find the area of a rhombus one of whose side measures 20 cm and one of whose diagonals is 24 cm
Answers
Given AC = 24cm.
Then AO = 12cm.
Let BO = x and AB = 20cm.
By Pythagoras theorem, we know that
c^2 = a^2 + b^2
20^2 = 12^2 + x^2
400 = 144 + x^2
x^2 = 400 - 144
x^2 = 256
x = 16cm.
Diagonal BD = 2 * 16
= 32cm.
We know that Area of the rhombus = 1/2 * (product of diagonals)
= 1/2 * 24 * 32
= 12 * 32
= 384 sq.cm.
Hope this helps!
The area of the rhombus is \bold{384\mathrm{cm}^{2}}384cm
2
Given:
One side of a rhombus = 20cm and one diagonal = 24 cm
To Find:
The area of a rhombus
Solution:
In the rhombus ABCD,
Let, AB = 20 cm
And AC = 24 cm
From the diagram,
AO + CO = 24 cm
AO = 12 cm [∵ AO = CO]
In \Delta \mathrm{AOB},ΔAOB,
AB = 20 cm and AO= 12 cm
Using Pythagoras Theorem,
\mathrm{BO}=\sqrt{20^{2}-12^{2}}\ \mathrm{cm}BO=
20
2
−12
2
cm
=\sqrt{400-144}\ \mathrm{cm}=
400−144
cm
=\sqrt{256}\ \mathrm{cm}=
256
cm
=16\ \mathrm{cm}=16 cm
Hence, BO = DO = 16 cm
∴ BD = 16 cm + 16 cm = 32 cm.
Area of a Rhombus =\frac{1}{2} \times \text { Diagonal } 1 \times \text { Diagonal } 2=
2
1
× Diagonal 1× Diagonal 2
=\frac{1}{2} \times A C \times B D=
2
1
×AC×BD
=\frac{1}{2} \times 24\ \mathrm{cm} \times 32\ \mathrm{cm}=
2
1
×24 cm×32 cm
=384\ \mathrm{cm}^{2}=384 cm