Question

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Find the altitudes of an quadrilateral triangle whose sides are 10cm long.

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Answer 1

Given : The length of the sides of an equilateral triangle is 10cm.

To Find : Find the altitudes of an quadrilateral triangle ?

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Solution : Let the quadrilateral traingle be x.

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$\underline{\frak{As ~we~ know ~that~:}}$

• $\underset{\blue{\sf Pythagoras\ Theorem}}{\underbrace{\boxed{\sf{\pink{(H)^2~=~(B)^2~+~(P)^2}}}}}$

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Where,

• H = Hypotenuse = AB => 10cm
• B = Base = BD => 5cm
• P = Perpendicular = (altitude) => AD

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$\pmb{\sf{\underline{According~ to~ the~ Given~ Question~:}}}$

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$\qquad{\sf:\implies{(AD)^2~=~(BD)^2~+~(AD)^2}}$

$\qquad{\sf:\implies{(10cm)^2~=~(5cm)^2~+~(AD)^2}}$

$\qquad{\sf:\implies{100cm^2~=~25cm^2~=~(AD)^2}}$

$\qquad{\sf:\implies{AD^2~=~100cm^2~-~25cm^2}}$

$\qquad{\sf:\implies{AD^2~=~75cm^2}}$

$\qquad{\sf:\implies{AD^2~=~\sqrt{75cm}}}$

$\qquad{\sf:\implies{AD~=~\sqrt{3~×~5~×~5cm}}}$

$\qquad{\sf:\implies{AD~=~\sqrt{3~×~(5)^2cm}}}$

$\qquad:\implies{\underline{\boxed{\frak{\pink{\pmb{AD~=~5\sqrt{3cm}}}}}}}$★

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Hence,

$\therefore\underline{\sf{The ~altitudes~ of~ an~ quadrilateral ~triangle~\bf{\underline{\pmb{5\sqrt{3cm}}}}}}$