Question

ABCD is a rhombus in which length of diagonal is 48 cm and side is 25 cm. Find the sum
of the length of diagonals.​

Answer 1

Answer:

14 + 48 = 62

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Step-by-step explanation:

Answer 2

$\huge\sf\pink{Answer}$

☞ Your Answer is 62 cm

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$\huge\sf\blue{Given}$

✭ One side of a Rhombus is 25 cm

✭ Diagonal of a Rhombus is 48 cm

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$\huge\sf\gray{To \:Find}$

◈ The sum of the length of the Diagonals

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$\huge\sf\purple{Steps}$

We know that,

$\underline{\boxed{\sf \dfrac{d_2}{2} = \sqrt{a^2-(\dfrac{d_1}{2})^2}}}$

◕ $\sf d_1 = 48 cm$

◕ $\sf a = 25 cm$

◕ $\sf d_2 = ?$

Substituting the given values,

➝ $\sf \dfrac{d_2}{2} = \sqrt{a^2-(\dfrac{d_1}{2})}$

➝ $\sf \dfrac{d_2}{2} = \sqrt{25^2-(\dfrac{48}{2})^2}$

➝ $\sf \dfrac{d_2}{2} = \sqrt{625-24^2}$

➝ $\sf \dfrac{d_2}{2} = \sqrt{625-576}$

➝ $\sf \dfrac{d_2}{2} = \sqrt{49}$

➝ $\sf \dfrac{d_2}{2} = 7$

➝ $\sf d_2 = 7 \times 2$

➝ $\sf \red{d_2 = 14 \ cm}$

Sum of the two diagonals will be,

➳ $\sf 48+14$

➳ $\sf\orange{Sum \ of \ Diagonals = 62 \ cm}$

$\sf\star\: Diagram \:\star$

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(1,3)\qbezier(3,0)(3,0)(4,3)\qbezier(1,3)(1,3)(4,3)\qbezier(3,0)(0,0)(0,0)\qbezier(0,0)(0,0)(4,3)\qbezier(1,3)(3,0)(3,0)\put(0.6,1.2){\sf 48 cm}\put(2,3,2){\sf 25 cm}\put(1.9,1.7){ \bf O }\put(-0.3,-0.2){ \sf A }\put(3.1,-0.2){ \sf B }\put(4,3){ \sf C }\put(0.7,3){ \sf D }\end{picture}$

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