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

- Think of a square with a side length you know. How does its area relate to that length? - Only a nonnegative value makes sense for a measured length.

1. If the hypotenuse has length \(c\), the area of its square is \(c^2=36\,\text{cm}^2\). 2. Length is positive, so \(c=6\,\text{cm}\).

\(6\,\text{cm}\).

- 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 fold produces two right triangles. - Which sides meet at the square's right angle? - Find the positive square root, then round to tenths.

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

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

- 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) \(b=12\)

- Both sides of the square form legs of equal length. - The diagonal is opposite the right angle. - After finding a side length, how could you check it against the diagonal?

1. A diagonal forms a right triangle with both legs equal to \(a\), so \(a^2+a^2=20^2\). 2. Thus \(2a^2=400\) and \(a^2=200\). 3. The side length is \(a=\sqrt{200}\,\text{ft}\approx14.14\,\text{ft}\).

a) \(a^2+a^2=20^2\). b) \(a=\sqrt{200}\,\text{ft}\approx14.14\,\text{ft}\).

- 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) \(b=2\,\text{m}\)

- Find the side opposite the right-angle mark. - Use the two labeled legs to determine the hypotenuse. - Could you use the exact unknown length in an area calculation without rounding it first?

1. The hypotenuse satisfies \(x^2=6^2+4^2=52\), so \(x=\sqrt{52}\,\text{cm}\). 2. The rectangular face is \(8\,\text{cm}\) long and \(\sqrt{52}\,\text{cm}\) wide. 3. Its area is \(8\sqrt{52}\,\text{cm}^2\).

Hypotenuse \(\sqrt{52}\,\text{cm}\); face area \(8\sqrt{52}\,\text{cm}^2\).

- Identify the side across from the right angle. - How must that side's square relate to the other two squares? - Why might the exact answer and a rounded decimal look different?

1. The missing side is opposite the right angle, so it is the hypotenuse. 2. \(x^2=6^2+9^2=117\). 3. Thus \(x=\sqrt{117}\,\text{cm}\approx10.82\,\text{cm}\).

\(x=\sqrt{117}\,\text{cm}\approx10.82\,\text{cm}\).

- Which side lies opposite the right-angle mark? - Does \(x\) label that side or another side? - Should the missing side be shorter or longer than the side marked \(17\,\text{cm}\)? - How can you rearrange the triangle's square relationship to isolate the unknown?

1. The hypotenuse is \(17\,\text{cm}\) and one leg is \(9\,\text{cm}\). 2. \(x^2=17^2-9^2=208\). 3. Thus \(x=\sqrt{208}\,\text{cm}\approx14.42\,\text{cm}\).

\(x=\sqrt{208}\,\text{cm}\approx14.42\,\text{cm}\).

- 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, \(x=\sqrt{449}\approx21.2\,\text{dm}\).

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

- Think about the right triangle inside the square. - What are the two leg lengths? - Solve the squared-length equation for its positive root.

1. The diagonal and two neighboring sides form a right triangle with both legs \(a\). 2. The Pythagorean theorem gives \(d^2=a^2+a^2=2a^2\). 3. Since lengths are positive, \(d=\sqrt{2a^2}\).

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

- 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, \(AC=\sqrt{170}\,\text{cm}\approx13.04\,\text{cm}\).

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

- Identify the hypotenuse in the rectangle's right triangle. - Which square-length equation connects the diagonal with the two sides? - Check that your result is shorter than the diagonal before rounding.

1. A rectangle's diagonal is the hypotenuse of a right triangle with the rectangle's sides as legs. 2. If the other side is \(b\), then \(b^2=19^2-7^2=312\). 3. Therefore, \(b=\sqrt{312}\,\text{cm}\approx17.66\,\text{cm}\).

\(b\approx17.66\,\text{cm}\).

- What kind of angle do the window's two edges form? - Which part of the window would be opposite that angle in a triangle?

1. A window corner is a right angle. The width and height form the legs of a right triangle, and the diagonal brace is its hypotenuse. 2. Therefore, \(d^2=9^2+12^2\).

\(d^2=9^2+12^2\), because the diagonal crosses a right triangle formed by the window's width and height.

- Which side is opposite the right angle? - How are the squares of the three side lengths related? - Should the unknown length be longer or shorter than either leg?

1. The legs are \(8\,\text{cm}\) and \(15\,\text{cm}\); \(BC\) is opposite the right angle. 2. \(BC^2=8^2+15^2=64+225=289\). 3. \(BC=17\,\text{cm}\).

\(BC^2=8^2+15^2=289\), so \(BC=17\,\text{cm}\).

- Which side is opposite the right angle? - Does \(AC\) name the longest side or a shorter side? - How do the squares of both legs relate to the hypotenuse's square? - How could you check whether a proposed leg is too long?

1. \(BC=13\,\text{cm}\) is the hypotenuse and \(AB=5\,\text{cm}\) is a leg. 2. \(AC^2+5^2=13^2\), so \(AC^2=169-25=144\). 3. Therefore, \(AC=12\,\text{cm}\). Subtraction isolates the missing leg's square.

\(AC=12\,\text{cm}\); subtract the known leg's square from the hypotenuse's square.

- Add the squares of the legs first. - An exact square root need not be a whole number. - Use the third decimal place to round to hundredths.

