Free math worksheets

- Identify which side is opposite the right angle. - Put the square of the hypotenuse on one side of the Pythagorean equation.

Since \(x\) is the hypotenuse, the required equation is \(3^2+4^2=x^2\).

\(3^2+4^2=x^2\)

- Decide which given length is the hypotenuse. - In the Pythagorean theorem, the hypotenuse square is the term by itself.

The ladder is the hypotenuse, so its square must equal the sum of the squares of the two legs. Therefore, equation A is correct.

A. \(x^2+8^2=10^2\)

- 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 part a, \(d=\sqrt{1^2+1^2}=\sqrt{2}\), which is irrational. 2. For part b, \(d=\sqrt{3^2+4^2}=\sqrt{25}=5\), which is rational. 3. Part 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, \(d=\sqrt{13.69}=3.7\,\text{m}\).

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

- The crease divides the square into two right triangles. - Identify which two segments are the legs of one of those triangles. - Include the side-length equation in your response before evaluating the square root.

1. The diagonal is the hypotenuse of a right triangle whose legs are both \(20\,\text{cm}\), so \(d^2=20^2+20^2\). 2. Thus \(d=\sqrt{800}=20\sqrt{2}\,\text{cm}\approx28.3\,\text{cm}\).

\(d^2=20^2+20^2\), so \(d\approx28.3\,\text{cm}\).

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

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

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

- Treat the square's diagonal as the hypotenuse of one right triangle. - Both legs have the same unknown length, so your equation should contain the same variable twice. - Simplify the radical before rounding.

1. A diagonal divides the square into a right triangle with legs \(a\) and \(a\) and hypotenuse \(20\), so the required equation is \(a^2+a^2=20^2\). 2. Therefore, \(2a^2=400\), so \(a^2=200\). 3. Thus \(a=\sqrt{200}=10\sqrt{2}\,\text{ft}\approx14.14\,\text{ft}\).

a) \(a^2+a^2=20^2\). b) \(a=10\sqrt{2}\,\text{ft}\approx14.14\,\text{ft}\).

- Use one of the right triangles created by the diagonal. - Keep \(a^2\) in your work; that quantity is also the square's area. - Your answer must include the side-length equation, not only the final area.

1. The diagonal forms a right triangle with legs \(a\) and \(a\), so \(a^2+a^2=60^2\). 2. Thus \(2a^2=3600\), so \(a^2=1800\). 3. The square's area is \(a^2\), so the area is \(1800\,\text{in.}^2\).

\(a^2+a^2=60^2\), so the area is \(1800\,\text{in.}^2\).

- Model the display diagonal as the hypotenuse of a right triangle. - Keep the squared side lengths separate before adding them. - Compare the calculated diagonal with the limit only after rounding as requested.

1. The display's diagonal is the hypotenuse of a right triangle with legs \(8\,\text{in.}\) and \(13\,\text{in.}\). 2. Thus \(d^2=8^2+13^2=64+169=233\). 3. Therefore, \(d=\sqrt{233}\,\text{in.}\approx15.26\,\text{in.}\). 4. Since \(15.26\,\text{in.}>15\,\text{in.}\), the tablet exceeds the sleeve's size limit.

No. The display diagonal is \(\sqrt{233}\,\text{in.}\approx15.26\,\text{in.}\), which is longer than \(15\,\text{in.}\).

- 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. For part a, \(b=\sqrt{37^2-12^2}=\sqrt{1225}=35\,\text{cm}\). 2. For part b, \(b=\sqrt{2.5^2-1.5^2}=\sqrt{4}=2\,\text{m}\).

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

- Identify the side opposite the marked right angle in the end view. - What relationship connects the three side lengths of a right triangle? - After finding the missing side, use it as one dimension of the rectangular face.

1. The hypotenuse \(x\) satisfies \(x^2=6^2+4^2=52\), so \(x=\sqrt{52}=2\sqrt{13}\,\text{cm}\). 2. The rectangular face has dimensions \(2\sqrt{13}\,\text{cm}\) by \(8\,\text{cm}\), so its area is \(16\sqrt{13}\,\text{cm}^2\).

Hypotenuse: \(2\sqrt{13}\,\text{cm}\); rectangular face area: \(16\sqrt{13}\,\text{cm}^2\).

- Decide whether the unknown side is a leg or the hypotenuse. - Use the two displayed leg lengths to form the squared-length relationship. - Simplify the radical before finding the requested decimal approximation.

