADED PART = AREA OF SHADED PART!

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Answer 2

☞︎︎︎ Solution :

★ Given :

• AO = 12cm

• BO = 5cm

• AC = 15cm

• BC = 14cm

★ Solving :

↣ Area of △AOB

$\large \: \: \: \: \: \: \: = \dfrac{1}{2} \times \rm oa \times ob$

$\large \: \: \: \: \: \: \: = \dfrac{1}{2} \times \rm 12 \times 5$

$\large \: \: \: \: \: \: \: = \rm 30c {m}^{2}$

★ Also,

$\large \rm \rightarrow \: ab {}^{2} = o {a}^{2} + ob {}^{2}$

$\large \rm \: \: \: \: \: \: \: \: \: \: : ⇒ \: ab{}^{2} = 12^{2} + {5}^{2}$

$\large \rm \: \: \: \: \: \: \: \: \: \: : ⇒ \: ab{}^{2} = 144^{} + {25}^{}$

$\large \rm \: \: \: \: \: \: \: \: \: \: : ⇒ \: ab{}^{2} = 169$

$\large \rm \: \: \: \: \: \: \: \: \: \: : ⇒ \: ab{}^{} = \sqrt{ 169 \: }$

$\large \bold{ \therefore \rm \: \: ab = 13cm}$

★ Finding s with formula -

$\large \looparrowright\boxed{\rm \: s = \dfrac{a + b + c}{2} }$

$\: \: \: \: \large \rm \: \: : ⇒ \: s = \dfrac{14 + 15 + 13}{2}$

$\: \: \: \: \large \rm \: \: : ⇒ \: s = \dfrac{42}{2}$

$\large \bold{ \therefore \rm \: \: s = 21cm}$

★ Area of triangle ABC

• Using Herons Formula :

$\:$

$\boxed{ \large \rm = \: \sqrt{s(s - a)(s - b)(s - c)} }$

$\:$

• Solving :

$\:$

$\rm \large : ⇒ \: \sqrt{21(21 - 14)(21 - 15)(21 - 13)}$

$\:$

$\large\rm \: \: \: \: \: \: \: \: \: \: \: = \sqrt{21 \times 7 \times 6 \times 8}$

$\:$

$\large\rm \: \: \: \: \: \: \: \: \: \: \: = \sqrt{7056}$

$\\ \large\rm \: \: \: \: \: \: \: \: \: \: \therefore \: area \: = 84 {cm}^{2}$

$\:$

★ area of shaded region :

$\:$

= 84cm² - 30cm²

$\:$

Area of shaded region = 54cm²