1. By the Pythagorean theorem, \(c^2=1^2+1^2=2\). 2. The exact length is \(\sqrt{2}\,\text{cm}\). 3. Since \(\sqrt{2}\approx1.4142\), the length rounds to \(1.41\,\text{cm}\).

Exactly \(\sqrt{2}\,\text{cm}\); approximately \(1.41\,\text{cm}\).

- Does adding the two walks give the route length or the shortest distance home? - What angle do east and north directions form? - Think about the direct path across the triangle.

1. The sum \(6+8=14\,\text{m}\) is how far Aria walked along the two streets, not her straight-line distance from the start. 2. The east and north parts meet at a right angle, making the straight path a hypotenuse. 3. Its length is \(\sqrt{6^2+8^2}=10\,\text{m}\).

\(14\,\text{m}\) is the total distance walked. The direct path satisfies \(d^2=6^2+8^2\), so its length is \(10\,\text{m}\).

- Where is the farthest point of a square lawn from its center? - What simpler lengths could help you find that distance? - Why might it help to keep the exact length until after you subtract?

1. The horizontal and vertical distances from the center to a corner are both \(20\,\text{ft}\), so \(r^2=20^2+20^2=800\). 2. Therefore, \(r=\sqrt{800}\,\text{ft}\approx28.28\,\text{ft}\). 3. The extra reach is \((\sqrt{800}-20)\,\text{ft}\approx8.28\,\text{ft}\).

a) \(r^2=20^2+20^2\); \(r=\sqrt{800}\,\text{ft}\approx28.28\,\text{ft}\). b) \((\sqrt{800}-20)\,\text{ft}\approx8.28\,\text{ft}\).

- 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.

- Sketch the pole, rope, and ground as the sides of a triangle. - What would be true of those three side lengths if the base angle were \(90^\circ\)? - Compare the original triangle with one drawn for an upright pole. Which two side lengths would remain the same?

1. If the pole were upright, its first \(4.00\,\text{m}\), the \(3.30\,\text{m}\) ground distance, and the rope would form a right triangle. 2. The sum of the leg squares is \(4.00^2+3.30^2=26.89\), while the current rope square is \(5.20^2=27.04\). They differ, so the angle at the pole base is not right. 3. For an upright pole, the rope would measure \(\sqrt{26.89}\,\text{m}\approx5.19\,\text{m}\).

The pole is not perpendicular to the ground. The upright-pole rope length would be \(\sqrt{26.89}\,\text{m}\approx5.19\,\text{m}\).

- 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 rectangle's diagonal makes two right triangles. - Find the missing length before finding the area. - Do not round the length until the final step.

1. The diagonal is the hypotenuse, so \(b^2=15^2-7^2=176\). 2. The length is \(b=\sqrt{176}\,\text{yd}\approx13.3\,\text{yd}\). 3. Use the unrounded length for area: \(A=7\sqrt{176}\,\text{yd}^2\approx92.9\,\text{yd}^2\).

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

- 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 leg appears in both triangles. - Use the first triangle to find \(k^2\). - The second triangle needs \(k^2\), not a rounded value of \(k\).

1. In the first triangle, \(k^2=13^2-7^2=120\), so \(k=\sqrt{120}\,\text{cm}\). 2. In the second triangle, \(x^2=k^2+9^2=120+81=201\). 3. Therefore, \(x=\sqrt{201}\,\text{cm}\).

a) \(k=\sqrt{120}\,\text{cm}\). b) \(x=\sqrt{201}\,\text{cm}\).

- Find the shared side from the first right triangle. - Each triangle's area is half the product of its legs. - Combine the two areas before approximating the shared side.

1. The shared side satisfies \(s^2=17^2-9^2=208\), so \(s=\sqrt{208}\,\text{yd}\). 2. The triangle areas sum to \(A=\frac{1}{2}(9s)+\frac{1}{2}(11s)=10s\). 3. Thus \(A=10\sqrt{208}\,\text{yd}^2\approx144.2\,\text{yd}^2\).

\(10\sqrt{208}\,\text{yd}^2\approx144.2\,\text{yd}^2\).

- 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) \(V=864\,\text{cm}^3\)

- 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 \(39\,\text{in.}\).

- Equal volumes let you find the cube's edge from the prism volume. - A space diagonal combines three perpendicular edge lengths. - Subtract before rounding the difference.

1. The prism volume is \(16\cdot8\cdot4=512\,\text{cm}^3\), so \(s^3=512\) and \(s=8\,\text{cm}\). 2. In the prism, first find a face diagonal: \(d_f^2=16^2+8^2=320\). The face diagonal and third edge form a right triangle, so the space diagonal is \(\sqrt{320+4^2}=\sqrt{336}\,\text{cm}\approx18.33\,\text{cm}\). 3. In the cube, first find a face diagonal squared: \(8^2+8^2=128\). Then its space diagonal is \(\sqrt{128+8^2}=\sqrt{192}\,\text{cm}\approx13.86\,\text{cm}\). 4. Subtract the unrounded lengths: \(\sqrt{336}-\sqrt{192}\approx4.47\,\text{cm}\).

a) \(s=8\,\text{cm}\). b) Prism diagonal \(\sqrt{336}\,\text{cm}\approx18.33\,\text{cm}\); cube diagonal \(\sqrt{192}\,\text{cm}\approx13.86\,\text{cm}\); difference \(\sqrt{336}-\sqrt{192}\,\text{cm}\approx4.47\,\text{cm}\).