Question

which pairs of triangles are similar? check all that apply.
□ △abc ~ △def
□ △def ~ △ghi
□ △ghi ~ △abc
□ △ghi ~ △jkl
□ △jkl ~ △abc
(diagrams: △abc with right angle at c, ac=14, bc=20; △def with right angle at f, df=8, ef=10; △ghi with right angle at i, hi=15, gi=12; △jkl with right angle at l, lj=7, lk=10)

Explanation:

To determine similar triangles, we check the ratios of corresponding sides (since all are right triangles, we can use the Side - Side - Side (SSS) similarity criterion or check the ratios of the two legs and the hypotenuse).

Step 1: Analyze $\triangle ABC$ and $\triangle DEF$

For $\triangle ABC$ (right - angled at $C$), the legs are $AC = 14$ and $BC = 20$. For $\triangle DEF$ (right - angled at $F$), the legs are $DF=8$ and $EF = 10$.
Calculate the ratios of the legs: $\frac{AC}{DF}=\frac{14}{8}=\frac{7}{4}$ and $\frac{BC}{EF}=\frac{20}{10} = 2$. Since $\frac{7}{4}
eq2$, the triangles are not similar.

Step 2: Analyze $\triangle DEF$ and $\triangle GHI$

For $\triangle DEF$ (right - angled at $F$), legs $DF = 8$, $EF=10$. For $\triangle GHI$ (right - angled at $I$), legs $HI = 15$, $GI = 12$.
Calculate the ratios: $\frac{DF}{GI}=\frac{8}{12}=\frac{2}{3}$ and $\frac{EF}{HI}=\frac{10}{15}=\frac{2}{3}$. Now, let's find the hypotenuses.
For $\triangle DEF$, hypotenuse $DE=\sqrt{8^{2}+10^{2}}=\sqrt{64 + 100}=\sqrt{164}=2\sqrt{41}$.
For $\triangle GHI$, hypotenuse $GH=\sqrt{15^{2}+12^{2}}=\sqrt{225 + 144}=\sqrt{369}=3\sqrt{41}$.
The ratio of hypotenuses $\frac{DE}{GH}=\frac{2\sqrt{41}}{3\sqrt{41}}=\frac{2}{3}$. Since the ratios of all three corresponding sides are equal ($\frac{2}{3}$), $\triangle DEF\sim\triangle GHI$ (by SSS similarity).

Step 3: Analyze $\triangle GHI$ and $\triangle ABC$

For $\triangle GHI$ (right - angled at $I$), legs $HI = 15$, $GI = 12$. For $\triangle ABC$ (right - angled at $C$), legs $AC = 14$, $BC = 20$.
$\frac{HI}{AC}=\frac{15}{14}$, $\frac{GI}{BC}=\frac{12}{20}=\frac{3}{5}$. Since $\frac{15}{14}
eq\frac{3}{5}$, the triangles are not similar.

Step 4: Analyze $\triangle GHI$ and $\triangle JKL$

For $\triangle GHI$ (right - angled at $I$), legs $HI = 15$, $GI = 12$. For $\triangle JKL$ (right - angled at $L$), legs $JL = 7$, $KL = 10$.
$\frac{HI}{KL}=\frac{15}{10}=\frac{3}{2}$, $\frac{GI}{JL}=\frac{12}{7}$. Since $\frac{3}{2}
eq\frac{12}{7}$, the triangles are not similar.

