Question

the area of a rhombus whose one side and one diagonal measure 20 cm and 24 cm respectively is​

Answer 1

$\underline\mathfrak\pink{Diagram:-}$

$\underline\mathfrak\pink{Given:-}$

• the rhombus is one side and one diagonal measure 20 cm and 24 cm.

$\large\underline\mathfrak{To \: find:-}$

• Find the area of the rhombus ....?

$\large\underline\mathfrak\pink{Solutions:-}$

Area of ∆ADC

$\: \: \: \: \: \sqrt{s \: (s \: - \: a) \: (s \: - \: b) \: (s \: - \: c)}$

where,

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{a \: + \: b \: + \: c}{2}$

• a ⇢ 24 cm

• b ⇢ 20 cm

• c ⇢ 20 cm

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{{24} \: + \: {20} \: + \: {20}}{2}$

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{64}{2}$

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: {32} \: cm$

Now, area of ∆ ADC

$\: \: \: \: \: \leadsto \: \: \sqrt{{32} \: ({32} \: - \: {24}) \: ({32} \: - \: {20}) \: ({32} \: - \: {20})}$

$\: \: \: \: \: \leadsto \: \: \sqrt{{32} \: \times \: {(8)} \: \times \: {(12)} \: \times \: {(12)}}$

$\: \: \: \: \: \leadsto \: \: \sqrt{36864}$

$\: \: \: \: \: \leadsto \: \: {192} \: {cm}^{2}$

Now, area of rhombus is 2 × ∆ ADC because ∆ ADC is equal to ∆ ADB.

• Area of rhombus

= 192 × 2

= 384 cm²

Hence, area of rhombus is 384cm²

Answer 2

$\huge{ \underline{ \bold{ᴀɴsᴡᴇʀ....{ \heartsuit}}}}$

the rhombus is one side and one diagonal measure 20 cm and 24 cm.

\large\underline\mathfrak{To \: find:-}

Tofind:−

Find the area of the rhombus ....?

\large\underline\mathfrak\pink{Solutions:-}

Solutions:−

Area of ∆ADC

\: \: \: \: \: \sqrt{s \: (s \: - \: a) \: (s \: - \: b) \: (s \: - \: c)}

s(s−a)(s−b)(s−c)

where,

\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{a \: + \: b \: + \: c}{2}⇝S=

2

a+b+c

a ⇢ 24 cm

b ⇢ 20 cm

c ⇢ 20 cm

\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{{24} \: + \: {20} \: + \: {20}}{2}⇝S=

2

24+20+20

\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{64}{2}⇝S=

2

64

\: \: \: \: \: \leadsto \: \: S \: \: = \: \: {32} \: cm⇝S=32cm

Now, area of ∆ ADC

\: \: \: \: \: \leadsto \: \: \sqrt{{32} \: ({32} \: - \: {24}) \: ({32} \: - \: {20}) \: ({32} \: - \: {20})}⇝

32(32−24)(32−20)(32−20)

\: \: \: \: \: \leadsto \: \: \sqrt{{32} \: \times \: {(8)} \: \times \: {(12)} \: \times \: {(12)}}⇝

32×(8)×(12)×(12)

\: \: \: \: \: \leadsto \: \: \sqrt{36864}⇝

36864

\: \: \: \: \: \leadsto \: \: {192} \: {cm}^{2}⇝192cm

2

Now, area of rhombus is 2 × ∆ ADC because ∆ ADC is equal to ∆ ADB.

Area of rhombus

= 192 × 2

= 384 cm²

Hence, area of rhombus is 384cm