If the area of a rhombus of perimeter 32 cm is 36 sq. cm, then what is the sum of the lengths (in cm) of its diagonals?

To solve this problem, we need to use the formulas for the area and perimeter of a rhombus.

Step 1: The perimeter of a rhombus is given by 4a, where a is the length of a side. Given that the perimeter is 32 cm, we can find the length of a side as follows: a = 32/4 = 8 cm

Step 2: The area of a rhombus is given by (d1d2)/2, where d1 and d2 are the lengths of the diagonals. Given that the area is 36 sq. cm, we can write the equation as follows: 36 = (d1d2)/2

Step 3: We also know that in a rhombus, the diagonals are perpendicular bisectors of each other. Therefore, by Pythagoras' theorem, we have: (d1/2)^2 + (d2/2)^2 = a^2 Substituting the value of a from step 1, we get: (d1/2)^2 + (d2/2)^2 = 8^2 (d1^2 + d2^2)/4 = 64 d1^2 + d2^2 = 256

Step 4: We need to find the sum of the lengths of the diagonals, which is d1 + d2. By squaring this, we get: (d1 + d2)^2 = d1^2 + 2d1d2 + d2^2 Substituting the value of d1^2 + d2^2 from step 3, we get: (d1 + d2)^2 = 256 + 2*(2*36) (d1 + d2)^2 = 256 + 144 (d1 + d2)^2 = 400

Step 5: Taking the square root of both sides, we get: d1 + d2 = 20 cm

So, the sum of the lengths of the diagonals of the rhombus is 20 cm.