1. The unknown side is the hypotenuse, and the legs are \(6\,\text{cm}\) and \(9\,\text{cm}\). 2. Thus \(x^2=6^2+9^2=117\). 3. Therefore, \(x=\sqrt{117}=3\sqrt{13}\,\text{cm}\approx10.82\,\text{cm}\).

\(x=3\sqrt{13}\,\text{cm}\approx10.82\,\text{cm}\).

- Identify the side opposite the right-angle mark before deciding which length is the hypotenuse. - Isolate the square of the unknown leg before taking a square root. - Simplify the exact radical before rounding.

1. The hypotenuse is \(17\,\text{cm}\), the known leg is \(9\,\text{cm}\), and the other leg is \(x\). 2. Thus \(x^2=17^2-9^2=289-81=208\). 3. Therefore, \(x=\sqrt{208}=4\sqrt{13}\,\text{cm}\approx14.42\,\text{cm}\).

\(x=4\sqrt{13}\,\text{cm}\approx14.42\,\text{cm}\).

- Express both displayed lengths in the same unit before calculating. - Which displayed side is opposite the marked right angle? - Isolate the square of the unknown side before taking a square root.

1. Convert the hypotenuse to centimeters: \(1.1\,\text{dm}=11\,\text{cm}\). 2. The right-triangle relation gives \(x^2+6^2=11^2\), so \(x^2=85\). 3. Thus \(x=\sqrt{85}\approx9.2\,\text{cm}\).

\(x\approx9.2\,\text{cm}\)

- Put the two displayed leg lengths in the same unit. - Identify which side is opposite the marked right angle. - Keep the square-root value unrounded until the final answer.

1. Express both legs in decimeters: \(0.7\,\text{m}=7\,\text{dm}\). The legs are \(7\,\text{dm}\) and \(20\,\text{dm}\). 2. The hypotenuse satisfies \(x^2=7^2+20^2=449\). 3. Therefore, \(x=\sqrt{449}\approx21.2\,\text{dm}\).

\(x\approx21.2\,\text{dm}\)

- What triangle is formed when a square is divided along a diagonal? - How are the two legs and the diagonal related in 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. The side-length relationship gives \(d^2=a^2+a^2=2a^2\). 3. Since a length is positive, \(d=\sqrt{2a^2}=a\sqrt{2}\).

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

- Identify the side opposite the right-angle mark. - Keep the two decimal squares separate until you add them. - Round only after taking the square root.

1. The side \(x\) is the hypotenuse, and the legs have lengths \(4.2\,\text{cm}\) and \(7.1\,\text{cm}\). 2. Thus \(x^2=4.2^2+7.1^2=17.64+50.41=68.05\). 3. Therefore, \(x=\sqrt{68.05}\,\text{cm}\approx8.25\,\text{cm}\).

\(x\approx8.25\,\text{cm}\).

- A rectangle's diagonal forms a right triangle with two adjacent sides. - Use the two dimensions shown as perpendicular legs. - Round after taking the square root.

1. Two adjacent sides and diagonal \(AC\) form a right triangle. 2. Thus \(AC^2=11^2+7^2=170\). 3. Therefore, \(AC=\sqrt{170}\,\text{cm}\approx13.04\,\text{cm}\).

\(AC\approx13.04\,\text{cm}\).

- Decide which displayed segment is the hypotenuse. - Subtract the square of the known leg from the square of the diagonal. - Take the positive square root and round only at the end.

1. The diagonal is the hypotenuse of a right triangle formed by the rectangle's two side lengths. 2. Let the unknown side be \(b\). Then \(b^2=19^2-7^2=361-49=312\). 3. Therefore, \(b=\sqrt{312}=2\sqrt{78}\,\text{cm}\approx17.66\,\text{cm}\).

The other side is approximately \(17.66\,\text{cm}\) long.

- From the center of the square to a corner, determine the two perpendicular component lengths. - Your required equation should use those two components as the legs and the spray radius as the hypotenuse. - For part b, compare the corner-reaching radius with the current radius.

1. From the center to a corner, the horizontal and vertical distances are each \(20\,\text{ft}\), so the required equation is \(r^2=20^2+20^2\). 2. Thus \(r=\sqrt{800}=20\sqrt{2}\,\text{ft}\approx28.28\,\text{ft}\). 3. The additional radius is \(20\sqrt{2}-20=20(\sqrt{2}-1)\,\text{ft}\approx8.28\,\text{ft}\).

a) \(r^2=20^2+20^2\), so \(r=20\sqrt{2}\,\text{ft}\approx28.28\,\text{ft}\). b) \(20(\sqrt{2}-1)\,\text{ft}\approx8.28\,\text{ft}\).

