Question

13. The lengths of the diagonals of a rhombus
are 24 cm and 32 cm. The perimeter of
the rhombus is ...........

Answer 1

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${\underline{\overline{\bf{\mid{Solution:-}\mid}}}}$

•Let ABCD be a rhombus

with diagonal AC= 24 cm& DB= 32 cm

•By using property :-

Diagonal of a rhombus bisect each other perpendicularly

•°• AO= OC & BO= DO

OA = 12cm

DO= 16cm

•Now, In the figure ∆AOD is right angled triangle by the property and <AOD = 90°

By using Pythagoras theorem,

AD²= OA²+ OD²

AD²= 12²+16²

AD² = 144+256

AD²= 400

AD = √400

${\boxed{\bf{AD= 20 cm}}}$

•Now, perimeter of Rhombus= 4×side

(4 × 20)cm

$\huge{\underline{\overline{\bf{\mid{80 cm}\mid}}}}$

Answer 2

Given :-----

• Lengths of diagonals of a rhombus are 24cm and 32cm..

To Find :---

• Perimeter of Rhombus ?

Formula used :-------

• Diagonals of a rhombus bisect each other at 90°.
• Perimeter of Rhombus is Equal to 4 × side .

Calculation :------

Since diagonals of rhombus bisect each other at 90° , so by Pythagaras Theoram we can say that ,,

$\green{( \frac{D_1}{2} )^{2} + ( \frac{D_2}{2} )^{2} = (side)^{2} }$

Putting D1 as 24cm and D2 as 32 cm we get,,,

$\red\leadsto \: { (\frac{24}{2}) }^{2} + ( \frac{32}{2})^{2} = (side)^{2} \\ \\ \red\leadsto \: {12}^{2} + {16}^{2} = (side)^{2} \\ \\ \red\leadsto \: 144 + 256 = (side)^{2} \\ \\ \red\leadsto \: 400 = (side)^{2} \\ \\ \red\leadsto \: side \: = \sqrt{400} \\ \\ \red{\red\leadsto \: side \: = \: 20}$

Hence, Perimeter of Rhombus will be = 4 × side ..

$\pink{\large\boxed{\boxed{\bold{perimeter = 4 \times 20 = 80cm}}}}$

$\large\underline\textbf{Hope it Helps You.}$