CEMC Cayley, 2020

Geometry Difficulty 3.8 AMC 10/12 Find the answer Canada

In the diagram, points $S$ and $T$ are on sides $QR$ and $PQ$ , respectively, of $\triangle PQR$ so that $PS$ is perpendicular to $QR$ and $RT$ is perpendicular to $PQ$ .

If $PT=1$ , $TQ=4$ , and $QS=3$ , what is the length of $SR$ ?

$3$
$\frac{11}{3}$
$\frac{15}{4}$
$\frac{7}{2}$
$4$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Since $PT=1$ and $TQ=4$ , then $PQ = PT+TQ=1+4=5$ .

$\triangle PSQ$ is right-angled at $S$ and has hypotenuse $PQ$ .

We can thus apply the Pythagorean Theorem to obtain $PS^2 = PQ^2 - QS^2 = 5^2 - 3^2 = 16$ .

Since $\text{PS gt;0}$ , then $PS=4$ .

Consider $\triangle PSQ$ and $\triangle RTQ$ .

Each is right-angled and they share a common angle at $Q$ . Thus, these two triangles are similar.

This tells us that $\dfrac{PQ}{QS} = \dfrac{QR}{TQ}$ .

Using the lengths that we know, $\dfrac{5}{3} = \dfrac{QR}{4}$ and so $QR = \dfrac{4 \cdot 5}{3} = \dfrac{20}{3}$ .

Finally, $SR = QR - QS = \dfrac{20}{3} - 3 = \dfrac{11}{3}$ .

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