- Compare the length of the two-side route with the straight corner-to-corner route. - What right triangle is formed by the park's two side lengths and its diagonal? - Keep the diagonal unrounded until the final subtraction.

1. The route along two sides is \(110+150=260\,\text{yd}\). 2. The direct route is the rectangle's diagonal, so its length is \(\sqrt{110^2+150^2}=\sqrt{34{,}600}\approx186.0\,\text{yd}\). 3. The difference is \(260-\sqrt{34{,}600}\approx73.989\,\text{yd}\), which rounds to \(74.0\,\text{yd}\).

The direct route is \(74.0\,\text{yd}\) 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. The longest side is \(c=24\,\text{cm}\). Since \(15^2+20^2=625\) while \(24^2=576\), the triangle is not a right triangle. 2. 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 \(15^2+20^2=625\ne576=24^2\). b) \(c=25\,\text{cm}\)

- Treat the beams as legs and the diagonal as the possible hypotenuse. - Compare the sum of the squares of the beam lengths with the square of the measured diagonal. - For part c, think about which side is unknown after the other leg and the diagonal are fixed.

1. \(12^2+16^2=400\), while \(20.5^2=420.25\). Because the values are not equal, the beams do not form 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. Let the unknown beam length be \(a\). Then \(a^2=20.5^2-16^2=164.25\), so \(a=\sqrt{164.25}\,\text{ft}\approx12.82\,\text{ft}\).

a) The beams do not form an exact right angle because \(12^2+16^2=400\ne420.25=20.5^2\). b) \(20\,\text{ft}\) c) Approximately \(12.82\,\text{ft}\)

- A rectangle's diagonal and two adjacent sides form a right triangle. - Find the unknown side before using the rectangle formulas. - Keep the exact side length until the final rounding of each requested quantity.

1. The diagonal is the hypotenuse of a right triangle formed by two sides of the rectangle. Thus \(b^2=13^2-6^2=133\), so \(b=\sqrt{133}\approx11.5\,\text{cm}\). 2. The area is \(A=6\sqrt{133}\approx69.2\,\text{cm}^2\). 3. The perimeter is \(P=2(6+\sqrt{133})\approx35.1\,\text{cm}\).

\(b\approx11.5\,\text{cm}\), \(A\approx69.2\,\text{cm}^2\), and \(P\approx35.1\,\text{cm}\).

- Use the gate's area and width to determine its other dimension first. - The diagonal brace and the two side lengths form a right triangle. - Round only after finding the diagonal from the exact side lengths.

1. The gate's height is \(35\div5=7\,\text{ft}\). 2. The brace is the rectangle's diagonal, so its length is \(\sqrt{5^2+7^2}=\sqrt{74}\approx8.602\,\text{ft}\). 3. Rounded to the nearest tenth, the brace should be \(8.6\,\text{ft}\) long.

The brace should be \(8.6\,\text{ft}\) long.

- Express all three side lengths in the same unit before comparing them. - Which side is longest after the conversion? - How can the three squared side lengths tell you whether the triangle is right? - If the triangle is right, which two sides can serve as the base and height?

1. Convert \(300\,\text{cm}\) to \(3\,\text{m}\). The longest side is \(3\,\text{m}\). 2. The squares of the two shorter side lengths sum to \((\sqrt{2})^2+(\sqrt{7})^2=2+7=9\), and the square of the longest side is \(3^2=9\). Therefore, the triangle is right. 3. The two shorter sides are the legs, so the area is \(\frac{1}{2}(\sqrt{2})(\sqrt{7})=\frac{\sqrt{14}}{2}\,\text{m}^2\).

The triangle is right, and its area is \(\frac{\sqrt{14}}{2}\,\text{m}^2\).

- 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 \(26.89\ne27.04\), 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 \(\sqrt{26.89}\approx5.1856\), 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. 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. Rope A satisfies \(9^2+12^2=225=15^2\), 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 \(5^2+12^2=169=13^2\), 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 \(60\div12=5\), scaling each side by \(5\) gives \(15\), \(20\), 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 \(60\div30=2\), scaling each side by \(2\) gives \(10\), \(24\), 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 in each part. - Make sure the measurements in a single calculation use the same unit. - What equation relates the three side lengths of any right triangle? - Delay rounding until the final side length in each part.

