The Pythagorean Theorem: practice solutions, Grade 8 – download the PDF

1 Hypotenuse of a 6-8 triangle ★★★

The hypotenuse is \(c\), so \(c^2 = 6^2 + 8^2 = 36 + 64 = 100\).

\(c = \sqrt{100} = 10\).

The hypotenuse is 10 cm long.

2 A longer triple ★★★

\(c^2 = 7^2 + 24^2 = 49 + 576 = 625\), so \(c = \sqrt{625} = 25\).

The hypotenuse is 25 in long.

3 Find a missing leg ★★★

Subtract, since the hypotenuse is known: \(a^2 = 26^2 - 10^2 = 676 - 100 = 576\).

\(a = \sqrt{576} = 24\).

The other leg is 24 m long.

4 True or false? ★★★

She is not right. She added the legs instead of adding their squares.

\(c^2 = 4^2 + 5^2 = 16 + 25 = 41\), so \(c = \sqrt{41} \approx 6.40\) cm.

Also, a hypotenuse is always shorter than the sum of the legs, so 9 cm is too long. False: the hypotenuse is about 6.40 cm.

5 Three hypotenuses ★★★

1. \(c^2 = 64 + 225 = 289\), so \(c = 17\).
2. \(c^2 = 81 + 1600 = 1681\), so \(c = 41\).
3. \(c^2 = 144 + 1225 = 1369\), so \(c = 37\).

6 A hypotenuse that is not a whole number ★★★

\(c^2 = 3^2 + 5^2 = 9 + 25 = 34\).

Exact value: \(c = \sqrt{34}\) ft. Since \(5.83^2 \approx 33.99\), we get \(c \approx 5.83\) ft.

7 Right triangle or not? ★★★

1. The longest side is 15. \(9^2 + 12^2 = 81 + 144 = 225 = 15^2\). Yes, it is a right triangle.
2. The longest side is 10. \(7^2 + 8^2 = 49 + 64 = 113\), but \(10^2 = 100\). No, it is not a right triangle.

8 The leaning ladder ★★★

The ladder is the hypotenuse (15 ft) and the distance on the ground is one leg (9 ft).

\(h^2 = 15^2 - 9^2 = 225 - 81 = 144\), so \(h = 12\).

The ladder reaches 12 ft up the wall (about 3.66 m).

9 A shortcut across a field ★★★

Diagonal: \(d^2 = 24^2 + 32^2 = 576 + 1024 = 1600\), so \(d = 40\) m.

Along two sides: \(24 + 32 = 56\) m.

Savings: \(56 - 40 = 16\). The shortcut saves 16 m.

10 Height of an isosceles triangle ★★★

The height splits the base into two parts of 5 cm and forms a right triangle with hypotenuse 13 cm.

\(h^2 = 13^2 - 5^2 = 169 - 25 = 144\), so \(h = 12\) cm.

Area \(= \dfrac{1}{2} \times 10 \times 12 = 60\). The height is 12 cm and the area is 60 cm².

11 Distance between two points ★★★

The legs are \(9 - 1 = 8\) and \(17 - 2 = 15\).

\(AB = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\).

\(AB = 17\) units.

12 Distance with negative coordinates ★★★

The horizontal change is \(8 - (-4) = 12\) and the vertical change is \(-2 - 3 = -5\), which has length 5.

\(PQ = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\).

\(PQ = 13\) units.

13 A triangle with decimal sides ★★★

The longest side is 2.5 cm. \(1.5^2 + 2^2 = 2.25 + 4 = 6.25\) and \(2.5^2 = 6.25\).

The equation holds, so the triangle is right. The right angle is opposite the 2.5 cm side.

14 Flight of a drone ★★★

1. The east and north legs are perpendicular, so \(d^2 = 120^2 + 50^2 = 14{,}400 + 2{,}500 = 16{,}900\) and \(d = 130\) m.
2. Total: \(120 + 50 + 130 = 300\). The trip is 300 m long.

15 Will the rod fit in the box? ★★★

1. Base diagonal: \(\sqrt{12^2 + 16^2} = \sqrt{400} = 20\) in. Space diagonal: \(\sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29\) in.
2. The longest straight rod is 29 in. Since \(28 \lt 29\), the 28 in rod fits. Since \(30 \gt 29\), the 30 in rod does not fit.

16 The broken tree ★★★

The standing part has length \(x\), and the fallen part has length \(16 - x\) and is the hypotenuse.

\(x^2 + 8^2 = (16 - x)^2\), so \(x^2 + 64 = 256 - 32x + x^2\).

The \(x^2\) terms cancel: \(64 = 256 - 32x\), so \(32x = 192\) and \(x = 6\).

Check: the fallen part is 10 ft and \(6^2 + 8^2 = 100 = 10^2\). The tree broke 6 ft above the ground.

17 Area and perimeter from a hypotenuse ★★★

Other leg: \(\sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24\) cm.

Perimeter: \(7 + 24 + 25 = 56\) cm.

Area: \(\dfrac{1}{2} \times 7 \times 24 = 84\) cm².

18 A right angle in the coordinate plane ★★★

\(AB^2 = 4^2 + 3^2 = 25\). \(AC^2 = (-3)^2 + 4^2 = 25\). \(BC^2 = (-3 - 4)^2 + (4 - 3)^2 = 49 + 1 = 50\).

Since \(AB^2 + AC^2 = 50 = BC^2\), the converse applies. The triangle is right, with the right angle at \(A\). It is also isosceles because \(AB = AC = 5\).

19 Diagonal of a square ★★★

The two sides and the diagonal form a right triangle: \(s^2 + s^2 = 14^2\), so \(2s^2 = 196\) and \(s^2 = 98\).

Exact side: \(s = \sqrt{98} = 7\sqrt{2} \approx 9.90\) cm.

The area is \(s^2 = 98\) cm², half of the square of the diagonal.

20 Height of an equilateral triangle ★★★

The height cuts the base into two segments of 5 cm: \(h^2 = 10^2 - 5^2 = 100 - 25 = 75\).

\(h = \sqrt{75} = 5\sqrt{3} \approx 8.66\) cm.

Area \(= \dfrac{1}{2} \times 10 \times 8.66 \approx 43.30\) cm². The height is about 8.66 cm and the area is about 43.30 cm².

21 Distance in 3D coordinates ★★★

The differences are \(3 - 1 = 2\), \(6 - 2 = 4\) and \(7 - 3 = 4\).

\(PQ = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6\).

\(PQ = 6\) units.