Q192
ArithmeticGeometry & Mensuration
A ladder is kept against a wall. The top of the ladder touches the wall at a height of 270 cm from the ground. The base of the ladder is 54 cm away from the wall on the ground. Find the length of the ladder.
- a.$\sqrt{53682}$ cm.
- b.$\sqrt{68164}$ cm.
- c.$\sqrt{75816}$ cm.
- d.$\sqrt{82547}$ cm.
Answer: (C) $\sqrt{75816}$ cm.
The wall, the ground and the ladder form a right-angled triangle with the ladder as the hypotenuse. Length $=\sqrt{270^2+54^2}=\sqrt{72900+2916}=\sqrt{75816}$ cm. Since $270=5\times54$, this equals $54\sqrt{26}\approx275.3$ cm. A quick unit-digit check helps: $270^2$ ends in 0 and $54^2$ ends in 6, so the sum must end in 6 - only $\sqrt{75816}$ fits.