Free Math Worksheets

- Use the Pythagorean theorem. - When is the square root of a natural number rational? - Look for familiar integer side-length combinations in right triangles.

1. For a), \(d = \sqrt{1^2 + 1^2} = \sqrt{2}\), which is irrational. 2. For b), \(d = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\), which is rational. 3. Case b) is a counterexample to the claim because a rectangle with integer side lengths \(3\) and \(4\) has rational diagonal length \(5\). Therefore, the claim is false.

a) \(d = \sqrt{2}\), which is irrational. b) \(d = 5\), which is rational. The claim is false; a \(3\)-by-\(4\) rectangle has diagonal length \(5\).

- A diagonal divides a rectangle into two right triangles. - Use the rectangle’s side lengths as the legs.

1. The side lengths and diagonal form a right triangle. 2. By the Pythagorean theorem, \(d^2=1.2^2+3.5^2=1.44+12.25=13.69\). 3. Therefore, \(d=\sqrt{13.69}=3.7\,\text{m}\).

\(d=3.7\,\text{m}\)

- A diagonal divides a square into two isosceles right triangles. - Use the two side lengths as the legs.

1. The diagonal is the hypotenuse of an isosceles right triangle with legs \(20\,\text{cm}\). 2. By the Pythagorean theorem, \(d^2=20^2+20^2=800\). 3. Thus \(d=\sqrt{800}=20\sqrt{2}\,\text{cm}\approx 28.3\,\text{cm}\).

The crease is approximately \(28.3\,\text{cm}\) long.

- Squaring a square root returns the number under the radical. - Rearrange the Pythagorean theorem for the unknown side. - Identify the hypotenuse before substituting.

1. Part a: \(c^2=(\sqrt{13})^2+(\sqrt{23})^2=13+23=36\), so \(c=6\). 2. Part b: \(b^2=15^2-9^2=225-81=144\), so \(b=12\).

a) \(c=6\) b) \(b=12\)

- Draw the diagonal and identify the right triangle. - The two legs have the same length. - Simplify the radical before rounding.

1. A diagonal divides a square into two congruent right triangles. The square’s sides are the legs, and the diagonal is the hypotenuse. 2. Therefore, \(a^2+a^2=20^2\), so \(2a^2=400\) and \(a^2=200\). 3. Thus \(a=\sqrt{200}=10\sqrt{2}\,\text{ft}\approx 14.14\,\text{ft}\).

1. The diagonal divides the square into two right triangles. 2. \(a=10\sqrt{2}\,\text{ft}\approx 14.14\,\text{ft}\)

- A diagonal divides the square into two right triangles. - You can solve directly for \(a^2\), which is also the area.

1. If the side length is \(a\), then \(60^2=a^2+a^2=2a^2\). 2. Therefore, \(a^2=\frac{3600}{2}=1800\). 3. Since the square’s area is \(a^2\), the area is \(1800\,\text{in.}^2\).

The area is \(1800\,\text{in.}^2\).

- What geometric shape models the display? - How are the side lengths of a rectangle related to its diagonal? - Compare the diagonal you calculate with the sleeve's limit.

1. Model the display as a rectangle with side lengths \(9\,\text{in.}\) and \(12\,\text{in.}\). 2. Use the Pythagorean theorem: \(d^2=9^2+12^2\). 3. Simplify: \(d^2=81+144=225\), so \(d=15\,\text{in.}\). 4. Because \(15\,\text{in.}>14\,\text{in.}\), the tablet does not meet the sleeve's size limit.

No. The display diagonal is \(15\,\text{in.}\), which is longer than the sleeve's \(14\,\text{in.}\) limit.

- Recall the relationship among the three side lengths of a right triangle. - Identify the hypotenuse before substituting values. - Rearrange the formula when a leg, rather than the hypotenuse, is unknown.

1. Rearrange the Pythagorean theorem to solve for the missing leg: \(b=\sqrt{c^2-a^2}\). 2. For a), \(b=\sqrt{37^2-12^2}=\sqrt{1369-144}=\sqrt{1225}=35\,\text{cm}\). 3. For b), \(b=\sqrt{2.5^2-1.5^2}=\sqrt{6.25-2.25}=\sqrt{4}=2\,\text{m}\).

a) \(b=35\,\text{cm}\) b) \(b=2\,\text{m}\)

- Decide whether the unknown side is a leg or the hypotenuse. - Use the relationship among the squared side lengths of a right triangle. - The hypotenuse is opposite the right angle.

1. The unknown side is the hypotenuse, and the legs are \(5\,\text{cm}\) and \(12\,\text{cm}\). 2. Apply the Pythagorean theorem: \(x^2=5^2+12^2=25+144=169\). 3. Since a length is positive, \(x=\sqrt{169}=13\,\text{cm}\).

