Question

Find the area of a rhombus each side of which measure $20cm$ and one of the diagonal is $24cm$.

Answer 1

R.E.F image

Find Area of Rhombus

Side = 20 cm

one diagonal is 24 cm

$\to \square ABCD$ BC = 24 cm, so, $BM=MC=\frac{24}{2}=12cm$

$\to$ diagonals $AD=AM+MD$

$\to In△AMC,$ using Pythagoras theorem

$A{C}^2=A{M}^2+M{C}^2$

$(20{)}^2=A{M}^2+(12{)}^2$

$A{M}^2=(20{)}^2+(12{)}^2$

$=400-144$

$A{M}^2=256$

$AM=\sqrt{256}$

$AM=16cm$

$\to AM=MD=16cm$

$\to$ Area of Rhombus $=\frac{1}{2}\times$ Product of diagonal

$=\frac{1}{2}\times (AD.BC)$

$=\frac{1}{2}\times \left(\right(16+16\left).24\right)$

$=\frac{1}{2}\times 32\times 24$

$=384c{m}^2$