Question

area of rhombus is 840 cm². if the perimeter of the rhombus is 148 cm, then find the sum of the length of its two diagonals​

Answer 1

AnswEr :

• Area of Rhombus is 840 cm²
• Perimeter of Rhombus is 148 cm
• Find the Sum of Diagonals.

$\small \boxed{\bf{Area \: of \: Rhombus = \frac{1}{2} \times D_1 \times D_2}}$

$\leadsto \sf{Area = \dfrac{1}{2} \times D_1 \times D_2}$

$\leadsto \sf{840 = \dfrac{1}{2} \times D_1 \times D_2}$

$\leadsto \sf{840 \times 2 = D_1 \times D_2}$

$\leadsto \sf{ D_1 \times D_2 = 1680 \: cm^{2}}$

$\small \boxed{ \bf{Perimeter \: of \: Rhombus = 4 \times Side}}$

$\leadsto\sf{Perimeter = 4 \times Side}$

$\leadsto\sf{148 = 4 \times Side}$

$\leadsto\sf{ \cancel\dfrac{148}{4} =Side}$

$\leadsto\sf{ Side = 37\: cm}$

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$\small \boxed{ \bf{(Side)^{2} = \dfrac{(Diagonal_1)^{2} + (Diagonal_2)^{2} }{4}}}$

$\longrightarrow \sf{(Side)^{2} = \dfrac{(D_1)^{2} + (D_2)^{2} }{4}}$

$\longrightarrow \sf{(37)^{2} = \dfrac{(D_1)^{2} + (D_2)^{2} }{4}}$

$\longrightarrow \sf{1369 = \dfrac{(D_1)^{2} + (D_2)^{2} }{4}}$

$\longrightarrow \sf{1369 \times 4 = (D_1)^{2} + (D_2)^{2}}$

$\longrightarrow \sf{ (D_1)^{2} + (D_2)^{2} = 5476}$

• (a + b)² = a² + b² + 2ab
• therefore a² + b² = (a + b)² - 2ab

$\longrightarrow \sf{ (D_1 + D_2)^{2} - (2 \times D_1 \times D_2) = 5476}$

$\longrightarrow \sf{ (D_1 + D_2)^{2} - (2 \times 1680) = 5476}$

$\longrightarrow \sf{ (D_1 + D_2)^{2} - 3360= 5476}$

$\longrightarrow \sf{ (D_1 + D_2)^{2} = 5476 + 3360}$

$\longrightarrow \sf{ (D_1 + D_2)^{2} = 8836}$

$\longrightarrow \sf{D_1 + D_2 = \sqrt{8836} }$

$\longrightarrow \sf{D_1 + D_2 =94\: cm }$

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∴ Sum of Diagonals of Rhombus is 94 cm.

Answer 2

Given :-----

• Area of rhombus = 840cm²
• Perimeter of rhombus = 148cm

To Find :---

• Sum of length of its two diagonals .

Concept used :----

• Area of Rhombus = 1/2 × D1 × D2
• Perimeter of Rhombus = 4 × side
• Diagonals bisect each other perpendicularly..
• a² + b² = (a + b)² - 2ab
• Pythagoras Theoram

Solution :------

1/2 × d1 × d2 = 840

d1 × d2 = 1680 -------------------- Equation (1)

Now,

4 × side = 148

side = 148/4 = 37 cm .

Now, see image , since ∆ COD is Rt. Angled ∆ , Right angle at 0.

in ∆COD we have,

DC = Side of Rhombus = 37cm

CO = Half of diagonal = d1/2

DO = d2/2

and , by pythagoras theoram we have ,

DC² = CO² + D0²

Putting values we get,

${37}^{2} = ( \frac{d1}{2} )^{2} + ( \frac{d2}{2} )^{2} \\ \\ 1369 = \frac{d1^{2} + {d2}^{2} }{4} \\ \\ d1^{2} + {d2}^{2} = 1369 \times 4 = 5476$

Now, we know that, a² + b² = (a + b)² - 2ab

so, we can write,

d1² + d2² = (d1+d2)² - 2×d1×d2

so,

(d1+d2)² - 2×d1×d2 = 5476

putting value of Equation (1) Now, we get,

→ (d1+d2)² - 2×1680 = 5476

→ (d1+d2)² = 5476 + 3840

→ (d1+d2)² = 8836

→ (d1+d2) = √(8836)

→ (d1+d2) = 94 cm.

So, the sum of length of both Diagonals is 94cm ..