\(x=13\,\text{cm}\)

- Identify the hypotenuse before writing the equation. - Rearrange the Pythagorean equation when a leg is unknown. - Subtract the square of the known leg from the square of the hypotenuse.

1. The hypotenuse is \(25\,\text{cm}\), the known leg is \(7\,\text{cm}\), and the other leg is \(x\). 2. Apply the Pythagorean theorem: \(x^2+7^2=25^2\). 3. Solve for \(x^2\): \(x^2=625-49=576\). 4. Since a length is positive, \(x=\sqrt{576}=24\,\text{cm}\).

\(x=24\,\text{cm}\)

- Express all side lengths in the same unit before calculating. - Choose the unit that avoids unnecessary decimals. - Rearrange the equation so that the square of the unknown side is isolated.

1. Convert the hypotenuse to centimeters: \(1\,\text{dm}=10\,\text{cm}\). 2. Apply the Pythagorean theorem: \(x^2+6^2=10^2\). 3. Solve: \(x^2=100-36=64\), so \(x=\sqrt{64}=8\,\text{cm}\).

\(x=8\,\text{cm}\)

- Make sure both given lengths use the same unit. - Identify the hypotenuse. - Add the squares of the legs to find the square of the hypotenuse.

1. Express both legs in decimeters: \(0.7\,\text{m}=7\,\text{dm}\). The legs are \(7\,\text{dm}\) and \(24\,\text{dm}\). 2. Apply the Pythagorean theorem: \(x^2=7^2+24^2=49+576=625\). 3. Since a length is positive, \(x=\sqrt{625}=25\,\text{dm}\).

\(x=25\,\text{dm}\)

- What triangle is formed when a square is divided along a diagonal? - Apply the Pythagorean theorem to that right triangle. - Simplify the radical as far as possible.

1. Two adjacent sides and the diagonal form a right triangle with legs \(a\) and \(a\). 2. By the Pythagorean theorem, \(d^2=a^2+a^2=2a^2\). 3. Since a length is positive, \(d=\sqrt{2a^2}=a\sqrt{2}\).

\(d=a\sqrt{2}\)

- Identify the hypotenuse and the two legs. - Use the Pythagorean theorem. - Check whether the unknown is a leg or the hypotenuse.

1. The side \(x\) is the hypotenuse, and the legs have lengths \(4.8\,\text{cm}\) and \(3.6\,\text{cm}\). 2. Apply the Pythagorean theorem: \(x^2=4.8^2+3.6^2=23.04+12.96=36\). 3. Therefore, \(x=\sqrt{36}=6\,\text{cm}\).

\(x=6\,\text{cm}\)

- A diagonal divides a rectangle into two right triangles. - Use the two side lengths as the legs.

1. Two adjacent sides and the diagonal form a right triangle. 2. Apply the Pythagorean theorem: \(AC^2=12^2+5^2=169\). 3. Therefore, \(AC=13\,\text{cm}\).

\(AC=13\,\text{cm}\)

- The side lengths and diagonal form a right triangle. - Rearrange the Pythagorean theorem to find a leg.

1. The diagonal is the hypotenuse of a right triangle formed by the rectangle’s side lengths. 2. Let the unknown side be \(b\). Then \(b^2=17^2-8^2=289-64=225\). 3. Therefore, \(b=15\,\text{cm}\).

The other side is \(15\,\text{cm}\) long.

- The direct route is the hypotenuse of a right triangle. - Find both route lengths separately, then subtract.

1. The route along two sides is \(120+160=280\,\text{yd}\). 2. The direct route is the diagonal: \(d^2=120^2+160^2=40{,}000\), so \(d=200\,\text{yd}\). 3. The difference is \(280-200=80\,\text{yd}\).

The direct route is \(80\,\text{yd}\) shorter.

- Recall how the converse of the Pythagorean theorem relates the three side lengths. - Compare the sum of the squares of the two shorter sides with the square of the longest side. - For the right-triangle case, use \(c\) as the hypotenuse in the Pythagorean theorem.

1. a) The longest side is \(c = 24\,\text{cm}\). Compare \(a^2 + b^2\) with \(c^2\): \(15^2 + 20^2 = 225 + 400 = 625\), while \(24^2 = 576\). Since \(625 \ne 576\), the triangle is not a right triangle. 2. b) For \(c\) to be the hypotenuse, \(c^2 = 15^2 + 20^2 = 625\). Therefore, \(c = \sqrt{625} = 25\,\text{cm}\).

a) The triangle is not a right triangle because \(15^2 + 20^2 = 625 \ne 576 = 24^2\). b) \(c = 25\,\text{cm}\).

