Geometry Similar Triangle Theorems Worksheet Solutions (PDF)

\(\frac{4}{6} = \frac{6}{EC}\)

\(EC = \frac{6 \cdot 6}{4} = 9\)

EC = 9

\(\frac{x}{10} = \frac{6}{15}\)

\(15x = 60\)

\(x = 4\)

x = 4

Triangle ADE is similar to triangle ABC with ratio \(\frac{8}{20} = 0.4\).

\(\frac{6}{BC} = 0.4\)

\(BC = 15\)

BC = 15

\(\frac{5}{7.5} = \frac{2}{3}\) and \(\frac{4}{6} = \frac{2}{3}\).

The sides are divided in the same ratio, so DE is parallel to BC.

Yes

\(\frac{6}{9} = \frac{2}{3} = 0.667\) but \(\frac{5}{8} = 0.625\).

The ratios are not equal, so DE is not parallel to BC.

No

\(\frac{x + 1}{6} = \frac{x + 3}{9}\)

\(9x + 9 = 6x + 18\)

\(3x = 9\)

\(x = 3\)

\(AD = 4\) and \(AE = 6\).

x = 3; AD = 4; AE = 6

\(\frac{9}{6} = 1.5\) and \(\frac{12}{8} = 1.5\).

The included right angles are congruent, so the triangles are similar by SAS~ with scale factor 1.5.

Yes, SAS~, k = 1.5

Triangle 2: \(\frac{10}{4} = \frac{12.5}{5} = \frac{15}{6} = 2.5\), so it is similar.

Triangle 3: \(\frac{8}{4} = \frac{10}{5} = 2\) but \(\frac{13}{6} \neq 2\), so it is not.

Only triangle 2

Hypotenuse: \(\sqrt{6^{2} + 8^{2}} = 10\).

Leg relationship: \(6^{2} = AD \cdot 10\), so \(AD = 3.6\), and \(8^{2} = DB \cdot 10\), so \(DB = 6.4\).

Altitude: \(h^{2} = 3.6 \cdot 6.4 = 23.04\), so \(h = 4.8\).

Hypotenuse 10; h = 4.8; AD = 3.6; DB = 6.4

\(\sqrt{4 \cdot 9} = \sqrt{36} = 6\)

6

\(h^{2} = 4 \cdot 9 = 36\)

\(h = 6\)

h = 6

\(a^{2} = 9 \cdot 25 = 225\), so \(a = 15\).

\(b^{2} = 16 \cdot 25 = 400\), so \(b = 20\).

15 and 20

\(15^{2} + 20^{2} = 225 + 400 = 625 = 25^{2}\).

\(pc + qc = 9 \cdot 25 + 16 \cdot 25 = (9 + 16) \cdot 25 = 625\).

Both equal 625

Triangle ADE is similar to triangle ABC with ratio \(\frac{6}{15} = 0.4\).

\(DE = 0.4 \cdot 20 = 8\)

DE = 8

Scale factor \(\frac{15}{6} = 2.5\).

Perimeter: \(2.5 \cdot 28 = 70\). Area ratio: \(2.5^{2} = 6.25\).

70; area ratio 6.25

\(\frac{3}{5} = \frac{x}{10}\)

\(x = 6\)

x = 6

The light ray makes equal angles with the ground, so the two right triangles are similar by AA.

\(\frac{h}{20} = \frac{6}{4}\)

\(h = 30\)

30 feet

The crossbar and the floor make similar triangles with ratio \(\frac{4}{10} = 0.4\).

\(0.4 \cdot 5 = 2\)

2 feet

Hypotenuse: \(\sqrt{15^{2} + 20^{2}} = 25\).

Segments: \(15^{2} = 25p\) gives \(p = 9\), and \(20^{2} = 25q\) gives \(q = 16\).

\(h = \sqrt{9 \cdot 16} = 12\)

12 feet

Area ratio: \(\left( \frac{5}{3} \right)^{2} = \frac{25}{9}\).

\(54 \cdot \frac{25}{9} = 150\)

150 square yards