Question

7. The area of a rhombus is 21 cm. If the perimeter of
the rhombus is 40 cm, then find the sum of its
diagonals (in cm).

Answer 1

AnswEr :

• Area of Rhombus is 21 cm
• Perimeter of Rhombus is 40 cm
• Find the Sum of Diagonals.

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

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

$\leadsto \sf{21 = \dfrac{1}{2} \times D_1 \times D_2}$

$\leadsto \sf{21 \times 2 = D_1 \times D_2}$

$\leadsto \sf{ D_1 \times D_2 = 42 \: cm^{2}}$

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

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

$\leadsto\sf{40 = 4 \times Side}$

$\leadsto\sf{ \cancel\dfrac{40}{4} =Side}$

$\leadsto\sf{ Side = 10 \: cm}$

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

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

$\longrightarrow \sf{(10)^{2} = \dfrac{(D_1)^{2} + (D_2)^{2} }{4}}$

$\longrightarrow \sf{100 = \dfrac{(D_1)^{2} + (D_2)^{2} }{4}}$

$\longrightarrow \sf{100 \times 4 = (D_1)^{2} + (D_2)^{2}}$

$\longrightarrow \sf{ (D_1)^{2} + (D_2)^{2} = 400}$

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

$\longrightarrow \sf{ (D_1 + D_2)^{2} - (2 \times D_1 \times D_2) = 400}$

$\longrightarrow \sf{ (D_1 + D_2)^{2} - (2 \times 42) = 400}$

$\longrightarrow \sf{ (D_1 + D_2)^{2} - 84= 400}$

$\longrightarrow \sf{ (D_1 + D_2)^{2} = 400 + 84}$

$\longrightarrow \sf{ (D_1 + D_2)^{2} = 484}$

$\longrightarrow \sf{D_1 + D_2 = \sqrt{484} }$

$\longrightarrow \sf{D_1 + D_2 =22 \: cm }$

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∴ Sum of Diagonals of Rhombus is 22 cm.