AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to

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9.1 — Easy An angle in a semicircle is a right angle, so both $	riangle APB$ and $	riangle AQB$ are right-angled at $P$ and $Q$. Apply Pythagoras in $	riangle APB$ to find $AP$ (hence $AQ$), then again in $	riangle AQB$ to get $QB$. A B P Q &#9654; Replay Radius $=5\Rightarrow$ diameter $AB=10$. Given $PB=6$ and $AP=2\,AQ$. Let $AQ=x$, so $AP=2x$. 1 Right angle at $P$. $AB$ is a diameter, so $\angle APB=90^\circ$. Pythagoras in $	riangle APB$: $$\begin{aligned} &AB^2=AP^2+PB^2\ &\Rightarrow\ 10^

CAT 2019 Slot 1 — QA Question 27

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