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

Correct — A. Two triangles, two areas, then subtract.

ΔABC has sides 10, 20, 20, so it is isosceles. Drop the height onto the 10 cm base, which splits it into 5 and 5.
Height = √(20² − 5²) = √375 = 5√15
Area = ½ × 10 × 5√15 = 25√15 = 25 × 3.87 = 96.75 cm²

ΔPQR is right-angled with hypotenuse 13 and one perpendicular 12, so the other leg is √(169 − 144) = 5 — the 5-12-13 triple.
Area = ½ × 12 × 5 = 30 cm²

Difference = 96.75 − 30 = 66.75 cm² → option (a)

Why the others are wrong

• (b)
  53.58 — ΔPQR is fixed at 30 cm² by the legs 12 and 5, so 53.58 would need ΔABC to be 83.58. Its area is 25√15, which with the printed √15 = 3.87 is 96.75.
• (c)
  36.57 — 36.57 would need ΔABC = 66.57 cm², not 96.75. The height is √(400 − 25) = √375 = 5√15, giving 25 × 3.87 once the ½ × base is applied.
• (d)
  70.05 — Close enough to be dangerous: 70.05 needs ΔABC = 100.05, i.e. √15 taken as 4.002. The Note prints 3.87, and using it gives 96.75.

Neither triangle needs Heron's formula, and both give up their area to a single perpendicular.

In an isosceles triangle the height to the unequal side bisects it, so with equal sides a and base b the height is √(a² − b²⁄4). For 20, 20 and 10 that is √(400 − 25) = √375 = 5√15, and the area is ½ × 10 × 5√15 = 25√15.

In a right triangle the two legs are themselves a base and a height, so the area is half their product. Recognising 5-12-13 saves the Pythagoras step entirely.

The word 'difference' is the last instruction: compute both, then subtract the smaller.

Key facts

• In an isosceles triangle with equal sides a and base b, the height to the base is √(a² − b²/4).
• For sides 20, 20 and base 10 that height is 5√15 and the area is 25√15.
• 5-12-13 is a Pythagorean triple, so hypotenuse 13 with one leg 12 leaves the other leg 5.
• A right triangle's area is half the product of its two legs, not half the hypotenuse times a leg.

Common traps

• Halving the wrong side and using √(20² − 10²) for the height
• Taking ½ × 13 × 12 for ΔPQR, treating the hypotenuse as a base
• Substituting a remembered √15 = 3.873 instead of the paper's 3.87

No directly related past PYQ was found.