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CAT 2019 Slot 2 — QA Question 20

Geometric CentersEasy

Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is

Answer & solution

  • 10

  • B

    8√2

  • C

    6√2​​​​​​​

  • D

    5

Solution

Easy

A right angle at AA means AA lies on the circle with BCBC as diameter (Thales). The foot PP is on BCBC, and APAP is largest when AA is at the top of the semicircle — directly above the centre — giving APAP equal to the radius.

B C A P AP = 10
1

Locate AA on a semicircle. Since ∠BAC=90∘\angle BAC=90^{\circ}, vertex AA lies on the circle whose diameter is the hypotenuse BC=20BC=20, so the radius is 1010.

r=BC2=202=10\begin{aligned} &r=\frac{BC}{2}=\frac{20}{2}=10 \end{aligned}
2

Maximise the altitude. APAP is the distance from AA to the diameter; it is greatest when AA is the highest point of the circle, i.e. directly above the centre, where that distance equals the radius.

APmax⁡=r=10 cm\begin{aligned} &AP_{\max}=r=10\ \text{cm} \end{aligned}
APmax⁡=10 cmAP_{\max}=10\ \text{cm}
CAT 2019 Slot 2 QA Q20: Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the — Solution | TheCATExam