Similarity in Right Triangles: Practice

Squaring the leg ratio and dividing their geometric-mean relations gives

The whole contains five equal parts, so $AH=8$ cm and $HB=32$ cm.

Taking the positive root gives $CH=16$ cm. Halving the whole assumes equal pieces; $8$ is only one piece, and $16\sqrt5$ is the longer leg.

The leg relation first determines the whole hypotenuse.

Thus $AB=28$ cm and $HB=21$ cm. Use the two pieces.

Therefore $CH=7\sqrt3$ cm. Using $AB$ instead of $HB$ returns the given leg $14$, while using $CA$ as a piece gives $7\sqrt2$. The other leg is $14\sqrt3$.

The whole hypotenuse is $YZ=18$ cm. Leg $XY$ touches the $3$ cm piece at $Y$.

Thus $XY=3\sqrt6$ cm. The altitude is $3\sqrt5$ cm, and the other leg is $3\sqrt{30}$ cm.

Divide the leg and altitude relations. The positive factor $AH$ cancels.

The given length ratio has squared ratio $25/16$, so

Thus $HB=16$ cm. The requested piece is

Therefore $AH=9$ cm. Using the unsquared ratio would give the incorrect piece $HB=20$.

The exact measurements must satisfy the altitude relation.

But the recorded square is $49$ and the recorded product is $60$. They differ, so no right triangle with the stated altitude can have all three measurements.

An altitude can meet unequal pieces. Merely lying between their lengths or producing right angles does not replace the required correspondence.

Each small triangle has a $40^\circ$ endpoint angle and a right angle at $H$. Its remaining angle is $50^\circ$.

The small triangles therefore share angle set $40^\circ$, $50^\circ$, $90^\circ$ and are similar by AA. The whole has angle set $40^\circ$, $40^\circ$, $100^\circ$, so it is similar to neither small triangle.

The original triangle's right-angle hypothesis is essential to the three-triangle result.

The two small angles at $C$ make the whole right angle.

Thus $\angle HCB$ measures $27^\circ$. The angle is not a copy or half of $\angle ACH$; $117^\circ$ would be its supplement, not its complement.

Divide the two leg relationships.

Thus the squared leg ratio is $16:9$. Both lengths are positive, so taking square roots gives

The piece order is reversed because $BC$ touches $HB$, while $CA$ touches $AH$.

The leg relations give $a^2=cq$ and $b^2=cp$. Divide them using positive lengths.

Cancel the common nonzero factor $c$.

Similarity gives proportional sides; it does not require equal sides or a midpoint.

For each triangle, the ratio of squared legs equals the ratio of the corresponding pieces. Since $18>8$ and $24>6$, compare longer leg with shorter leg.

For the first triangle,

This ratio is $3/2$. For the second, the ratio is $\sqrt{24/6}=2$. Different within-triangle side ratios rule out similarity, even though both altitudes are $12$ cm.

The ratios $9:4$ and $4:1$ are squared ratios, not leg ratios.

Write $p=AH$ and $q=HB$. The leg relation and altitude relation give

Subtract the second equality from the first. The terms $pq$ cancel.

Thus $p^2=36$, so the positive length $AH$ is $6$ cm.

The value $\sqrt2$ comes from taking the square root of $10-8$ instead of subtracting squared lengths. The value $\sqrt{164}$ adds the squares. The value $25/2$ divides the leg square by the altitude, which is not the relevant leg relationship.

The ordered matching is $A$ to $A$, $B$ to $C$, and $C$ to $H$. The original side $BC$ matches $CH$, while $AB$ matches $AC$.

Keeping original length over image length gives

Using $AH$ as the match for $BC$, or reversing only one ratio, breaks the correspondence.