Question

The length of the diagonals of a rhombus are 24cm and 32cm, then the length of the altitude of the rhombus is..
a)12cm b)12.8 c)19cm d)19.2cm​

Answer 1

Answer :

• Altitude of the rhombus is 19.2cm
• Option (d)

Given :

• The length of the Diagonal of a rhombus are 24cm and 32cm

To find :

• Length of the altitude of the rhombus is

Solution :

Given,

The length of the Diagonal of a rhombus are 24cm and 32cm so,

As we know that,

• Area of rhombus = 1/2 d1d2

Where,

• d1 is 24cm
• d2 is 32cm

⤁ Area of rhombus = 1/2 d1d2

⤁ Area of rhombus = 1/2 × 24 × 32

⤁ Area of rhombus = 12 × 32

⤁ Area of rhombus = 384cm²

Hence , Area of rhombus is 384cm²

Finding the side :

We know That,

Using the Pythagoras theorem,

• Side = (1/2 d1) + (1/2 d2)

⤁ Side = √(12)² + (16)²

⤁ Side = √144 + 256

⤁ side = √400

⤁ side = 20cm

Hence , Side is 20cm

Find the length of the altitude of rhe rhombus :

As we know that,

• Area of rhombus = base × altitude

• Let the altitude be a
• base is 20cm
• Area of rhombus is 384cm²

⤁ 384 × 20 × a

⤁a = 384/20

⤁ a = 19.2cm

Hence, Altitude of the rhombus is 19.2cm

Option (d)

Answer 2

To find : length of altitude of the rhombus

Given :  length of the diagonals of a rhombus are $24\ cm$ and $32\ cm$

Solution :  

• Area of rhombus  $=\frac{1}{2} \mathrm{~d}_{1} \mathrm{~d}_{2}=\frac{1}{2} \times 24 \times 32=384 \mathrm{~cm}^{2}$
• Now , using pythagoras theorem ,we get

        $\begin{array}{l}\text { side }^{2}=\left(\frac{1}{2} \mathrm{~d}_{1}\right)^{2}+\left(\frac{1}{2} \mathrm{~d}_{2}\right)^{2} \\=12^{2}+16^{2}=144+256=400 \\\text { Side }=20 \mathrm{~cm}\end{array}$

• Area of rhombus = base x altitude

       $384=20 \times { altitude }$  

       $altitude = \frac{384}{20} \\= 19.2\ cm$

∴ altitude of rhombus is $19.2\ cm$ .