The diagonals of a rhombus are 12cm and 16cm. What is its perimeter? Is it 20 (according to the proof that the sum of the squares of the diagonals of a rhombus is equal to the sum of the squares of its sides) or is it 40?
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To find this, you need to solve the problem by considering four right angled triangles. Since the diagonals bisect at right angles, the length of 1/2 of diagonal would be 6 cm and 8 cm respectively.
Since 6cm and 8cm meet at right angles, the hypotenuse is sqrt (6^2+8^2) = sqrt(36+64) or sqrt(100) = 10.
Hence, the side of the rhombus is 10, and the perimeter would be 4*10 = 40.
Thus, the perimeter is 40 cm.
Since 6cm and 8cm meet at right angles, the hypotenuse is sqrt (6^2+8^2) = sqrt(36+64) or sqrt(100) = 10.
Hence, the side of the rhombus is 10, and the perimeter would be 4*10 = 40.
Thus, the perimeter is 40 cm.
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