Question 592·Hard·Math

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In right triangle $\triangle PQR$ , $\angle Q$ is a right angle. The altitude from $Q$ meets hypotenuse $\overline{PR}$ at $S$ . If $QS=6$ units and $RS=4$ units, what is the length, in units, of $\overline{PQ}$ ?

An altitude from a right triangle’s right-angle vertex to its hypotenuse splits that side into two pieces. The altitude squared equals the product of those pieces. Find the missing piece first, then use the Pythagorean theorem in a smaller triangle to find the requested leg. Watch out for finding the other leg instead.

Hints

Hint 1
The altitude creates two smaller right triangles with matching angles. Because they’re similar, the altitude squared equals the product of the two pieces of the hypotenuse. Which piece is missing?
Hint 2
Let $PS$ be the missing piece. Put the given lengths into $QS^2=PS\cdot RS$ and solve for $PS$ . That gives you a side of a smaller triangle, not $PQ$ yet.
Hint 3
In right triangle $PQS$ , $PQ$ is the hypotenuse, the side opposite the right angle at $S$ . Use both legs, $PS$ and $QS$ , in the Pythagorean theorem.

Step-by-step

Approach 1: Find the missing piece, then use the Pythagorean theorem
Step 1
Relate the altitude to the two pieces
Because $QS$ is an altitude, it meets $PR$ at a right angle. Each smaller triangle shares an acute angle with the original right triangle, so the triangles are similar, or the same shape. Their matching sides give the altitude relationship:
$QS^2=PS\cdot RS.$
Put in the given lengths:
$6^2=PS\cdot4.$
Step 2
Find the missing piece
In Desmos, type $6^2\sim4x_1$ , where $x_1$ stands for $PS$ . Use $x_1$ instead of $x$ so Desmos solves for a value, and use $\sim$ to tell it to fit the equation. Under PARAMETERS, it shows $x_1=9$ . So $PS=9$ , not $PQ$ .
Step 3
Use the smaller right triangle
The altitude makes angle $PSQ$ a right angle, so $PQ$ is the hypotenuse of triangle $PQS$ . The Pythagorean theorem says the squares of its two legs add to the square of its hypotenuse:
$PQ^2=PS^2+QS^2.$
Substitute the lengths:
$PQ^2=9^2+6^2.$
Step 4
Find the exact length
The choices are multiples of $\sqrt{13}$ , so type $\frac{\sqrt{9^2+6^2}}{\sqrt{13}}$ in Desmos to find the coefficient of $\sqrt{13}$ . It prints $3$ . A length is positive, so $PQ=3\sqrt{13}$ units. Choice B.

Approach 2: Use similarity again to find the leg
Step 1
Find the whole hypotenuse
With $PS=9$ already found, add the two pieces to get the whole hypotenuse:
$PR=PS+RS=9+4=13.$
Step 2
Match the sides of the similar triangles
The big triangle $PQR$ and small triangle $PQS$ share angle $P$ , and both have a right angle, so they’re similar. Their matching sides are $PR$ with $PQ$ , and $PQ$ with $PS$ . Keep those pairs in the same order:
$\frac{PQ}{PR}=\frac{PS}{PQ}.$
Step 3
Build the leg relationship
Cross-multiply the matching-side proportion:
$PQ^2=PR\cdot PS.$
Put in the whole hypotenuse and the piece next to $PQ$ :
$PQ^2=13(9).$
Step 4
Read the radical’s coefficient
Type $\frac{\sqrt{13\cdot9}}{\sqrt{13}}$ in Desmos. It prints $3$ , the coefficient of $\sqrt{13}$ . So $PQ=3\sqrt{13}$ units. Choice B.

Question 592·Hard·Math

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