Question

Find the area of rhombus whose side is 16 cm and whose altitude is 4 cm.

If one of its diagonals is 8 cm long, find the length of the other diagonal.​

Answer 1

☆Answer

$\sf \color{cyan} \: Area \: Of \: Rhombus = Base(side) \times Altitute \\ \sf \color{cyan}16 \times 4 = 64 \\ \sf \: 64 = \frac{1}{2} \times 8 \times d \\ \sf \: d = \frac{1 \times 8}{2 \times 64} \\ \sf \: d = \frac{8}{128} \\ d = 16 \: cm$

Hence, The area is 64cm² and the Diagonal is 16cm.

Answer 2

Answer:

❣Answer❣

➡Area of a Rhombus= Base × Alitude

= 16 × 4 =64

➡$64 = \frac{1}{2} \times 8 \times d$

➡$d = \frac{1 \times 8}{2 \times 64 }$

➡$d = \frac{8}{128}$

$d = 16cm$

$Hence ,\: the \: area \: is \: {64cm}^{2} and \: the \: diagonl \: is \: 16cm.$