Question

Calculate the shared region.

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Answer 1

       $\text{\huge{\underline{Solution:}}}$

       $\text{In the figure it is given that:}$

       $BD = 5 \text{ cm}$

       $AD = 12 \text{ cm}$

       $AC = 14 \text{ cm}$

       $BC = 15 \text{ cm}$

       $\text{Area of shaded region} = arADBC = ar\triangle{ABC} - ar\triangle ADB$

       $\text{In }\triangle{ADB}$

       $ar\triangle ADB= \text{Area of right triangle} = \dfrac12(\text{base})(\text{height})$

$\implies \text{base} = 12\text{ cm}, \text{height} = 5 \text{ cm}$

$\implies ar\triangle{ADB} = \dfrac12(12)(5)$

$\implies ar\triangle{ADB} = 6 \times 5$

$\implies ar\triangle ADB = 30 \text{ cm}^{2}$

       $\text{Applying pythagorus theorem:}$

$\implies AD^{2} + BD^{2} = AB^{2}$

$\implies (12)^{2} + (5)^{2} = AB^{2}$

$\implies 144 + 25 = AB^{2}$

$\implies 169 = AB^{2}$

$\implies \sqrt{169} = AB$

$\implies +13,-13 = AB$

       $\text{Since, length cannot be negative}$

$\implies 13 \text{ cm} = AB$

       $\text{In }\triangle{ABC}$

       $\text{Applying Heron's formula}$

$\implies \sqrt{s(s-a)(s-b)(s-c)} = \text{ Area of triangle}$

       $\text{In our case:}$

       $a = AB = 13 \text{ cm}$

       $b = AC = 14\text{ cm}$

       $c = BC = 15\text{ cm}$

       $s = \dfrac{a + b + c}{2} = \dfrac{13 + 14 + 15}{2} = \dfrac{42}{2} = 21 \text{ cm}$

$\implies ar\triangle{ABC}=\sqrt{21(21-13)(21-14)(21-15)}$

$\implies ar\triangle{ABC}=\sqrt{21(8)(7)(6)}$

$\implies ar\triangle{ABC}=\sqrt{7056}$

$\implies ar\triangle{ABC}=+ 84,-84$

       $\text{Since, area cannot be negative}$

$\implies ar\triangle{ABC}=84 \text{ cm}^{2}$

$\implies \text{Area of shaded region} = arADBC = ar\triangle{ABC} - ar\triangle ADB$

$\implies arADBC = 84 - 30$

$\implies \boxed{arADBC = 54 \text{ cm}^{2}}$