Free Math Worksheets

- Recall the side-length relationship in a \(30^\circ\text{-}60^\circ\text{-}90^\circ\) triangle. - The shortest side is opposite the \(30^\circ\) angle.

1. In a \(30^\circ\text{-}60^\circ\text{-}90^\circ\) triangle, the side opposite \(30^\circ\) is half the hypotenuse. 2. Therefore, \(c=2\cdot 4\,\text{cm}=8\,\text{cm}\).

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

- Find the measure of the third angle. - What type of triangle has two congruent angles? - What does that imply about the opposite sides?

1. The third angle is \(180^\circ-90^\circ-45^\circ=45^\circ\). 2. Because the two acute angles are congruent, the triangle is isosceles. 3. The legs opposite those congruent angles are equal, so the other leg is also \(6\,\text{cm}\).

The other leg is \(6\,\text{cm}\) because the triangle is a \(45^\circ\text{-}45^\circ\text{-}90^\circ\) triangle.

- Use the Pythagorean theorem. - Remember that the two legs of an isosceles right triangle are congruent.

1. Both legs are \(5\,\text{cm}\), so the Pythagorean theorem gives \(c^2=5^2+5^2=50\). 2. Therefore, \(c=\sqrt{50}=5\sqrt{2}\,\text{cm}\approx 7.07\,\text{cm}\).

\(c=5\sqrt{2}\,\text{cm}\approx 7.07\,\text{cm}\)

- Use the symmetry of a \(45^\circ\text{-}45^\circ\text{-}90^\circ\) triangle. - What happens to the hypotenuse when the altitude is drawn from the right angle?

1. The triangle is a \(45^\circ\text{-}45^\circ\text{-}90^\circ\) triangle. 2. By symmetry, the altitude to the hypotenuse bisects the hypotenuse into two \(5\,\text{cm}\) segments. 3. Each smaller triangle is also a \(45^\circ\text{-}45^\circ\text{-}90^\circ\) triangle, so its legs are equal. Therefore, \(h=5\,\text{cm}\).

\(h=5\,\text{cm}\)

- A square root does not distribute over addition. - Add the squares before taking the square root. - Check whether the proposed diagonal satisfies the triangle inequality.

1. Leon incorrectly distributed the square root over a sum. In general, \(\sqrt{a^2 + b^2} \ne a + b\). 2. Apply the Pythagorean theorem: \(d^2 = 12^2 + 9^2 = 144 + 81 = 225\). 3. Therefore, \(d = \sqrt{225} = 15\,\text{in}.\)

Leon incorrectly treated \(\sqrt{12^2 + 9^2}\) as \(12 + 9\). The correct diagonal is \(15\,\text{in}.\)

- Which lengths form the legs of a right triangle inside the rhombus? - Use half of each diagonal. - Apply the Pythagorean theorem.

1. The diagonals of a rhombus are perpendicular bisectors. Their half-lengths are \(2.4\,\text{cm}\) and \(0.7\,\text{cm}\). 2. A side is the hypotenuse of the resulting right triangle, so \(s^2=2.4^2+0.7^2=5.76+0.49=6.25\). 3. Therefore, \(s=\sqrt{6.25}=2.5\,\text{cm}\).

The side length is \(2.5\,\text{cm}\).

- Divide the trapezoid into a rectangle and a right triangle. - Find the short horizontal leg of the right triangle from the base lengths. - Apply the Pythagorean theorem to the right triangle.

1. The horizontal leg of the right triangle is the difference of the base lengths: \(5-2=3\,\text{m}\). 2. The vertical leg is the height, \(4\,\text{m}\). 3. Apply the Pythagorean theorem: \(s^2=3^2+4^2=9+16=25\), so \(s=5\,\text{m}\).

The slanted side has length \(s=5\,\text{m}\).

- The altitude of an equilateral triangle lands at the midpoint of the opposite side. - Use the Pythagorean theorem in one of the two right triangles.

1. The altitude of an equilateral triangle bisects the base, so one leg of the resulting right triangle is \(4.5\,\text{cm}\), and the hypotenuse is \(9\,\text{cm}\). 2. By the Pythagorean theorem, \(h^2=9^2-4.5^2=81-20.25=60.75\). 3. Therefore, \(h=\sqrt{60.75}=\frac{9\sqrt{3}}{2}\,\text{cm}\approx 7.79\,\text{cm}\).

The height is \(\frac{9\sqrt{3}}{2}\,\text{cm}\), or approximately \(7.79\,\text{cm}\).

- Draw the altitude to the base. - Determine the length of half the base. - Use the Pythagorean theorem in one of the right triangles.

1. In an isosceles triangle, the altitude to the base bisects the base, creating a right triangle with one leg \(5\,\text{cm}\) long and hypotenuse \(13\,\text{cm}\). 2. Apply the Pythagorean theorem: \(h^2+5^2=13^2\). 3. Thus \(h^2=169-25=144\), so \(h=12\,\text{cm}\).

The altitude is \(h=12\,\text{cm}\).

- What are the two acute angles in an isosceles right triangle? - What kinds of triangles are formed by drawing the altitude to the hypotenuse? - Use the symmetry of the original triangle.

1. An isosceles right triangle has two \(45^\circ\) angles. 2. The altitude from the right angle to the hypotenuse divides the triangle into two congruent \(45^\circ\text{-}45^\circ\text{-}90^\circ\) triangles. 3. In triangle \(ACD\), the legs are equal, so \(AD=CD=8\,\text{cm}\). 4. By symmetry, \(DB=8\,\text{cm}\). 5. Therefore, \(AB=AD+DB=8+8=16\,\text{cm}\).

