Similarity in Right Triangles: Practice

First recover the other hypotenuse piece.

Thus $HB=9$ cm. Now use both pieces.

The positive root gives $CH=3\sqrt7$ cm. Using the whole $16$ in place of the other piece gives $4\sqrt7$, which is the leg $CA$.

The given leg touches $AH$.

So $AH=3$ cm. The requested piece is the other one.

Therefore $HB=10$ cm. The altitude would be $\sqrt{30}$ cm; $39/10$ uses a piece where the whole hypotenuse belongs.

Use the altitude to recover the missing piece.

This gives $AH=8$ cm. Add the two pieces.

Thus the whole hypotenuse is $26$ cm. The $8$ is only one piece; adding or multiplying the original givens does not find the whole.

The hypotenuse is $RT$, whose pieces add to $16$ cm. Leg $RS$ touches $RD$, the $4$ cm piece.

Taking the positive root gives $RS=8$ cm. The altitude is $4\sqrt3$ cm, and the other leg is $8\sqrt3$ cm.

There are nine equal ratio parts in the whole $27$ cm, so each part is $3$ cm.

Use their product.

Thus $CH=3\sqrt{14}$ cm. The sum of the ratio numbers is not a length, and $\sqrt{14}$ omits the scale of the pieces.

A dilation multiplies every length by its factor. The geometric-mean relation agrees: each hypotenuse piece triples, so their product is multiplied by $9$.

Both lengths are positive, giving $h_{\text{new}}=3h_{\text{old}}$. The factor $9$ belongs to squared lengths.

The ordered similarity is

The original hypotenuse $AB$ matches the image hypotenuse $CB$. Image length divided by original length gives

The reciprocal reverses the direction. The other choices use sides that do not correspond in this mapping.

Both squared legs use the same positive whole hypotenuse $c$.

Multiplying by $c$ preserves order, and positive lengths have the same order as their squares. Thus the longer leg touches the longer piece: $a>b$ exactly when $q>p$.

The ratio $q/p$ is the ratio of squared legs. Taking its positive square root gives $a/b$.

The missing piece is the whole minus $AH$.

The altitude uses the two pieces.

The products $cp$ and $c(c-p)$ give the squares of the legs. The sum $p+c$ does not express the altitude relation.

The $4$ cm piece touches $CA$, not $BC$. The piece at $B$ is

Thus $HB=6$ cm, giving

The positive root is $BC=2\sqrt{15}$ cm. The wrong product gives $CA$; $\sqrt{24}$ is the altitude.

The equal pieces each have length $AB/2$. Their geometric mean is the same positive number.

Therefore $CH=AB/2$. No special-triangle ratio is needed; this follows directly from the altitude relation.

Substitute the two pieces into the altitude relationship.

Since $x$ is positive, $6x=18$, so $x=3$. Add the pieces.

Thus $AB=39$ cm. The $13$ omits the common scale, $18$ is the altitude, and $117$ uses $x^2$ in a length.