Question

Q.4. Find the area of the shaded region in figure.​

Answer 1

Answer:

$\large{ \green{ \underline{ \tt \: Given : - }}}$

$\sf \: Sides \: \: of \: \: triangle : -$

$\sf \triangle \: ABC \: \: \: \: \: \: \: \: \\ \\ \sf \: AC = 52 \: cm \\ \sf \: BC = 48 \: cm$

$\sf \triangle \: ABD \\ \\ \sf \: AD = 12 \: cm \\ \sf \: BD = 16 \: cm$

$\\ \sf \: For \: finding \: the \: side \: AB$

$\sf \: Taking \: \: \triangle ABD$

$\rm \: \bold{ Appling \: \: pythagoras \: \: theorem : - }$

$\sf \: ( {hypotenious)}^{2} = ( {base)}^{2} + ( {perpen}^{r}) {}^{2}$

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

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

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

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

$\implies \sf \: AB {}^{2} = {20}^{2}$

$\sf \: Taking \: \: Square \: \: root \: \: both \: \: sides : -$

$\implies \sf \: AB = 20 \: cm$

$\rm \: Now \: \: in \: \triangle ABC$

$\rm \: Appling \: \: Heron's \: Formula : -$

$\rm So \: first \: we \: need \: to \: find \: perimeter \: \\ \rm \: of \: \: \triangle ABC \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \: Perimeter \: of \: \triangle ABC = sum \: of \: all \: sides$

$\sf \: Perimeter (p) = 20 + 48 + 52$

$\sf \: Perimeter (p) = 120 \: cm$

$\sf \: Now \: \: semi \: \: perimeter (s) = \frac{perimeter}{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: = \frac{120}{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: = 60 \: cm$

$\sf \: Now \: \: area \: \: of \triangle ADB = \frac{1}{2} (base) \times (height)$

$\green{ \sf We \: \: are \: using \: this \: formula \: \: because \:} \\ \green{ \sf \: \triangle \: ADB \: is \: right \: \: triangle \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

$\sf \: Now \: \: area \: \: of \triangle ADB = \frac{1}{2} \times (12) \times (16) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf \:96 \: {cm}^{2}$

$\sf \: Area \: \: of \: \: \triangle \: ABC$

$\rm \: Appling \: \: Heron's \: Formula : -$

$\sf \: Area \: of \: \triangle ABC = \sqrt{s(s - a)(s - b)(s - c)}$

$\sf Here \: \: a,b,c \: \: are \: the \: \: sides \: \: of \: \: \triangle ABC$

$\sf \: a = 20cm \\ \\ \sf \: b = 48cm \\ \\ \sf \: c = 52 \: cm$

$\sf \: Hence \: \: the \: \: area : -$

$\sf = \sqrt{60(60 - 20)(60 - 48)(60 - 52)}$

$= \sf \sqrt{60 \: (40) \: (12) \: (8)}$

$= \sf \: \sqrt{60 \times 40 \times 12 \times 8} \\ \\ \sf \: = \sqrt{480 \times 480} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \: = 480 \: {cm}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \: Now \: area \: of \: shaded \: region : -$

$\sf \: = Area \: of \: \triangle ABC - Area \: of \: \triangle ABD$

$\sf \: = 480 - 96$

$\sf \: = 384 \: {cm}^{2}$

$\sf Now \: area \: of \: shaded \: region \: \green{\boxed{ \sf384 \: {cm}^{2}} }$

Thank you.