Question

1.If the length of one side and
one diagonal of a rhombus are
20 mand 24 m respectively,
then its area is *​

Answer 1

Answer:

Diagonal of rhombus bisect each other.

24/2=12cm

From Pythagoras theorem,

12^2+x^2=20^2

144+x^2=400

x^2=400-144

x=√256=16

Other diagonal=16(2)=32

Area=1/2×(32)×(24)

=384

Answer 2

AnswEr :

⠀⠀⠀⠀⠀If one side of the rhombus is 20 m and another diagonal measures 24 m then, the area of the rhombus would be 384 cm².

ExplanaTion :

$\sf Here \: \begin{cases} \sf S ide = 20 \: m \\ \\ \sf Diagonal = 24 \: m \: \end{cases}$

Let us assume ABCD as a rhombus,

we know that,

• All sides of the rhombus are equal .[ Property of a rhombus ]

• AB = BC = CD = DA = 20 m

Also, BD = 24 m [Diagonal]

In ΔABD,

$\star \: { \boxed{ \sf{ \purple{Semi \: perimeter = \dfrac{A + B +C}{2} }}}}$

Using the above formula,

$\sf : \implies \: \: Semi \: perimeter = \dfrac{20+ 20 + 24}{2}$

$\sf : \implies \: \: Semi \: perimeter = \dfrac{64}{2}$

$\sf : \implies \: \: \blue{Semi \: perimeter = 32}$

Now,

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$\star \: { \boxed{ \sf{ \red{Area \: of \: \triangle ABD = \sqrt{s(s - a)(s - b)(s - c)} }}}}$

Using the formula,

$:\implies \sf \: \: \: Area = \sqrt{32 \times (32 - 20)(32 - 20)(32 - 24)}$

$:\implies \sf \: \: \: Area = \sqrt{32 \times 12 \times 12 \times 8}$

$:\implies \sf \: \: \: Area = \sqrt{4 \times 8 \times 8 \times {12}^{2} }$

$:\implies \sf \: \: \: Area = \sqrt{ {2}^{2} \times {8}^{2} \times {12}^{2} }$

$:\implies \sf \: \: \: Area = 2 \times 8 \times 12$

$:\implies \sf \: \: \: { \green{Area = 192 \: {m}^{2}}}$

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$\: \: \: \: \: \: \: \: \: \: \star \: { \underline{ \sf{Area \: of \: rhombus :}}}$

$\sf : \implies \: \: Area (ABCD) = Area (ABD) + Area (BCD)$

$\sf : \implies \: \: Area (ABCD) = 192 + 192$

$\sf : \implies \: \: { \red{ Area (ABCD) = 384 \: {m}^{2}}}$

Therefore, Area of the rhombus is 384 cm².

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