Similarity in Right Triangles: Practice

The hypotenuse is opposite the right angle of the triangle being named. The whole triangle is right at $C$, so its hypotenuse is $AB$.

$BC$ and $CA$ are its legs. $CH$ is an interior altitude, not a side of the whole triangle.

The angle order in $ABC$ is $31^\circ$, $59^\circ$, $90^\circ$. In the small triangle, these angles are at $A$, $C$, and $H$, respectively.

The other orders put either the right angle or an acute angle in the wrong position.

In $ACH$, the angle at $C$ complements the $29^\circ$ angle at $A$.

The two angles at $C$ split the whole right angle.

Thus $\angle HCB$ measures $29^\circ$.

Since $H$ lies between $A$ and $B$, rays $AH$ and $AB$ coincide. Therefore $\angle CAH$ is the whole angle at $A$.

Together with the right angles at $C$ and $H$, this supplies two matching angle pairs for AA. Equal areas, equal hypotenuses, and a midpoint are not given or needed.

The altitude uses the two hypotenuse pieces.

Therefore $CH^2=144$. Taking the positive square root gives

Adding, averaging, or leaving the product as the length gives the other choices.

Leg $CA$ touches piece $AH$, so its square is the product of the whole hypotenuse and that piece.

The other piece is $29-9=20$ cm. The altitude would be $\sqrt{9\cdot20}=6\sqrt5$ cm. The sum $38$ and the square $261$ are not the requested leg length.

Leg $BC$ uses the piece $HB$ at endpoint $B$.

Thus

The altitude is $12$ cm because the other piece is $16$ cm. Adding the given lengths or keeping their product does not give $BC$.

The foot is inside the hypotenuse, so the pieces add.

Thus $AB=18$ cm. The square root of the product would give the altitude, not the whole hypotenuse.

The altitude relation is

Divide $25$ by $2$.

Multiplying the given lengths, subtracting them, or forgetting to divide by $2$ gives the other choices.

The ordered matching is $A$ to $C$, $C$ to $B$, and $H$ to $H$. Hence $AH$ matches $CH$, while $CH$ matches $BH$.

The first two wrong proportions would require equal hypotenuse pieces. The last introduces the whole hypotenuse into a small-triangle correspondence.

Multiply the two numbers, then take the positive square root.

The arithmetic mean is $14.5$, the product is $100$, and the difference is $21$. These are different operations.

The required construction is the perpendicular from the right-angle vertex to the hypotenuse.

A midpoint, a length equality involving $CH$, or equal legs does not by itself make this particular segment $CH$ perpendicular.