- Treat the beams as legs and the diagonal as the hypotenuse. - Compare the sum of the squares of the beam lengths with the square of the measured diagonal. - Rearrange the Pythagorean theorem for the unknown leg in part c.

1. Check the converse of the Pythagorean theorem: \(12^2+16^2=400\), while \(20.5^2=420.25\). Because the values are not equal, the angle is not a right angle. 2. For a right angle with beams \(12\,\text{ft}\) and \(16\,\text{ft}\), \(d=\sqrt{12^2+16^2}=20\,\text{ft}\). 3. Keeping the \(20.5\,\text{ft}\) diagonal and the \(16\,\text{ft}\) beam, let the other beam be \(a\). Then \(a^2=20.5^2-16^2=164.25\), so \(a\approx 12.82\,\text{ft}\).

a) The frame is not square because \(12^2+16^2=400\ne 420.25=20.5^2\). b) \(20\,\text{ft}\) c) Approximately \(12.82\,\text{ft}\)

- Use the diagonal and known side to find the missing side. - Then apply the rectangle area and perimeter formulas.

1. Find the missing side: \(b^2=17^2-15^2=64\), so \(b=8\,\text{cm}\). 2. The area is \(A=15\cdot 8=120\,\text{cm}^2\). 3. The perimeter is \(P=2(15+8)=46\,\text{cm}\).

The area is \(120\,\text{cm}^2\), and the perimeter is \(46\,\text{cm}\).

- Use the area and width to find the gate’s height. - Then use the Pythagorean theorem for the diagonal.

1. Find the height from the area: \(h=\frac{12}{3}=4\,\text{ft}\). 2. The brace is the rectangle’s diagonal, so \(d=\sqrt{3^2+4^2}=5\,\text{ft}\).

The brace should be \(5\,\text{ft}\) long.

- Convert all side lengths to the same unit. - Compare the sum of the squares of the two shorter sides with the square of the longest side. - If the triangle is right, use its legs as the base and height.

1. Convert \(250\,\text{cm}\) to \(2.5\,\text{m}\). 2. Check the converse of the Pythagorean theorem: \(0.7^2+2.4^2=0.49+5.76=6.25\), and \(2.5^2=6.25\). Therefore, the triangle is right. 3. The legs are \(0.7\,\text{m}\) and \(2.4\,\text{m}\), so \(A=\frac{1}{2}\cdot 0.7\cdot 2.4=0.84\,\text{m}^2\).

The triangle is right, and its area is \(0.84\,\text{m}^2\).

- What geometric figure is formed by the pole, the ground, and the taut rope? - Which segment would be the hypotenuse if the pole were perpendicular to the ground? - How can you find the third side length when the included angle is \(90^\circ\)?

1. If the pole is perpendicular to the ground, the pole segment and ground distance are the legs, and the rope is the hypotenuse. 2. Calculate \(4.00^2 + 3.30^2 = 16.00 + 10.89 = 26.89\). 3. Calculate \(5.20^2 = 27.04\). Since \(26.89 \ne 27.04\), the pole is not exactly perpendicular to the ground. 4. The required rope length is \(\sqrt{4.00^2 + 3.30^2} = \sqrt{26.89}\,\text{m}\). 5. Since \(\sqrt{26.89} \approx 5.1856\), the rope should be about \(5.19\,\text{m}\) long.

The pole is not exactly perpendicular. The required rope length is exactly \(\sqrt{26.89}\,\text{m}\), or approximately \(5.19\,\text{m}\).

- What condition must three lengths satisfy to form a closed triangle? - What relationship among the squared side lengths shows that a triangle is right? - Is equality in the triangle inequality enough to form a triangle? - Which side must be treated as the possible hypotenuse?

1. a) For Rope B, the two shorter lengths add to \(7 + 10 = 17\), which is less than \(18\). Therefore, Rope B cannot form a triangle. 2. b) Rope A satisfies \(9^2 + 12^2 = 81 + 144 = 225 = 15^2\), so it forms a right triangle. 3. Rope C gives \(10^2 + 10^2 = 200\), while \(14^2 = 196\), so it does not form a right triangle. 4. Rope D satisfies \(5^2 + 12^2 = 25 + 144 = 169 = 13^2\), so it forms a right triangle.

a) Rope B cannot form a triangle because \(7 + 10 < 18\). b) Ropes A and D form right triangles because \(9^2 + 12^2 = 15^2\) and \(5^2 + 12^2 = 13^2\).

- Recall a simple whole-number triple that satisfies the Pythagorean theorem. - Scaling all three sides of a right triangle by the same factor preserves the right angle. - The three side lengths must add to exactly \(60\).