1. For part a, \(c=\sqrt{9^2+14^2}=\sqrt{277}\approx16.6\,\text{cm}\). 2. For part b, \(b=\sqrt{1.8^2-0.9^2}=\sqrt{2.43}\approx1.6\,\text{dm}\). 3. For part c, convert \(80\,\text{mm}=8\,\text{cm}\). Then \(a=\sqrt{15^2-8^2}=\sqrt{161}\approx12.7\,\text{cm}\).

a) \(c\approx16.6\,\text{cm}\) b) \(b\approx1.6\,\text{dm}\) c) \(a\approx12.7\,\text{cm}\)

- A diagonal splits a rectangle into right triangles. - Determine the unknown side before computing the rectangle's area. - Keep the exact side length until the final rounding.

1. The diagonal is the hypotenuse, so \(b^2=15^2-7^2=176\). Thus \(b=\sqrt{176}=4\sqrt{11}\approx13.3\,\text{yd}\). 2. The area is \(A=7\sqrt{176}=28\sqrt{11}\approx92.9\,\text{yd}^2\).

\(b\approx13.3\,\text{yd}\) and \(A\approx92.9\,\text{yd}^2\).

- The screen's width, height, and diagonal form a right triangle. - Which of those three measurements is opposite the right angle? - Keep the unrounded width when using it in the area calculation.

1. Let the width be \(w\). Then \(w^2+5.9^2=6.8^2\), so \(w^2=46.24-34.81=11.43\). Therefore, \(w=\sqrt{11.43}\approx3.4\,\text{in.}\). 2. Using the unrounded width, the area is \(5.9\sqrt{11.43}\approx19.9469\,\text{in.}^2\), which rounds to \(19.9\,\text{in.}^2\).

a) \(3.4\,\text{in.}\) b) \(19.9\,\text{in.}^2\)

- The same segment \(k\) belongs to both right triangles. - Use the first triangle to determine what \(k^2\) must equal. - You can carry \(k^2\) into the second triangle without approximating \(k\).

1. In the first triangle, \(k^2=13^2-7^2=120\), so \(k=\sqrt{120}=2\sqrt{30}\,\text{cm}\). 2. In the second triangle, \(x^2=k^2+9^2=120+81=201\), so \(x=\sqrt{201}\,\text{cm}\).

a) \(k=2\sqrt{30}\,\text{cm}\) b) \(x=\sqrt{201}\,\text{cm}\)

- Determine the shared side before finding either triangle's area. - The shared side is a leg in the first right triangle. - Once \(s\) is known, it serves as a base or height in both area calculations. - Keep the shared length exact until the final approximation.

1. In the first triangle, \(s^2=17^2-9^2=208\), so \(s=\sqrt{208}=4\sqrt{13}\,\text{yd}\). 2. The first triangle's area is \(\frac{1}{2}\cdot9\cdot4\sqrt{13}=18\sqrt{13}\,\text{yd}^2\). 3. The second triangle's area is \(\frac{1}{2}\cdot11\cdot4\sqrt{13}=22\sqrt{13}\,\text{yd}^2\). 4. The total area is \(40\sqrt{13}\,\text{yd}^2\approx144.2\,\text{yd}^2\).

The total area is \(40\sqrt{13}\,\text{yd}^2\), or approximately \(144.2\,\text{yd}^2\).

- When the two points have no sideways offset, how are they positioned relative to line \(m\)? - The difference between the two given distances gives one component of \(\overline{RS}\). - For part b, what relationship connects perpendicular and sideways components to the full segment length?

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. A length of \(4\,\text{cm}\) is possible. The perpendicular component of \(\overline{RS}\) is \(3\,\text{cm}\). If the parallel component is \(x\), then \(x^2+3^2=4^2\), so \(x=\sqrt{7}\,\text{cm}\). Such an offset gives \(RS=4\,\text{cm}\).

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

- Combine the two perpendicular distances because the points are on opposite sides of \(g\). - The separation along \(g\) gives a second perpendicular component of the segment. - Compare the case with a nonzero parallel component to the case when that component is zero.

1. The perpendicular change from \(A\) to \(B\) is \(4+2=6\,\text{cm}\), while the component parallel to \(g\) is \(7\,\text{cm}\). 2. These perpendicular components form the legs of a right triangle with hypotenuse \(AB\), so \(AB^2=6^2+7^2=85\). 3. Therefore, \(AB=\sqrt{85}\,\text{cm}\approx9.22\,\text{cm}\). 4. The minimum \(6\,\text{cm}\) occurs only when the parallel component is \(0\). Here it is \(7\,\text{cm}\), so the segment must be longer.

a) \(AB=\sqrt{85}\,\text{cm}\approx9.22\,\text{cm}\). b) The points also differ by \(7\,\text{cm}\) parallel to \(g\), so their separation is the hypotenuse of a right triangle rather than the \(6\,\text{cm}\) perpendicular minimum.

