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Question
If each side of a rhombus is doubled, how much will its area increase?
Options
1.5 times
2 times
3 times
4 times
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Solution
3 times
Explanation:
Let the given rhombus be ABCD with side length a unit and length of diagonals AC and BD are d1 and d2 respectively.
In ΔAOD,
(AD)2 = (OA)2 + (OD)2 ...[Using pythagoras theorem]
`a^2 = (d_1/2)^2 + (d_2/2)^2`
`\implies a = sqrt(d_1^2/4 + d_2^2/4)`
`\implies a = 1/2 sqrt(d_1^2 + d_2^2)`
`\implies 2a = sqrt(d_1^2 + d_2^2)` ...(1)
Now, side length of rhombus is doubled.
i.e., 2a and let length of new diagonals are d’1 and d’2·
`\implies 2a = sqrt((d_1^’/2)^2 + (d_2^’/2)^2`
Using (1), we get
`d_1^’ = 2d_1` and `d_2^’ = 2d_2`
Area of rhombus ABCD = `1/2 xx d_1 xx d_2 = 1/2d_1d_2`
After doubling the side length, area of rhombus = `1/2 xx 2d_1 xx 2d_2 = 2d_1d_2`
∴ Increase in area = `2d_1d_2 - 1/2 d_1d_2`
= `3/2 d_1d_2`
= 3 × (Area of rhombus ABCD)
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