Mathematics
Two squares have sides x cm and (x + 5) cm. The sum of their areas is 697 sq. cm.
(i) Express this as an algebraic equation in x.
(ii) Solve this equation to find the sides of the squares.
Answer
(i) Given,
Two squares have sides x cm and (x + 5) cm.
Sum of areas of squares = 697 cm2
∴ x2 + (x + 5)2 = 697
⇒ x2 + x2 + 10x + 25 = 697
⇒ 2x2 + 10x + 25 - 697 = 0
⇒ 2x2 + 10x - 672 = 0
⇒ 2(x2 + 5x - 336) = 0
⇒ x2 + 5x - 336 = 0.
Hence, the required equation is x2 + 5x - 336 = 0.
(ii) Solving,
⇒ x2 + 5x - 336 = 0
⇒ x2 + 21x - 16x - 336 = 0
⇒ x(x + 21) - 16(x + 21) = 0
⇒ (x - 16)(x + 21) = 0
⇒ x - 16 = 0 or x + 21 = 0 [Using zero-product rule]
⇒ x = 16 or x = -21.
Since, length cannot be negative.
Thus, x = 16
Length of side of first square = x = 16 cm
Length of side of second square = x + 5 = 16 + 5 = 21 cm
Hence, length of sides of squares are 16 cm and 21 cm.
Related Questions
The hypotenuse of a right triangle is 20 m. If the difference between the lengths of other sides be 4 m, find the other sides.
The lengths of the sides of a right triangle are (2x − 1) m, (4x) m and (4x + 1) m, where x > 0. Find :
(i) the value of x,
(ii) the area of the triangle.
The area of a right-angled triangle is 96 m2. If its base is three times its altitude, find the base.
The lengths of the parallel sides of a trapezium are (x + 8) cm and (2x + 3) cm, and the distance between them is (x + 4) cm. If its area is 590 cm2, find the value of x.