1. A scaled \((3, 4, 5)\) triangle can have perimeter \(60\,\text{m}\). Since \(3+4+5=12\) and \(60 \div 12=5\), scale each side by \(5\) to get \(15\), \(20\), and \(25\). The perimeter is \(60\), and \(15^2+20^2=225+400=625=25^2\). 2. A scaled \((5, 12, 13)\) triangle also works. Since \(5+12+13=30\) and \(60 \div 30=2\), scale each side by \(2\) to get \(10\), \(24\), and \(26\). The perimeter is \(60\), and \(10^2+24^2=100+576=676=26^2\).

1. \(15\,\text{m}, 20\,\text{m}, 25\,\text{m}\), because \(15+20+25=60\) and \(15^2+20^2=25^2\). 2. \(10\,\text{m}, 24\,\text{m}, 26\,\text{m}\), because \(10+24+26=60\) and \(10^2+24^2=26^2\).

- Identify whether the missing side is a leg or the hypotenuse. - Convert measurements to the same unit before calculating. - Use the Pythagorean theorem in each part.

1. For a), \(c=\sqrt{15^2+20^2}=25\,\text{cm}\). 2. For b), \(b=\sqrt{1.7^2-0.8^2}=\sqrt{2.25}=1.5\,\text{dm}\). 3. For c), convert \(120\,\text{mm}=12\,\text{cm}\). Then \(a=\sqrt{13^2-12^2}=5\,\text{cm}\).

a) \(c=25\,\text{cm}\) b) \(b=1.5\,\text{dm}\) c) \(a=5\,\text{cm}\)

- A diagonal divides the rectangle into two right triangles. - Find the missing side before finding the area.

1. The diagonal is the hypotenuse, so \(b^2=13^2-5^2=144\). Thus \(b=12\,\text{yd}\). 2. The area is \(A=5\cdot 12=60\,\text{yd}^2\).

\(b=12\,\text{yd}\) and \(A=60\,\text{yd}^2\)

- The screen dimensions and diagonal form a right triangle. - After finding the width, clear the decimals in the ratio and reduce it.

1. Let the width be \(w\). Then \(w^2+6.0^2=6.8^2\), so \(w^2=46.24-36=10.24\). 2. Therefore, \(w=3.2\,\text{in.}\). 3. The ratio \(3.2:6.0\) becomes \(32:60\), which simplifies to \(8:15\).

a) \(3.2\,\text{in.}\) b) \(8:15\)

- Find the shared leg from the first triangle. - Then use that leg in the second triangle. - Identify the hypotenuse in each triangle.

1. In the first triangle, \(k^2=17^2-8^2=225\), so \(k=15\,\text{cm}\). 2. In the second triangle, \(x^2=15^2+20^2=625\), so \(x=25\,\text{cm}\).

\(x=25\,\text{cm}\)

- Find the shared side from the first right triangle. - Use the legs of each right triangle as its base and height. - Add the two areas.

1. In the first triangle, \(s^2=25^2-7^2=576\), so \(s=24\,\text{yd}\). 2. The first area is \(A_1=\frac{1}{2}\cdot 7\cdot 24=84\,\text{yd}^2\). 3. The second area is \(A_2=\frac{1}{2}\cdot 24\cdot 10=120\,\text{yd}^2\). 4. The total area is \(84+120=204\,\text{yd}^2\).

The total area is \(204\,\text{yd}^2\).

- The shortest segment between two parallel lines is perpendicular to them. - The difference between the two given distances is one leg of a right triangle. - For part b, use the Pythagorean theorem.

1. The least distance occurs when \(R\) and \(S\) lie on the same line perpendicular to \(m\). Their distances from \(m\) differ by \(5-2=3\,\text{cm}\), so the least possible length is \(3\,\text{cm}\). 2. Yes. If the component of \(\overline{RS}\) perpendicular to \(m\) is \(3\,\text{cm}\), choose a parallel component \(x\) so that \(x^2+3^2=4^2\). 3. This gives \(x^2=7\), so an offset of \(\sqrt{7}\,\text{cm}\) produces a segment of length \(4\,\text{cm}\).

a) \(3\,\text{cm}\) b) Yes. A sideways offset of \(\sqrt{7}\,\text{cm}\) gives \(RS=4\,\text{cm}\).

- The segment must cross line \(g\). - Add the two perpendicular distances to find the shortest possible route. - Think about when that shortest route is achieved.

1. Because \(A\) and \(B\) are on opposite sides of \(g\), the perpendicular component of \(\overline{AB}\) across the line is \(4+2=6\,\text{cm}\). 2. Any additional component parallel to \(g\) makes \(\overline{AB}\) longer than \(6\,\text{cm}\). Therefore, \(AB\ge 6\,\text{cm}\), so statement a is true and statement b is false. 3. Equality occurs only when there is no component parallel to \(g\). In that case, \(\overline{AB}\) follows one perpendicular line through \(g\), so statement c is true.

a) True. The least possible length is \(4+2=6\,\text{cm}\). b) False. \(AB\) cannot be less than \(6\,\text{cm}\). c) True. The minimum occurs only when \(\overline{AB}\perp g\).

