Question

In △PQR, QS is an altitude.
Solve for x and y.

Answer 1

If the altitude is drawn to the hypotenuse of a right triangle, the length of the altitude is the geometric mean between the lengths of the segments of the hypotenuse.

27/x = x/ 3

x^2 = 27*3

x = 9 which is SQ.

Answer 2

If the altitude is drawn to the hypotenuse of a right triangle, the length of either leg is the geometric mean between the hypotenuse and the segment of the hypotenuse adjacent to the leg.

27/y = y/30

y^2 = 27*30

y = 9√10

Answer 3

Suggestion: The theorems used above are corollaries from the theorem: If an altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to each other and to the given right triangle.

You can us that to find the missing values after wading through a miasma of similar triangles or just learn the two corollaries and be done with it. Either way, you have to know your theorems.

Answer 4

This can be solved using the pythagoras theorem.

For triangle PQS: (eqn1)
\[y ^{2}=27^{2}+x ^{2}\]

For triangle PQR:(eqn2)
\[30 ^{2}=y^{2}+QR^{2}\]

For triangle PQR:(eqn3)
\[QR ^{2}=x^{2}+3^{2}\]

Subtituting eqn3 in eqn2:
\[30 ^{2}=y^{2}+x^{2}+3^{2}\]

Removing the subject of formula, y:
\[30 ^{2}=27^{2}+x ^{2}+x^{2}+3^{2}\]
\[900=738+2x ^{2}\]
\[2x ^{2}=900-738=162\]
\[x ^{2}=81\]
\[x =9\]

Finding value of y:
\[y ^{2}=27^{2}+x ^{2}\]
\[y ^{2}=27^{2}+81=810\]
\[y=\sqrt{810}=\sqrt{81\times10}\]
\[y=9\sqrt{10}\]