Question

$\dag \underline\mathtt{question}$
Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.

$\pink{\dag}$ No copied answer✖​

Answer 1

Answer:

384cm²

Step-by-step explanation:

The diagonals of a rhombus bisect at right angles

Consider Δ AOD

Using the Pythagoras theorem

AD2=OD2+AO2

By substituting the values

202=OD2+122

On further calculation

OD2=400−144

By subtraction

OD2=256

By taking out the square root

OD = 256

So we get

OD = 16cm

We know that BD = 2OD

So we get

BC = 2 (16) = 32cm

We know that

Area of rhombus ABCD = 21×AC×BD

By substituting the values

Area of rhombus ABCD = 21×24×32

On further calculation

Area of rhombus ABCD = 384cm2

Therefore, the area of rhombus ABCD is 384cm2.

Answer 2

Solution:-

• The diagonals of a rhombus bisect at right angles

• Consider Δ AOD

By Using A theorem:-

$\: \: \: \: \: \sf \: \: \: \: \: \: \longrightarrow \: AD^2 =OD^2 +AO^2$

By substituting the values:-

$\: \: \: \: \: \: \: \sf \: \: \: \: \longrightarrow \: 20^2 =OD^2 +12^2$

On further calculation:-

• $\: \sf \: \: \: \longrightarrow \: OD^2 =400−144$

By subtraction

• $\: \: \: \longrightarrow\sf \: OD^2=256$

By taking out the square root

• OD = √256

So we get

• OD = 16cm

We know that BD = 2OD

So we get

• BC = 2× (16) = 32cm

We know that

$\sf \: Area \: of \: rhombus \: \: ABCD = \dfrac{1}{2} ×AC×BD$

By substituting the values

$\sf \: Area \: of \: rhombus \: \: ABCD = \frac{1}{2} ×24×32$

On further calculation

$\sf \: \: Area \: of \: rhombus \: \: \red{ \bold{ ABCD = 384cm^2}}$

$\sf \: \therefore \: \underline{ \underline{ \: the \: area \: of \: rhombus \: ABCD \: is}} \sf \: \: \longrightarrow \: \: \underline{\boxed{\purple{\bold{384cm^2}}}}$