- 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. The space diagonal satisfies \(d^2=a^2+b^2+c^2\). Thus, \(c=\sqrt{17^2-12^2-9^2}=\sqrt{64}=8\,\text{cm}\). 2. The volume is \(V=abc=12\cdot9\cdot8=864\,\text{cm}^3\).

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 \(4\), so \(AC=\sqrt{3^2+4^2}=5\) units. From \(B\) to \(C\), the horizontal change has magnitude \(3\) and the vertical change is \(4\), so \(BC=5\) units. 3. 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.

- Use the symmetry of the isosceles end to split the base into two equal parts. - The altitude, half the base, and one sloping side form a right triangle. - Use the sloping side as the width of each rectangular roof panel.

1. The altitude of the isosceles triangle bisects the \(2.4\,\text{m}\) base, so each half is \(1.2\,\text{m}\). 2. If \(s\) is a sloping side, then \(s^2=1.2^2+1.6^2=4\), so \(s=2.0\,\text{m}\). 3. Each roof panel is \(2.0\,\text{m}\) by \(3.0\,\text{m}\). The two panels have total area \(2\cdot2.0\cdot3.0=12.0\,\text{m}^2\).

Each sloping side is \(2.0\,\text{m}\); the two roof panels have total area \(12.0\,\text{m}^2\).

- Use the isosceles symmetry to determine how the altitude divides the base. - In one of the resulting right triangles, identify the hypotenuse and the unknown leg. - After finding the altitude, use the triangle area formula.

1. The altitude of an isosceles triangle bisects the \(6\,\text{cm}\) base, so each half is \(3\,\text{cm}\). 2. Using one half, \(h^2+3^2=5^2\), so \(h^2=16\) and \(h=4\,\text{cm}\). 3. The triangular end area is \(\frac{1}{2}\cdot6\cdot4=12\,\text{cm}^2\).

\(h=4\,\text{cm}\); triangular end area: \(12\,\text{cm}^2\)

- Compare the two parallel side lengths and use symmetry to determine the horizontal offset on one side. - The dashed height and that horizontal offset form the legs of a right triangle. - After finding one slanted side, use symmetry when adding the perimeter.

1. The two bases differ by \(10-4=6\,\text{cm}\). Symmetry splits this difference equally, so the horizontal leg beside either slanted side is \(3\,\text{cm}\). 2. The vertical leg is \(4\,\text{cm}\), so \(x^2=3^2+4^2=25\) and \(x=5\,\text{cm}\). 3. The perimeter is \(10+4+5+5=24\,\text{cm}\).

Each slanted side is \(5\,\text{cm}\); perimeter: \(24\,\text{cm}\)

- Compare the total height with the wall height to determine the roof's vertical rise. - Use symmetry to determine the horizontal run from one wall to the peak. - Those two distances are perpendicular legs for the slanted roof edge.

1. The roof rises \(8-5=3\,\text{cm}\) above the walls. Symmetry makes the horizontal run from a wall to the peak half of \(8\,\text{cm}\), or \(4\,\text{cm}\). 2. Thus \(x^2=4^2+3^2=25\), so \(x=5\,\text{cm}\). 3. The two equal roof edges have total length \(2\cdot5=10\,\text{cm}\).

Each slanted roof edge is \(5\,\text{cm}\); together they measure \(10\,\text{cm}\).

- Compare the full bottom width with the horizontal top segment to obtain one perpendicular change. - Compare the two vertical side lengths to obtain the other perpendicular change. - Use those changes as the legs of a right triangle for the slanted edge. - Add all five boundary edges after finding the missing one.

1. The horizontal change across the slanted section is \(120-80=40\,\text{cm}\), and the vertical change is \(50-20=30\,\text{cm}\). 2. Therefore, \(x^2=40^2+30^2=2500\), so \(x=50\,\text{cm}\). 3. The pentagonal end perimeter is \(120+20+50+80+50=320\,\text{cm}\).

\(x=50\,\text{cm}\); perimeter: \(320\,\text{cm}\)

- Convert the displayed measurements to the same unit. - Decide whether \(x\) is a leg or the hypotenuse from the right-angle mark. - Round only after taking the square root.