- How are the three edge lengths of a rectangular prism related to its space diagonal? - Rearrange the equation so that the missing edge length is the only unknown. - Which formula gives the volume of a rectangular prism?

1. Apply the Pythagorean theorem in three dimensions: \(d^2=a^2+b^2+c^2\). 2. Solve for \(c\): \(c=\sqrt{d^2-a^2-b^2}=\sqrt{17^2-12^2-9^2}=\sqrt{64}=8\,\text{cm}\). 3. Use \(V=abc\): \(V=12\cdot9\cdot8=864\,\text{cm}^3\).

a) \(c=8\,\text{cm}\) b) \(V=864\,\text{cm}^3\)

- A horizontal segment’s length is the difference of its x-coordinates. - Use the horizontal and vertical changes as the legs of a right triangle. - A triangle with two equal side lengths has a special classification.

1. Segment \(\overline{AB}\) is horizontal, so \(AB = 8 - 2 = 6\) units. 2. From \(A\) to \(C\), the horizontal change is \(3\) and the vertical change is \(4\). Therefore, \(AC = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\) units. 3. From \(B\) to \(C\), the horizontal change has magnitude \(3\) and the vertical change is \(4\). Therefore, \(BC = \sqrt{3^2 + 4^2} = 5\) units. 4. Since \(AC = BC\), the triangle is isosceles.

a) \(AB = 6\) units b) \(AC = 5\) units and \(BC = 5\) units c) The triangle is isosceles.

- First find the unknown side in the original triangle. - Determine the new hypotenuse after the change. - Repeat the calculation for the new triangle, then compare the two leg lengths.

1. For the original triangle, \(b=\sqrt{29^2-21^2}=\sqrt{841-441}=\sqrt{400}=20\,\text{cm}\). 2. The new hypotenuse is \(29+6=35\,\text{cm}\). 3. The new leg length is \(b_{\text{new}}=\sqrt{35^2-21^2}=\sqrt{1225-441}=\sqrt{784}=28\,\text{cm}\). 4. The increase is \(28-20=8\,\text{cm}\).

1. \(b=20\,\text{cm}\) 2. The leg increases by \(8\,\text{cm}\).

- Draw a sketch of the two paths. - What geometric figure is formed by the paths and the direct segment between the hikers? - First determine how far each hiker travels. - Use the relationship among the side lengths of a right triangle.

1. Hiker A travels \(4\cdot 2=8\,\text{mi}\). 2. Hiker B travels \(3\cdot 2=6\,\text{mi}\). 3. The two paths are perpendicular, so the hikers' separation is the hypotenuse of a right triangle. 4. Apply the Pythagorean theorem: \(d=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\,\text{mi}\).

The hikers are \(10\,\text{mi}\) apart after \(2\) hours.

- Place the lighthouse at the origin of a coordinate plane. - Determine each boat's position after \(2\) hours. - Find the horizontal and vertical separations between the boats. - Use those separations as the legs of a right triangle.

1. After \(2\) hours, the sailboat is \(12\cdot 2=24\,\text{mi}\) east of its starting point and remains \(15\,\text{mi}\) north of the lighthouse. 2. The motorboat is \(4\cdot 2=8\,\text{mi}\) north of the lighthouse. 3. The horizontal separation is \(24\,\text{mi}\), and the vertical separation is \(15-8=7\,\text{mi}\). 4. Use the Pythagorean theorem: \(d=\sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=25\,\text{mi}\).

The boats are \(25\,\text{mi}\) apart after \(2\) hours.

- Find the area of the right triangular base. - Use the Pythagorean theorem to find the third side. - Multiply the base perimeter by the prism length for lateral surface area. - Include both triangular bases for total surface area.

1. The triangular base area is \(B=\frac{1}{2}\cdot3\cdot4=6\,\text{cm}^2\). 2. The volume is \(V=6\cdot8=48\,\text{cm}^3\). 3. By the Pythagorean theorem, the hypotenuse is \(\sqrt{3^2+4^2}=5\,\text{cm}\). 4. The base perimeter is \(3+4+5=12\,\text{cm}\), so the lateral surface area is \(L=12\cdot8=96\,\text{cm}^2\). 5. The total surface area is \(S=2\cdot6+96=108\,\text{cm}^2\).

a) \(48\,\text{cm}^3\) b) \(96\,\text{cm}^2\) c) \(108\,\text{cm}^2\)

- Find the area of the triangular end. - Use the Pythagorean theorem to find each slanted side. - For surface area, include both triangular ends and all three rectangular faces. - The floor is one of the rectangular faces.

