Question

find the area of unshaded region
​

Answer 1

area of bigger triangle -area of right angled triangle
first, find the third side by Pythagoras theorem and then solve using 1/2*b*h

Answer 2

$\displaystyle\bf{Answer : - }$

$\displaystyle\bf{Given:-}$

$\sf{The \: side \: of \: the \: AD = 12cm}$

$\sf{The \: side \: of \: the \: BD = 16cm}$

$\sf{The \: side \: of \: the \: BC = 42cm}$

$\sf{The \: side \: of \: the \: AC = 34cm}$

$\displaystyle\bf{Formula's \: Used : - }$

$\sf{The \: area \: of \: the \: triangle = \frac{1}{2}×b×h}$

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

$\sf{The \: heron's \: formula, A = \sqrt {s(s-a)(s-b)(s-c)}}$

$\displaystyle\bf{Solution:-}$

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

$\sf{AB^{2} = 16^{2} + 12^{2}}$

$\sf{AB^{2} = 256 + 144}$

$\sf{AB^{2} =400 }$

$\sf{AB = \sqrt{400} }$

$\sf{ \implies AB = 20cm}$

$\sf{The \: area \: of \: the \: triangle = \frac{1}{2}×b×h}$

$\sf{ = \frac{1}{2} \times 12 \times 16}$

$\sf{ = 6 \times 16}$

$\sf{ = 96 \: cm {}^{2} }$

$\sf{The \: heron's \: formula, A = \sqrt {s(s-a)(s-b)(s-c)}}$

$\sf{s = \frac{a + b + c}{2} \implies \: \frac{34 + 42 + 20}{2} }$

$\sf{ = \frac{96}{2} \implies \: 48cm}$

$\sf{ A = \sqrt{s(s-a)(s-b)(s-c)}}$

$\sf{ = \sqrt {48(48-34)(48-42)(48-20)}}$

$\sf{ = \sqrt {48 \times 14 \times 6 \times 28}}$

$\sf{= \sqrt{8 \times 6 \times 2 \times 7 \times 6 \times 4 \times 7} }$

$\sf{ = \sqrt{8 \times 8 \times 7 \times 7 \times 6 \times 6} }$

$\sf {= 8 \times 7 \times 6}$

$\sf{ = 336 \: cm {}^{2} }$

$\sf{The \: unshaded \: region = Total \: area \: of \: triangle - \: shaded \: region}$

$\sf{= 336 \: cm ^{2} - 96 \: cm{}^{2} }$

$\sf{ = 240 \: cm {}^{2} }$

$\displaystyle \bf{Hence, \: The \: unshaded \: region \: is = {{ \red{240 \: cm {}^{2}}} }}$