Sum of the areas of two squares is 260 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Quadratic Equations
65 Likes

Question

Sum of the areas of two squares is 260 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

KnowledgeBoat · Accepted Answer

Let the sides of two squares be 'a' m and 'b' m. As we know that area of square = (side)2 and perimeter of square = 4 x side It is given that sum of the areas of two squares is 260 m2. ⇒ a2 + b2 = 260 .......... (1) And, the difference of their perimeters is 24 m. ⇒ 4a - 4b = 24 ⇒ 4(a - b) = 24 ⇒ a - b = ⇒ a - b = 6 ⇒ a = 6 + b .......... (2) Substituting the value of a in equation (1), we get ⇒ (6 + b)2 + b2 = 260 ⇒ 36 + b2 + 12b + b2 = 260 ⇒ 36 + 2b2 + 12b - 260 = 0 ⇒ 2b2 + 12b - 224 = 0 ⇒ 2(b2 + 6b - 112) = 0 ⇒ b2 + 6b - 112 = 0 ⇒ b2 + 14b - 8b - 112 = 0 ⇒ b(b + 14) - 8(b + 14) = 0 ⇒ (b + 14)(b - 8) = 0 ⇒ (b + 14) = 0 or (b - 8) = 0 ⇒ b = -14 or b = 8 Since, side of square cannot be negative. Side of one square = 8 m Side of other square = 6 + 8 = 14 m Thus, side of squares = 8 m and 14 m.

Mathematics

Sum of the areas of two squares is 260 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Quadratic Equations

Related Questions

The diagonal of a rectangular field is 16 m more than the shorter side. If the longer side is 14 m more than the shorter side, find the length and the breadth of the rectangular field.

A man walks 1 km/hr faster than his usual speed and covers a distance of 3 km in 15 minutes less time. Find his usual speed.

A bus travels at a certain speed for a distance of 75 km and then travels a distance of 90 km at a speed of 10 km/hr more than the first speed. If it takes 3 hours to complete the journey, find the original speed.