1. The triangular end area is \(B=\frac{1}{2}\cdot2.4\cdot1.6=1.92\,\text{m}^2\). 2. The volume is \(V=1.92\cdot3.0=5.76\,\text{m}^3\). 3. The triangle's altitude divides its base into two segments of \(1.2\,\text{m}\). By the Pythagorean theorem, each equal side is \(\sqrt{1.2^2+1.6^2}=2.0\,\text{m}\). 4. The triangular perimeter is \(2.4+2.0+2.0=6.4\,\text{m}\). 5. The total fabric area is \(S=2B+Pl=2\cdot1.92+6.4\cdot3.0=23.04\,\text{m}^2\).

a) \(5.76\,\text{m}^3\) b) \(23.04\,\text{m}^2\)

- Find the area of the triangular base first. - The altitude of an isosceles triangle bisects its base. - Use the Pythagorean theorem to find an equal side. - Use the base perimeter to find lateral surface area.

1. The triangular base area is \(B=\frac{1}{2}\cdot6\cdot4=12\,\text{cm}^2\). 2. The volume is \(V=12\cdot12=144\,\text{cm}^3\). 3. The altitude bisects the base into two \(3\,\text{cm}\) segments. By the Pythagorean theorem, each equal side is \(s=\sqrt{3^2+4^2}=5\,\text{cm}\). 4. The base perimeter is \(6+5+5=16\,\text{cm}\), so the lateral surface area is \(L=16\cdot12=192\,\text{cm}^2\). 5. The total surface area is \(S=2\cdot12+192=216\,\text{cm}^2\).

a) \(B=12\,\text{cm}^2\) and \(V=144\,\text{cm}^3\) b) \(s=5\,\text{cm}\), \(L=192\,\text{cm}^2\), and \(S=216\,\text{cm}^2\)

- Convert the measurements to the same unit. - The unknown side is a leg, not the hypotenuse. - Remember to take the square root after finding \(x^2\).

1. Convert the known leg to meters: \(10\,\text{dm}=1\,\text{m}\). The hypotenuse is \(2.6\,\text{m}\). 2. Apply the Pythagorean theorem: \(x^2+1^2=2.6^2\). 3. Solve: \(x^2=6.76-1=5.76\), so \(x=\sqrt{5.76}=2.4\,\text{m}\).

\(x=2.4\,\text{m}\)

- Decide whether \(x\) is a leg or the hypotenuse. - Convert all measurements to millimeters before calculating. - A calculator may help with the squares and square root.

1. Convert the known leg to millimeters: \(1.5\,\text{cm}=15\,\text{mm}\). The hypotenuse is \(39\,\text{mm}\). 2. Apply the Pythagorean theorem: \(x^2+15^2=39^2\). 3. Solve: \(x^2=1521-225=1296\), so \(x=\sqrt{1296}=36\,\text{mm}\).

\(x=36\,\text{mm}\)

- Identify the right angle and determine which side is the hypotenuse. - Compare Mia’s result with the ladder length. - When a leg is unknown, decide whether to add or subtract squared lengths.

1. The ladder is the hypotenuse, so a leg cannot be longer than the \(10\,\text{m}\) ladder. A height of \(11.7\,\text{m}\) is therefore impossible. 2. Mia added the square of the known leg instead of subtracting it from the square of the hypotenuse. 3. The correct equation is \(h^2=10^2-6^2=100-36=64\). 4. Therefore, \(h=\sqrt{64}=8\,\text{m}\).

Mia’s result is unreasonable because a leg cannot be longer than the hypotenuse. The correct height is \(h=8\,\text{m}\).

- First find the common vertical leg of the two right triangles. - Use the total base length for the larger triangle.

1. Let \(h\) be the vertical leg shared by the two right triangles. In the smaller triangle, \(h^2=12.5^2-3.5^2=156.25-12.25=144\), so \(h=12\,\text{cm}\). 2. The base of the larger triangle is \(3.5+1.5=5\,\text{cm}\). 3. Apply the Pythagorean theorem to the larger triangle: \(x^2=12^2+5^2=169\). Therefore, \(x=13\,\text{cm}\).

\(x=13\,\text{cm}\)

- The figure contains two right triangles that share a side. - Find the shared height from the left triangle first. - Then use that value in the right triangle to find \(x\).

1. Let \(h\) be the vertical segment shared by the two right triangles. In the left triangle, \(h^2+6^2=10^2\), so \(h^2=64\). 2. In the right triangle, \(x^2=h^2+15^2=64+225=289\). 3. Therefore, \(x=\sqrt{289}=17\,\text{cm}\).

