Question

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Answer 1

Step-by-step explanation:

unshaded area = ½x 16x 12

=8x12=96cm²

∆ABC area= ½x 52 x48

=52x24

=1248cm²

Shaded area= 1248-96

=1152cm²

Answer 2

Information provided with us:

• Length of BD is 16cm
• Length of AD is 12cm
• Length of AC is 52cm

What we have to calculate:

• We have to calculate the area of the shaded region of that given figure of triangle

Using Formulas:

Pythagoras theorem,

• $\large \boxed{ \sf{(Hypotenuse) {}^{2} = (base) {}^{2} + (height) {}^{2} }}$

Area of triangle,

• $\boxed{ \large {\sf{ \dfrac{1}{2} \times base \: \times \: height}}}$

Heron's formula,

• $\boxed{ \large {\sf{ \sqrt{s(s - a)(s - b)(s - c)} }}}$

Where,

• s is semi perimeter
• a, b, and c are the sides of triangle

Semi perimeter of triangle,

• $\boxed{ \large {\sf{ s \: = \: \frac{a + b + c}{2} }}}$

Performing Calculations:

______________

$\underline{ \underline{{\bf{Triangle \: ABD:- \: }}}}$

• Here we have to find out the hypotenuse of the triangle ABD so we would be using pythagoras theorem.

Here we have,

• Base (AD) is 16cm
• height (BD) is 12cm

Substituting the values in pythagoras theorem,

$\longrightarrow \: \sf{AB {}^{2} \: = \: (16) {}^{2} + (12) {}^{2} }$

$\longrightarrow \: \sf{AB {}^{2} \: = \: (16 \times 16)+ (12 \times 12) }$

$\longrightarrow \: \sf{AB {}^{2} \: = \: (256)+ (144) }$

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

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

$\longrightarrow \: \boxed{\purple{\bf{AB \: = \: 20cm}}}$

Finding out the area of triangle,

$\longrightarrow \: \sf{Area _{(triangle)} \: = \: \dfrac{1}{2} (16)(12)}$

$\longrightarrow \: \sf{Area _{(triangle)} \: = \: \dfrac{1}{2} \times 16 \times 12}$

$\longrightarrow \: \sf{Area _{(triangle)} \: = \: \dfrac{1}{ \cancel2} \times \cancel16 \times 12}$

$\longrightarrow \: \sf{Area _{(triangle)} \: = \: 8 \times 12}$

$\longrightarrow \: \boxed{\purple{\bf{Area _{(triangle)} \: = \: 96}}}$

• $\therefore \underline{\bf{Area \: of \: \triangle \: ABD \: is \: 96cm^{2}}}$

_____________

$\underline{ \underline{ \sf{\bf{Triangle \: ABC:- \: }}}}$

Finding out semi perimeter,

• Here we are calculating the semi perimeter (s) by substituting the values of sides a, b, and c in the formula of semi perimeter.

$: \longmapsto \: \sf{s_{(triangle)} \: = \: \dfrac{20 + 48 + 52}{2} }$

$: \longmapsto \: \sf{s_{(triangle)} \: = \: \dfrac{20 + 100 }{2} }$

$: \longmapsto \: \sf{s_{(triangle)} \: = \: \dfrac{120}{2} }$

$: \longmapsto \: \sf{s_{(triangle)} \: = \: \cancel\dfrac{120}{2} }$

$: \longmapsto \: \boxed{ \purple{\bf{s_{(triangle)} \: = \:60 }}}$

$\therefore \underline{\bf{Semi perimeter \: of \: \triangle \: ABC \: is \: 60cm}}$

Finding out area by using heron's formula,

• Substituting the values of s, a, b, and c in the heron's formula inorder to get its area.

$: \implies \: \tt{ \sqrt{60(60 - 20)(60 - 48)(60 - 52)} }$

$: \implies \: \tt{ \sqrt{60(40)(12)(8)} }$

$: \implies \: \tt{ \sqrt{60 \times (40) \times (12) \times (8)} }$

$: \implies \: \tt{ \sqrt{60 \times 40\times 12 \times 8} }$

$: \implies \: \tt{ \sqrt{2400\times 12 \times 8} }$

$: \implies \: \tt{ \sqrt{2400 \: \times \: 96} }$

$: \implies \: \tt{ \sqrt{230400} }$

$: \implies \: \underline{\red{\boxed{ \tt{ 480} }}}$

$\therefore \underline{\bf{Area \: of \: \triangle \: ABC \: is \: 480}}$

_________

$\underline{ \underline{ \sf{\bf{Shaded \: region:- \: }}}}$

• It would be calculated by subtracting the both the areas of ∆ABD and ∆ABC

Here we have,

• Area of ∆ABD = 96cm²
• Area of ∆ABC = 480cm²

$: \leadsto \: \bf{Shaded \: region \: area \: = \: 480 - 96}$

$: \leadsto \: \large{\pink{\boxed{\bf{Shaded \: region \: area \: = \: 384}}}}$

$\underline{ \tt{Henceforth, \: area \: of shaded \: region \: is \: 384cm {}^{2} </p><p>}}$