What is the difference in area (in cm²) of ΔABC having sides of 10 cm, 20 cm and 20 cm, and a right angled triangle ΔPQR with hypotenuse of 13 cm and one of the perpendiculars of 12 cm?

Note: √2 = 1.41, √3 = 1.73, √7 = 2.65, √13 = 3.61, √15 = 3.87, √21 = 4.58

Question

What is the difference in area (in cm²) of ΔABC having sides of 10 cm, 20 cm and 20 cm, and a right angled triangle ΔPQR with hypotenuse of 13 cm and one of the perpendiculars of 12 cm?

Note: √2 = 1.41, √3 = 1.73, √7 = 2.65, √13 = 3.61, √15 = 3.87, √21 = 4.58

Accepted Answer

Step 1: Calculate the area of ΔABC with sides 10 cm, 20 cm, and 20 cm using Heron's formula.
- Semi-perimeter, s = (10 + 20 + 20)/2 = 25 cm
- Area = √[s(s-a)(s-b)(s-c)] = √[25(25-10)(25-20)(25-20)] = √[25 × 15 × 5 × 5] = √(25 × 15 × 25) = √(9375)
- Simplify √9375: 9375 = 25 × 375 = 25 × 25 × 15 = 625 × 15
- √9375 = √(625 × 15) = 25 × √15
- Given √15 = 3.87, so area = 25 × 3.87 = 96.75 cm²

Step 2: Calculate the area of right angled triangle ΔPQR with hypotenuse 13 cm and one perpendicular 12 cm.
- Let the perpendicular sides be 12 cm and x cm.
- Using Pythagoras theorem: 12² + x² = 13²
- 144 + x² = 169
- x² = 25
- x = 5 cm
- Area = (1/2) × 12 × 5 = 30 cm²

Step 3: Find the difference in areas:
- Difference = 96.75 - 30 = 66.75 cm²

Therefore, the difference in area is 66.75 cm².

Answer

Answer

Answer