\(x=17\,\text{cm}\)

- Start with the smaller right triangle to find the shared vertical leg. - Use the total base length for the larger right triangle. - Apply the Pythagorean theorem again to find \(x\).

1. Let \(h\) be the vertical leg shared by the two right triangles. In the smaller triangle, \(h^2+7^2=25^2\), so \(h^2=625-49=576\). 2. The full base of the larger triangle is \(7+3=10\,\text{cm}\). 3. Apply the Pythagorean theorem to the larger triangle: \(x^2=576+10^2=676\). Therefore, \(x=26\,\text{cm}\).

\(x=26\,\text{cm}\)

- Use the Pythagorean theorem to find the missing leg. - Then use the two legs as the base and height.

1. Find the other leg \(b\): \(b^2=13^2-5^2=144\), so \(b=12\,\text{cm}\). 2. The area is \(A=\frac{1}{2}\cdot 5\cdot 12=30\,\text{cm}^2\).

The area is \(30\,\text{cm}^2\).

- Think of a circle centered in a square. - Find the part-to-whole percent using the two areas. - Divide the total dry area equally among the four corners. - Use the Pythagorean theorem from the center to a corner.

1. The square area is \(40\,\text{ft}\cdot40\,\text{ft}=1600\,\text{ft}^2\). 2. The watered area is \(\pi(20\,\text{ft})^2=400\pi\,\text{ft}^2\). The percent watered is \(\frac{400\pi}{1600}\cdot100\%=25\pi\%\approx78.54\%\). 3. The total dry area is \(1600-400\pi\,\text{ft}^2\), so one dry corner has area \(\frac{1600-400\pi}{4}\,\text{ft}^2\approx85.84\,\text{ft}^2\). 4. The required radius is the distance from the square's center to a corner. A right triangle has legs \(20\,\text{ft}\) and \(20\,\text{ft}\), so \(r=\sqrt{20^2+20^2}\,\text{ft}=20\sqrt{2}\,\text{ft}\approx28.28\,\text{ft}\).

a) \(1600\,\text{ft}^2\); approximately \(78.54\%\) watered b) Approximately \(85.84\,\text{ft}^2\) c) Approximately \(28.28\,\text{ft}\)

- Use the triangle inequality to limit the possible length of the longest side when the perimeter is \(10\). - List the whole-number triples that add to \(10\) after applying that limit. - For part b), look for a Pythagorean triple that is not a scaled \((3, 4, 5)\) triple. - Compare corresponding side ratios to decide whether two triangles are scaled copies.

1. a) Let the whole-number side lengths satisfy \(a \le b \le c\) and \(a+b+c=10\). The triangle inequality gives \(a+b>c\), so \(10-c>c\) and therefore \(c<5\). Also, because all three sides are at most \(c\), \(3c \ge 10\), so the whole number \(c\) must be at least \(4\). Thus, \(c=4\). 2. Then \(a+b=6\). With \(a \le b \le 4\), the only possibilities are \((a, b)=(2, 4)\) and \((a, b)=(3, 3)\). Neither works: \(2^2+4^2=20 \ne 16\), and \(3^2+3^2=18 \ne 16\). Therefore, no whole-number right triangle has perimeter \(10\). 3. b) The side lengths \(5\), \(12\), and \(13\) have perimeter \(30\). They form a right triangle because \(5^2+12^2=25+144=169=13^2\). 4. The ratio \(5:12:13\) is not a scaled version of \(3:4:5\), because the corresponding scale factors are not equal.

a) No whole-number right triangle has perimeter \(10\). The only possible longest side is \(4\), and the remaining possibilities \((2, 4, 4)\) and \((3, 3, 4)\) do not satisfy the Pythagorean theorem. b) The side lengths \(5\), \(12\), and \(13\) have perimeter \(30\) and satisfy \(5^2+12^2=13^2\). Their ratio is not \(3:4:5\).

- Subtract the bezel twice from each cabinet dimension. - Use the \(16:9\) ratio to determine which dimension limits the screen. - Find the diagonal with the Pythagorean theorem, then choose the greatest whole number that does not exceed it.

1. The maximum screen dimensions are \(36-2\cdot 1=34\,\text{in.}\) wide and \(24-2\cdot 1=22\,\text{in.}\) high. 2. At the maximum width, a \(16:9\) screen has height \(34\cdot\frac{9}{16}=19.125\,\text{in.}\), which is less than \(22\,\text{in.}\). Thus width is the limiting dimension. 3. The maximum screen diagonal is \(d=\sqrt{34^2+19.125^2}\approx 39.01\,\text{in.}\). 4. Therefore, the largest whole-number screen size that fits is \(39\,\text{in.}\).

The largest whole-number screen size is \(39\,\text{in.}\).

