Question

Q-1 Find the altitude of a trapezium, the sum of the lengths of whose base is 11 cm and whose area is 26cm cube?
Q-2Find the area of a rhombus whose diagonals are of measurements 6 cm and 8 cm?​

Answer 1

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☆The sum of the lengths of trapezium base is 11 cm and whose area is 26cm cube

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☆ The Altitude of the trapezium

$\dag$ Solution :

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☆ Let the altitude of the trapezium be h cm

$\dag$ We have ,

$: \implies \rm{Area \: of \: trapezium = {26 \: cm}^{2} }$

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$: \implies \rm{ \dfrac{1}{2} \times \bigg(Sum \: of \: the \: base \bigg)\times Altitude = 26}$

$: \implies \rm{ \dfrac{1}{2} \times 11\times Altitude = 26}$

$: \implies \rm{Altitude = \dfrac{26 \times 2}{11} } \: cm$

$: \implies \rm{Altitude = 4.72\: cm }$

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$\therefore \;$ Hence,the altitude of the trapezium is

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☆The diagonals of rhombus are of measurements 6 cm and 8 cm

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☆ The area of that rhombus

$\dag$ Solution :

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$: \implies \rm{Area \: of \: rhombus = \dfrac{d_{1}\times d_{2}}{2} }$

$\dag$ Where ,

$: \mapsto \: \rm{d_{1}\times d_{2}} \: are \: diagonals \: of \: the \: rhombus$

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☆ Substitute the given values in above formula and solve

$: \implies \rm{Area \: of \: rhombus = \dfrac{6\times 8}{2} }$

$: \implies \rm{Area \: of \: rhombus = \dfrac{48}{2} }$

$: \implies \rm{Area \: of \: rhombus = 24 \: {cm}^{2} }$

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$\therefore \;$ Hence,the area of the rhombus is

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