Question

if hypernews and base of a right angle are 10cm and 8cm are respectievly find its area.​

Answer 1

Appropriate Question:

• If Hypotenuse and Base of a right angle are 10cm and 8cm are respectively find its area.

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$\frak{Given}\begin{cases}\sf{\:\:\: Hypotenuse = 10 \ cm}\\\sf{\:\:\: Base = 8 \ cm}\end{cases}$

Need to find: Area of triangle.

☯ Let the Height be h cm.

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$\underline{\bigstar\:\boldsymbol{Using\: Pythagoras \ theorem\::}}$⠀⠀⠀

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$\star\;\boxed{\sf{\pink{(Hypotenuse)^2 = (Base)^2 + (Height)^2}}}$⠀

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$:\implies\sf (10)^2 = (8)^2 + (h)^2 \\\\\\:\implies\sf 100 = 64 + h^2 \\\\\\:\implies\sf 100 - 64 = h^2\\\\\\:\implies\sf 36 = h^2\\\\\\:\implies\sf h = \sqrt{36}\\\\\\:\implies{\underline{\boxed{\sf{\pink{ h = 6 \ cm}}}}}\:\bigstar$

⠀$\therefore\:{\underline{\sf{Hence, \ Height \ of \ the \ \triangle \ is \ \bf{6 \ cm}.}}}$⠀⠀

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$\star\:\boxed{\sf{\purple{Area_{\triangle} = \dfrac{1}{2} \times Height \times Base}}}\\\\\\:\implies\sf Area_{\triangle} = \dfrac{1}{\cancel{\: 2}} \times 6 \times \cancel{\:8} \\\\\\:\implies\sf Area_{\triangle} = 6 \times 4\\\\\\:\implies{\underline{\boxed{\frak{\pink{ Area_{\triangle} = 24 \ cm^2}}}}}\:\bigstar$

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$\therefore\:{\underline{\sf{Hence, \ area \ of \ the \ \triangle \ is \ \bf{24 \ cm^2}.}}}$⠀⠀⠀

Answer 2

Step-by-step explanation:

$\frak Given = \begin{cases} &\sf{Hypotenuse\ of\ the\ right\ angled\ triangle\ =\ 10cm.} \\ &\sf{Base\ of\ the\ right\ angled\ triangle\ =\ 8cm.} \end{cases}$

To find:- We have to find the area of the right angled triangle ?

☯️ In this question, the height of the triangle is not given. So, to find the area of the triangle first let us find out the height.

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$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf : \implies {Pythagoras\ Theorem\ =\ a^2\ +\ b^2\ =\ c^2.}$

Here:-

• a, is for height.
• b, is for base.
• c, is for hypotenuse.

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$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}}$

$\sf : \implies {a^2\ +\ b^2\ =\ c^2} \\ \\ \sf : \implies {a^2\ +\ 8^2\ =\ 10^2} \\ \\ \sf : \implies {a^2\ +\ 64\ =\ 100} \\ \\ \sf : \implies {a^2\ =\ 100\ -\ 64} \\ \\ \sf : \implies {a^2\ =\ 36} \\ \\ \sf : \implies {a\ =\ \sqrt{36}} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf a\ =\ 6cm.}}}}\bigstar$

Hence:-

$\sf \therefore {\underline{The\ required\ height\ of\ the\ triangle\ is\ 6cm.}}$

● Now, finding the area of the triangle:-

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$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf : \implies {Area\ of\ triangle\ =\ \dfrac{1}{2}\ \times\ b\ \times\ h.}$

Here:-

• b, is for base.
• h, is for height.

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$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}}$

$\sf : \implies {Area\ =\ \dfrac{1}{2}\ \times\ b\ \times\ h} \\ \\ \sf : \implies {\dfrac{1}{\cancel 2}\ \times\ \cancel 8\ \times\ 6} \\ \\ \sf : \implies {4\ \times\ 6} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf Area\ =\ 24cm^2.}}}}\bigstar$

Hence:-

$\sf \therefore {\underline{The\ required\ area\ is\ 24cm^2.}}$