Given:
Triangle ABC is right-angled at B.
AB + BC exceeds AC by 10 units.
Perimeter of the triangle = 60 units.
Formula Used:
Perimeter of a triangle = AB + BC + AC.
Area of a right-angled triangle = (1/2) × AB × BC.
Pythagoras theorem: AC2 = AB2 + BC2.
Calculation:
Let AB = x, BC = y, and AC = z.
⇒ x + y + z = 60 ...(i)
⇒ x + y = z + 10 ...(ii)
From (ii), substitute z = x + y - 10 into (i):
⇒ x + y + (x + y - 10) = 60
⇒ 2(x + y) - 10 = 60
⇒ 2(x + y) = 70
⇒ x + y = 35 ...(iii)
From (ii), z = x + y - 10 = 35 - 10 = 25.
Using Pythagoras theorem:
⇒ z2 = x2 + y2
⇒ 252 = x2 + y2
⇒ 625 = x2 + y2 ...(iv)
From (iii), y = 35 - x. Substitute y = 35 - x into (iv):
⇒ 625 = x2 + (35 - x)2
⇒ 625 = x2 + (1225 - 70x + x2)
⇒ 625 = 2x2 - 70x + 1225
⇒ 2x2 - 70x + 600 = 0
Divide through by 2:
⇒ x2 - 35x + 300 = 0
Factorize:
⇒ (x - 20)(x - 15) = 0
⇒ x = 20 or x = 15
If x = 20, then y = 35 - 20 = 15.
If x = 15, then y = 35 - 15 = 20.
In both cases, AB = 20 and BC = 15.
Area of the triangle:
⇒ Area = (1/2) × AB × BC
⇒ Area = (1/2) × 20 × 15
⇒ Area = 150 square units.
∴ The correct answer is 150 square units.