Question

the sides of a rhombus are 5cm each and one diagonal is 8cm. Calculate the length of the other diagonal and the area of the rhombus
it's answer is 6cm,24cm^2​

Answer 1

★ Given:-

• Side of Rhombus = 5cm

• Length of diagonal₁ = 8cm

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★ To Find:-

• Length of other diagonal

• Area of rhombus

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★ Formula/Concept Used:-

• Diagonal bisects each other perpendicularly

• Pythagorean theorem ( h² = b² + p² )

• Area of rhombus = ½ × diagonal₁ × diagonal₂

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★ Solution:-

Let half the length of second diagonal be x

As two diagonals and a side form right angled triangle ;

By pythagorean theoream:

⇒ 5² = 4² + x²

⇒ 25 = 16 + x²

⇒ x² = 25 - 16

⇒ x² = 9

⇒ x = √9

⇒ x = √3×3

∴ x = 3cm

Length of other diagonal = 2x = 2×3 = 6cm

Area of rhombus

= ½ × diagonal₁ × diagonal₂

= ½ × 8 × 6

= 48 ÷ 2

= 24cm²

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★ Answer:-

• Length of other diagonal = 6cm
• Area of rhombus = 24cm²

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Answer 2

AnswEr-:

• $\underline {\mathrm {\star{\pink{The\:Area \:of\:Rhombus \;is\:24\:cm^{2}\:.}}}}$

• $\underline {\mathrm {\pink{\star{ The\:Diagonal _{2} \:or \:Diagonal \:of\:Rhombus \:is\:6\:cm.}`}}}$

Explanation-:

$\mathrm {\bf{ Given -:}}$

• The each sides of a rhombus are of 5cm .

• The one of its Diagonal is 8 cm .

$\mathrm {\bf{ To\:Find -:}}$

• The Length of the other Diagonal.

• The Area of a Rhombus.

$\mathrm {\bf{\dag{Solution \:of\:Question \:-:}}}$

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$\mathrm {\bf{\dag{ Finding \: \:\:Other\:Diagonal \:of\:Rhombus \:-:}}}\:$

As , We know that ,

• Diagonals of Rhombus bisectes each other and they are Perpendicular.

And,

• As , Half of Two Diagonals form Right Angled Triangle in Rhombus.

$\mathrm {\bf{ Let's \:Assume -:}}$

• Side of Rhombus as Base Hypotenuse of Right angled triangle = 5cm

• Half of Diagonal 1 = Base of Right angled triangle = $\dfrac{8}{2} = 4\:cm$

• Half of Diagonal 2 = Perpendicular of Right angled triangle = x cm

As , We know that ,

Pythagoras Theorem-:

• $\underline{\boxed{\star{\sf{\blue{ Hypotenuse^{2}\: = \:Base^{2} \: + \: Perpendicular ^{2} }}}}}$

Now , By Putting known Values in Pythagoras Theorem-:

• $\qquad \quad \qquad \quad \longmapsto {\mathrm { 5^{2} = 4^{2} + Perpendicular^{2} }}$

• $\qquad \quad \qquad \quad \longmapsto {\mathrm { 25 = 16 + Perpendicular^{2} }}$

• $\qquad \quad \qquad \quad \longmapsto {\mathrm { 25 -16 = Perpendicular^{2} }}$

• $\qquad \quad \qquad \quad \longmapsto {\mathrm { 9 = Perpendicular^{2} }}$

• $\qquad \quad \qquad \quad \longmapsto {\mathrm { \sqrt {9}= Perpendicular }}$

• $\qquad \quad \qquad \quad \underline {\boxed{\pink{\mathrm { 3 cm = Perpendicular }}}}$

Then Putting x = 3 ,

• Half of Diagonal 2 = Perpendicular of Right angled triangle = x = 3 cm

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As , We know that ,

• Half of Diagonal 2 or Other Diagonal = 3 cm

Then ,

• Diagonal 2 or Other Diagonal of Rhombus = 3 × 2 = 6 cm

Hence ,

• $\underline {\mathrm {\pink{\star{ The\:Diagonal _{2} \:or \:Diagonal \:of\:Rhombus \:is\:6\:cm.}`}}}$

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$\mathrm {\bf{\dag{ Finding \: \:\:Area \:of\:Rhombus \:-:}}}\:$

As , We know that,

• $\underline{\boxed{\star{\sf{\red{ Area_{(Rhombus)}  \: = \dfrac{1}{2} \times Diagonal _{1} \times Diagonal _{2} \: sq.units }}}}}$

• $\mathrm{\bf{\blue {Here \:\: -:}}} \begin{cases} \sf{\blue{The\:Diagonal _{1} \:or\:one\:Diagonal \:of \:the\:Rhombus \: \:is\:= \frak{8\:cm}}} & \\\\\sf{\pink{The\:Diagonal _{2} \:or\:other\:Diagonal \:of \:the\:Rhombus \: \:is\:= \frak{\:6\:cm}}} \end{cases}$

Now By Putting known Values in Formula for Area of Rhombus-:

• $\qquad \quad \qquad \quad \longmapsto {\mathrm { \dfrac{1}{2} \times 8 \times 6 }}$

• $\qquad \quad \qquad \quad \longmapsto {\mathrm { \dfrac{1}{\cancel {2}} \times \cancel {8} \times 6 }}$

• $\qquad \quad \qquad \quad \longmapsto {\mathrm { 4 \times 6 }}$

• $\qquad \quad \qquad \quad \underline {\boxed{\pink{\mathrm { Area = 24\:cm^{2} }}}}$

Hence ,

• $\underline {\mathrm {\star{\pink{The\:Area \:of\:Rhombus \;is\:24\:cm^{2}\:.}}}}$

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$\large { \boxed {\mathrm |\:\:{\underline {More \:To\:Know\:-:}}\:\:|}}$

• Area of Rectangle = Length × Breadth sq.units

• Area of Square = Side × Side sq.units

• Area of Triangle = ½ × Base × Height sq.units

• Area of Trapezium = ½ × Height × ( a +b ) or Sum of Parallel sides sq.units

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