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Question

Find the area of the rhombus one side of which measures 20cm and one of whose diagonals is 24cm.
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Solution

The diagonals of a rhombus bisect at right angles
Consider $$\Delta$$ AOD
Using the Pythagoras theorem
$$AD^2 = OD^2 + AO^2$$
By substituting the values
$$20^2 =OD^2 + 12^2$$
On further calculation
$$OD^2 = 400 - 144$$
By subtraction
$$OD^2 = 256$$
By taking out the square root
OD = $$\sqrt{256}$$
So we get
OD = 16cm
We know that BD = 2OD
So we get
BC = 2 (16) = 32cm
We know that
Area of rhombus ABCD = $$\dfrac{1}{2} \times AC \times BD$$
By substituting the values
Area of rhombus ABCD = $$\dfrac{1}{2} \times 24 \times 32$$
On further calculation
Area of rhombus ABCD = $$384 cm^2$$
Therefore, the area of rhombus ABCD is $$384 cm^2.$$

Mathematics
RS Agarwal
Standard IX

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