Question

If the area of a rhombus is 48cm Sq. and one of its diagonal is 6cm, find its altitude. ​

Answer 1

$\large\bf\underline \blue{Given:-}$

• Area of rhombus = 48cm²
• One diagonal = 48cm

$\large\bf\underline \blue{To \: find:-}$

• Altitude of Rhombus.

$\huge\bf\underline \purple{Solution:-}$

Let ABCD be the given rhombus, with Diagonal AC = 6cm.

★ Area of rhombus =1/2 × Diagonal 1 × Diagonal 2.

$: \implies \rm \: 48 = AC \times BD \times \frac{1}{2} \\ \\ : \implies \rm \:48 = \cancel 6 \times \frac{1}{ \cancel2} \times BD \\ \\ : \implies \rm \: \cancel\frac{48}{3} = BD \\ \\ : \implies \bf \:BD = 16cm \\ \\ \large \bf\:{So} \: the \: other \: diagonal \: = 16cm \\ \\ \rm \: We \: know \: that \: diagonals \: bisect \: each \\ \rm \: other \: at \: right \: angle. \\ \bf\:So \\ : \implies \rm \:OB = \frac{1}{2} BD \\ \\ : \implies \rm \:OB = \frac{1}{2} \times 16\\ \\ : \implies \rm \:OB = \cancel \frac{16}{2} \\ \\ : \implies \bf \:OB = 8cm$

Now , IN triangle ∆BOC,

• By Pythagoras theorem:-

$\rm \: \: \: \: \: \: \: \: BC {}^{2} = OB {}^{2} + OC {}^{2} \\ \\ : \implies \rm \: BC {}^{2} = {8}^{2} + {3}^{2} \\ \\ : \implies \rm \: BC {}^{2} = 64 + 9 \\ \\ : \implies \rm \: BC = \sqrt{73}$

Now,

• Area of rhombus = base × altitude.

$: \implies \rm \: Base \: \times Altitude \: = 48 \\ \\ : \implies \rm \: \sqrt{73} \times Altitude = 48 \\ \\ : \implies \rm \:Altitude = \frac{48}{ \sqrt{73} } \\ \\ : \implies \rm \:Altitude = \frac{48}{ \sqrt{73} } \times \frac{ \sqrt{73} }{ \sqrt{73} } \\ \\ : \implies \rm \:Altitude = \frac{48 \sqrt{73} }{73}\\\\ : \implies \bf \:Altitude = 5.6$