1. Convert the known leg to meters: \(10\,\text{dm}=1\,\text{m}\). The hypotenuse is \(2.7\,\text{m}\). 2. The right-triangle relation gives \(x^2+1^2=2.7^2\), so \(x^2=6.29\). 3. Therefore, \(x=\sqrt{6.29}\approx2.51\,\text{m}\).

\(x\approx2.51\,\text{m}\)

- Put the displayed measurements in millimeters before calculating. - Use the right-angle mark to identify the hypotenuse. - Keep the square root unrounded until the requested precision.

1. Convert the known leg to millimeters: \(1.5\,\text{cm}=15\,\text{mm}\). The hypotenuse is \(41\,\text{mm}\). 2. The right-triangle relation gives \(x^2+15^2=41^2\), so \(x^2=1456\). 3. Therefore, \(x=\sqrt{1456}\approx38.2\,\text{mm}\).

\(x\approx38.2\,\text{mm}\)

- 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 two right triangles share the same vertical segment. - What can the left triangle tell you about the square of that shared length? - Carry that squared value into the second triangle before rounding anything.

1. Let \(h\) be the vertical segment shared by the two right triangles. In the left triangle, \(h^2+6^2=11^2\), so \(h^2=85\). 2. In the right triangle, \(x^2=h^2+13^2=85+169=254\). 3. Therefore, \(x=\sqrt{254}\approx15.9\,\text{cm}\).

\(x\approx15.9\,\text{cm}\)

- Identify which labeled side is the hypotenuse before finding the missing leg. - Keep the missing leg in radical form until you use it as a base or height. - Round the area, not the missing leg, at the final step.

1. Let the unlabeled leg be \(b\). Then \(b^2=15^2-8^2=225-64=161\), so \(b=\sqrt{161}\,\text{cm}\). 2. The area is \(A=\frac{1}{2}\cdot8\cdot\sqrt{161}=4\sqrt{161}\,\text{cm}^2\approx50.75\,\text{cm}^2\).

The area is approximately \(50.75\,\text{cm}^2\).

- Express all four side lengths using the common shorter length. - After finding the side lengths, drop perpendiculars from the endpoints of the shorter base. - How much of the longer base extends beyond the shorter base on each side? - Use a leg of the trapezoid as the hypotenuse of one of the resulting right triangles.

1. Let \(x\) be the common length of each leg and the shorter base. Then the longer base is \(3x\). The perimeter equation is \(x+x+x+3x=24\), so \(6x=24\) and \(x=4\). 2. The side lengths are therefore \(4\,\text{cm}\), \(4\,\text{cm}\), \(4\,\text{cm}\), and \(12\,\text{cm}\). 3. In an isosceles trapezoid, dropping perpendiculars from the shorter base splits the extra base length equally. Each horizontal leg of the resulting right triangles is \(\frac{12-4}{2}=4\,\text{cm}\). 4. If \(h\) is the trapezoid's height, the Pythagorean theorem gives \(h^2+4^2=4^2\). Thus \(h^2=0\), so \(h=0\). 5. A height of zero means the figure collapses into a line segment, so no nondegenerate trapezoid with positive area exists.

The longer base is \(12\,\text{cm}\), and the other three sides are \(4\,\text{cm}\) each. No nondegenerate trapezoid exists because the Pythagorean theorem gives height \(0\,\text{cm}\).

- 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. For part a, let \(a\le b\le c\) be whole-number side lengths with \(a+b+c=10\). The triangle inequality gives \(a+b>c\), so \(10-c>c\) and \(c<5\). Also \(3c\ge10\), so \(c\ge4\). Thus \(c=4\). 2. Then \(a+b=6\), giving only \((a,b)=(2,4)\) or \((3,3)\). Neither works: \(2^2+4^2=20\ne16\), and \(3^2+3^2=18\ne16\). Therefore, no whole-number right triangle has perimeter \(10\). 3. For part b, the side lengths \(5\), \(12\), and \(13\) have perimeter \(30\), and \(5^2+12^2=169=13^2\). The ratio \(5:12:13\) is not a scaled version of \(3:4:5\), so it meets the requirement.

a) No whole-number right triangle has perimeter \(10\); the only possible longest side is \(4\), and the resulting triples do not satisfy the Pythagorean theorem. b) \(5,12,13\) works because \(5+12+13=30\) and \(5^2+12^2=13^2\); its 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 relationship for each space diagonal. - Round only after finding the difference.

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

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.