Question

ABCD is a rhombus with each side of length 10cm and one diagonal of length 10cm. find the area of rhombus.

Answer 1

Heya, find the answer in the attached image.

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Answer 2

Answer:

Let ABCD be a rhombus which is divided by the diagonal BD = 10 on two equal parts. Clearly, triangles ABD and CBD are congruent.

Also triangles have equal perimeters.

s= $\sf\frac{AB+BD+AD}{2}=$$\sf\frac{10+10+10}{2}=$ $\sf\frac{30}{2}=15cm$

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Area of ∆ABD = $\sf\sqrt\red{s(s-a)(s-b)(s-c)}$

$\sf\sqrt\red{15(15-10)(15-10)(15-10)cm^2}$

$\rightarrow$$\sf\sqrt\red{15\times5\times5\times5\:cm^2}$

$\rightarrow$$\sf\sqrt\red{3\times5\times\times5\times5\times5}$

$\implies$25$\sf\sqrt\red{3}$cm²

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Area of ∆ABD = Area of ∆CBD = 25√3 cm²

so, area of rhombus ABCD = 2 $\times$ Area of ∆ABD

$\rightarrow$$\sf{2\times25\sqrt3}$cm²

= 50√3 cm²