- First find the volume of the rectangular prism. - How can you recover a cube's edge length from its volume? - Use the three-dimensional Pythagorean theorem for each space diagonal. - Round only after finding the difference.

1. The rectangular prism has volume \(V=16\cdot8\cdot4=512\,\text{cm}^3\). 2. For the cube, \(s^3=512\), so \(s=\sqrt[3]{512}=8\,\text{cm}\). 3. The rectangular prism's space diagonal is \(d_r=\sqrt{16^2+8^2+4^2}=\sqrt{336}\approx18.33\,\text{cm}\). 4. The cube's space diagonal is \(d_c=\sqrt{8^2+8^2+8^2}=\sqrt{192}=8\sqrt{3}\approx13.86\,\text{cm}\). 5. The difference is \(\sqrt{336}-\sqrt{192}\approx4.47\,\text{cm}\).

a) \(s=8\,\text{cm}\) b) Rectangular prism: \(d_r\approx18.33\,\text{cm}\); cube: \(d_c\approx13.86\,\text{cm}\). The rectangular prism's diagonal is approximately \(4.47\,\text{cm}\) longer.

- Use the trapezoid area formula. - Use symmetry to split the difference between the parallel sides equally. - Apply the Pythagorean theorem to one side triangle. - Use the base perimeter to find lateral surface area.

1. The trapezoidal base area is \(B=\frac{10+4}{2}\cdot4=28\,\text{cm}^2\). 2. The volume is \(V=28\cdot15=420\,\text{cm}^3\). 3. The difference between the parallel sides is \(10-4=6\,\text{cm}\). Symmetry gives a horizontal offset of \(3\,\text{cm}\) on each side. 4. By the Pythagorean theorem, each slanted side is \(s=\sqrt{3^2+4^2}=5\,\text{cm}\). 5. The base perimeter is \(10+5+4+5=24\,\text{cm}\), so the lateral surface area is \(L=24\cdot15=360\,\text{cm}^2\). 6. The total surface area is \(S=2\cdot28+360=416\,\text{cm}^2\).

a) \(B=28\,\text{cm}^2\) and \(V=420\,\text{cm}^3\) b) Each slanted side is \(5\,\text{cm}\), and \(S=416\,\text{cm}^2\).

- Split the house-shaped base into a rectangle and a triangle. - Use symmetry and the Pythagorean theorem to find a roof edge. - Multiply the base perimeter by the prism length for lateral surface area. - Include both pentagonal bases.

1. The rectangular part has area \(8\cdot5=40\,\text{cm}^2\). The roof triangle has height \(8-5=3\,\text{cm}\) and area \(\frac{1}{2}\cdot8\cdot3=12\,\text{cm}^2\). Thus, \(B=52\,\text{cm}^2\). 2. The volume is \(V=52\cdot15=780\,\text{cm}^3\). 3. Half the roof triangle has legs \(4\,\text{cm}\) and \(3\,\text{cm}\). By the Pythagorean theorem, each slanted roof edge is \(\sqrt{4^2+3^2}=5\,\text{cm}\). 4. The base perimeter is \(8+5+5+5+5=28\,\text{cm}\), so the lateral surface area is \(L=28\cdot15=420\,\text{cm}^2\). 5. The total surface area is \(S=2\cdot52+420=524\,\text{cm}^2\).

a) \(52\,\text{cm}^2\) b) \(780\,\text{cm}^3\) c) Each slanted roof edge is \(5\,\text{cm}\). d) \(L=420\,\text{cm}^2\) and \(S=524\,\text{cm}^2\)

- Decompose the pentagonal end into rectangles and a triangle. - Use the Pythagorean theorem to find the slanted edge. - Use the end area for volume and its perimeter for lateral surface area. - Convert cubic centimeters to liters and square centimeters to square meters.

1. Decompose the end into an \(80\,\text{cm}\times50\,\text{cm}\) rectangle, a \(40\,\text{cm}\times20\,\text{cm}\) rectangle, and a right triangle with legs \(40\,\text{cm}\) and \(30\,\text{cm}\). 2. The end area is \(B=4000+800+\frac{1}{2}\cdot40\cdot30=5400\,\text{cm}^2\). 3. The volume is \(V=5400\cdot100=540{,}000\,\text{cm}^3=540\,\text{L}\). 4. The slanted edge has length \(\sqrt{40^2+30^2}=50\,\text{cm}\). 5. The end perimeter is \(120+20+50+80+50=320\,\text{cm}\), so the lateral surface area is \(L=320\cdot100=32{,}000\,\text{cm}^2\). 6. The total surface area is \(S=2\cdot5400+32{,}000=42{,}800\,\text{cm}^2=4.28\,\text{m}^2\).

a) \(540\,\text{L}\) b) \(4.28\,\text{m}^2\)