Step 5: Analyze $\triangle JKL$ and $\triangle ABC$

For $\triangle JKL$ (right - angled at $L$), legs $JL = 7$, $KL = 10$. For $\triangle ABC$ (right - angled at $C$), legs $AC = 14$, $BC = 20$.
$\frac{JL}{AC}=\frac{7}{14}=\frac{1}{2}$, $\frac{KL}{BC}=\frac{10}{20}=\frac{1}{2}$. Now, find the hypotenuses.
For $\triangle JKL$, hypotenuse $JK=\sqrt{7^{2}+10^{2}}=\sqrt{49 + 100}=\sqrt{149}$.
For $\triangle ABC$, hypotenuse $AB=\sqrt{14^{2}+20^{2}}=\sqrt{196 + 400}=\sqrt{596}=2\sqrt{149}$.
The ratio of hypotenuses $\frac{JK}{AB}=\frac{\sqrt{149}}{2\sqrt{149}}=\frac{1}{2}$. Since the ratios of all three corresponding sides are equal ($\frac{1}{2}$), $\triangle JKL\sim\triangle ABC$ (by SSS similarity). Also, since $\triangle DEF\sim\triangle GHI$ and $\triangle JKL\sim\triangle ABC$, we can also note that $\triangle DEF\sim\triangle GHI$ and $\triangle JKL\sim\triangle ABC$ (and by transitivity, but we can check directly). Wait, also, let's re - check $\triangle DEF$ and $\triangle GHI$ and $\triangle JKL$ and $\triangle ABC$.

Wait, for $\triangle JKL$ and $\triangle ABC$:
Legs: $\frac{JL}{AC}=\frac{7}{14}=\frac{1}{2}$, $\frac{KL}{BC}=\frac{10}{20}=\frac{1}{2}$. Hypotenuse ratio: $\frac{JK}{AB}=\frac{\sqrt{7^{2}+10^{2}}}{\sqrt{14^{2}+20^{2}}}=\frac{\sqrt{149}}{\sqrt{596}}=\frac{\sqrt{149}}{2\sqrt{149}}=\frac{1}{2}$. So they are similar.

For $\triangle DEF$ and $\triangle GHI$:
Legs: $\frac{DF}{GI}=\frac{8}{12}=\frac{2}{3}$, $\frac{EF}{HI}=\frac{10}{15}=\frac{2}{3}$. Hypotenuse: $\frac{DE}{GH}=\frac{\sqrt{8^{2}+10^{2}}}{\sqrt{15^{2}+12^{2}}}=\frac{\sqrt{164}}{\sqrt{369}}=\frac{2\sqrt{41}}{3\sqrt{41}}=\frac{2}{3}$. So they are similar.

Also, since $\triangle JKL\sim\triangle ABC$ (as we saw) and let's check $\triangle DEF$ and $\triangle JKL$? No, but the pairs that work are $\triangle DEF\sim\triangle GHI$, $\triangle JKL\sim\triangle ABC$, and also, since similarity is transitive, but from the options:

The correct pairs are $\triangle DEF\sim\triangle GHI$, $\triangle JKL\sim\triangle ABC$. Wait, also, let's check the ratios again for $\triangle ABC$ and $\triangle JKL$:

$AC = 14$, $JL=7$; $BC = 20$, $KL = 10$; $AB=\sqrt{14^{2}+20^{2}}=\sqrt{196 + 400}=\sqrt{596}$, $JK=\sqrt{7^{2}+10^{2}}=\sqrt{49+100}=\sqrt{149}$. So $\frac{AC}{JL}=\frac{14}{7} = 2$, $\frac{BC}{KL}=\frac{20}{10}=2$, $\frac{AB}{JK}=\frac{\sqrt{596}}{\sqrt{149}}=\frac{\sqrt{4\times149}}{\sqrt{149}} = 2$. So $\triangle ABC\sim\triangle JKL$ (which is the same as $\triangle JKL\sim\triangle ABC$).

For $\triangle DEF$ and $\triangle GHI$:

$DF = 8$, $GI = 12$; $EF = 10$, $HI = 15$; $DE=\sqrt{8^{2}+10^{2}}=\sqrt{164}$, $GH=\sqrt{12^{2}+15^{2}}=\sqrt{369}$. $\frac{DF}{GI}=\frac{8}{12}=\frac{2}{3}$, $\frac{EF}{HI}=\frac{10}{15}=\frac{2}{3}$, $\frac{DE}{GH}=\frac{\sqrt{164}}{\sqrt{369}}=\frac{2\sqrt{41}}{3\sqrt{41}}=\frac{2}{3}$. So $\triangle DEF\sim\triangle GHI$.

Answer:

$\triangle DEF\sim\triangle GHI$, $\triangle JKL\sim\triangle ABC$ (the corresponding checkboxes are $\triangle DEF\sim\triangle GHI$ and $\triangle JKL\sim\triangle ABC$)