Question

if the area of a rhombus is 48 cm2 and one of its diagonals is 6 cm, find its altitude.

Answer with explanation. ...

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Answer 1

Step-by-step explanation:

Area of Rhombus = 48cm²

diagonal 1 = 6cm

let be another diagonal be ' y '

Formula

A = 1/2 ×d¹d²

48 = 1/2×6×y

48 = 3y

y = 48/3

y = 16

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Answer 2

$\huge \sf \red \bigstar \: {\pink{\underbrace{GIVEN}}} \: \red \bigstar$

$\sf \: area \: of \: rhombus = 48 cm^{2}$

$\\ \sf \: one \: of \: its \: diagonal = 6 \: cm$

$\\ \sf \: altitude = \: ?$

$\\ \huge \red \bigstar \sf \: {\pink{ \underbrace{SOLUTION}}} \: \red \bigstar$

➳ Let PQRS be a rhombus whose diagonal PR = 6 cm.

➳ Now, Area of rhombus PQRS given = 48 cm2

$\\ \sf \: ➳ \: \dfrac{1}{2} \times \: PR \: \times SQ = 48 {cm}^{2} \\ </p><p> \\ \sf \: ➳ \: \dfrac{1}{ \cancel{2}} \times { \cancel6} \: cm \times SQ = 48 {cm}^{2} \\ \\ \sf \: ➳ \: SQ = \frac{ \cancel{48} \: cm^{2} }{ \cancel{3} \: cm} = 16 \: cm$

➳ Since diagonals of a rhombus bisect each other at right angle, we have

$\\ \\ ➳ \sf \: OP = \dfrac{1}{2} \: \\ \\ \sf➳ \: PR = \frac{1}{ \cancel{2} } \times { \cancel6} \: cm \: = 3 \: cm \\ \\ \sf \: ➳ \: OQ = \dfrac{1}{2} \\ \\ \sf \:➳ \: SQ = \dfrac{1}{ \cancel{2} } \times { \cancel{16}} \: cm \: = 8 \: cm$

➳ Also, ∆POQ is a right triangle, where right ∆POQ = 90°

$\\ \sf \: ∴ \: PQ = \sqrt{OP ^{2}+OQ^{2}}</p><p> \sf \sqrt{(3^{2} + (8)^{2}</p><p>$