(C) 2183.The perimeter of a right angled triangle is 72 cm and its are

1.	(C) 2183.The perimeter of a right angled triangle is 72 cm and its area is 216 cm2. Find the sum of thelengths of its perpendicular sides (in cm)(A) 36(C) 42(B) 32(D) 50
Answer» Let the two perpendicular sides be x and y. xy/2 = 216, or xy = 432 …(1) The hypotenuse = (x^2+y^2)^0.5. The perimeter 72 = x + y + (x^2+y^2)^0.5, or (x^2+y^2)^0.5 = 72 -x-y. Squaring both sides x^2+y^2 = (72 -x-y)^2 = 5184 + x^2 + y^2 - 144x - 144y + 2xy, or 5184 - 144x - 144y + 2xy = 0 …(2). Put xy = 432 from (1) on (2) to get 5184 - 144x - 144y + 2*432 = 0, or 6048 - 144x - 144y = 0 42 - x - y = 0 Therefore, the sum of the lengths of perpendicular sides = 42 cm. Check: x + y = 42. Or y = 42-x xy/2 = 216 (42-x)x = 432, or x^2 - 42x + 432 = 0 (x-24)(x-18) = 0 Hence the two perpendicular sides are 24 cm and 18 cm,and the hypotenuse = (24^2+18^2)^0.5=(576+324)^0.5 = 900^0.5 = 30 cm. Like my answer if you find it useful!

(C) 2183.The perimeter of a right angled triangle is 72 cm and its area is 216 cm2. Find the sum of thelengths of its perpendicular sides (in cm)(A) 36(C) 42(B) 32(D) 50

Answer»

Let the two perpendicular sides be x and y.

xy/2 = 216, or

xy = 432 …(1)

The hypotenuse = (x^2+y^2)^0.5. The perimeter

72 = x + y + (x^2+y^2)^0.5, or

(x^2+y^2)^0.5 = 72 -x-y. Squaring both sides

x^2+y^2 = (72 -x-y)^2 = 5184 + x^2 + y^2 - 144x - 144y + 2xy, or

5184 - 144x - 144y + 2xy = 0 …(2). Put xy = 432 from (1) on (2) to get

5184 - 144x - 144y + 2*432 = 0, or

6048 - 144x - 144y = 0

42 - x - y = 0

Therefore, the sum of the lengths of perpendicular sides = 42 cm.

Check: x + y = 42. Or y = 42-x

xy/2 = 216

(42-x)x = 432, or

x^2 - 42x + 432 = 0

(x-24)(x-18) = 0

Hence the two perpendicular sides are 24 cm and 18 cm,and the hypotenuse = (24^2+18^2)^0.5=(576+324)^0.5 = 900^0.5 = 30 cm.

Like my answer if you find it useful!