\(AB=16\,\text{cm}\)

- Decompose the trapezoid into a rectangle and a triangle. - Identify the special right triangle. - Use the difference of the base lengths.

1. Draw the perpendicular from \(C\) to \(AB\), meeting \(AB\) at \(E\). 2. Quadrilateral \(AECD\) is a rectangle, so \(AE = CD = 12\,\text{cm}\). 3. Therefore, \(EB = AB - AE = 18 - 12 = 6\,\text{cm}\). 4. Triangle \(EBC\) is a \(45^{\circ}\text{-}45^{\circ}\text{-}90^{\circ}\) triangle, so its legs are congruent. Thus \(CE = EB = 6\,\text{cm}\). 5. Since \(AD = CE\), the trapezoid’s height is \(6\,\text{cm}\).

The height \(AD\) is \(6\,\text{cm}\).

- Draw the square's diagonals and locate the common center. - The incircle touches the side midpoints, and the circumcircle passes through the vertices.

1. The inradius is half the side length: \(r=\frac{8}{2}=4\,\text{cm}\). 2. The square's diagonal is \(8\sqrt{2}\,\text{cm}\), and the circumradius is half the diagonal. 3. Therefore, \(R=\frac{8\sqrt{2}}{2}=4\sqrt{2}\,\text{cm}\approx 5.66\,\text{cm}\).

\(R\approx 5.66\,\text{cm}\) and \(r=4.00\,\text{cm}\)

- What happens to the shorter leg when the triangle is reflected? - What kind of triangle has three equal side lengths? - What are the angle measures in an equilateral triangle? - How does the reflection split an angle of the larger triangle?

1. Reflecting the triangle across leg \(b\) forms a larger triangle with two sides of length \(c\). Its base consists of two segments of length \(a\), so the base has length \(2a\). 2. Since \(c = 2a\), all three sides of the larger triangle have length \(2a\). The larger triangle is equilateral. 3. Every angle in an equilateral triangle measures \(60^{\circ}\). Therefore, \(\beta = 60^{\circ}\). 4. Angle \(\alpha\) is half of a \(60^{\circ}\) angle, so \(\alpha = 30^{\circ}\).

a) \(\alpha = 30^{\circ}\) and \(\beta = 60^{\circ}\). b) The reflection forms an equilateral triangle because each side has length \(c = 2a\).

- Rewrite the given equation so that it has the form “leg squared plus leg squared equals hypotenuse squared.” - The right angle is opposite the hypotenuse. - In an isosceles right triangle, the two legs have equal lengths.

1. Squaring \(b = \sqrt{c^2 - a^2}\) gives \(b^2 = c^2 - a^2\), which rearranges to \(a^2 + b^2 = c^2\). Therefore, \(c\) is the hypotenuse and \(\angle C\) is the right angle. 2. Substitute the given values: \(b = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15\). Thus, \(b = 15\,\text{cm}\). 3. In an isosceles right triangle, the legs are equal, so \(a = b = \ell\). Then \(\ell^2 + \ell^2 = c^2\), or \(2\ell^2 = c^2\). Because side lengths are positive, \(c = \ell\sqrt{2}\), equivalently \(\ell = \frac{c}{\sqrt{2}}\).

a) \(\angle C\) is the right angle. b) \(b = 15\,\text{cm}\). c) \(c = \ell\sqrt{2}\), equivalently \(2\ell^2 = c^2\).

- Apply the power rules to \((3x)^2\) and \((4x)^2\). - Use the two legs to write the area of the right triangle. - The ratio \(\frac{z}{x}\) asks how many times as large \(z\) is as \(x\).

1. Expand and combine like terms: \(z^2=9x^2+16x^2=25x^2\). 2. Since \(x>0\) and \(z>0\), \(z=5x\), so \(\frac{z}{x}=5\). 3. The area is \(\frac{1}{2}(3x)(4x)=6x^2\). Set \(6x^2=54\), giving \(x^2=9\) and \(x=3\,\text{cm}\). 4. Then \(z=5x=15\,\text{cm}\).

a) \(z^2=25x^2\) b) \(\frac{z}{x}=5\) c) \(x=3\,\text{cm}\) and \(z=15\,\text{cm}\)

- Express the area using the two congruent legs. - Find the leg length before using the Pythagorean theorem. - Give the hypotenuse in exact radical form and as a decimal approximation.

1. Let each leg have length \(a\). The area is \(\frac{1}{2}a^2\), so \(\frac{1}{2}a^2=32\). 2. Then \(a^2=64\), and because a length is positive, \(a=8\,\text{cm}\). 3. The hypotenuse satisfies \(c^2=8^2+8^2=128\), so \(c=8\sqrt{2}\,\text{cm}\approx 11.31\,\text{cm}\).

The legs are each \(8\,\text{cm}\), and the hypotenuse is \(8\sqrt{2}\,\text{cm}\approx 11.31\,\text{cm}\).

- Draw the altitude and determine half the base. - Use the Pythagorean theorem to find one congruent side. - Then add all three side lengths.

1. The altitude bisects the base, so each half is \(8\,\text{cm}\). 2. Each congruent side is the hypotenuse of a right triangle with legs \(8\,\text{cm}\) and \(15\,\text{cm}\): \(s^2=8^2+15^2=289\), so \(s=17\,\text{cm}\). 3. The perimeter is \(16+2\cdot 17=50\,\text{cm}\).

The perimeter is \(50\,\text{cm}\).

- Write a separate Pythagorean equation for each specified hypotenuse. - Use the distributive property or a binomial-square pattern when expanding \((x+2)^2\). - Reject any solution that cannot represent a positive side length.

1. a) Set up the Pythagorean equation \(x^2 + (x+2)^2 = 10^2\). Expanding and simplifying gives \(2x^2 + 4x - 96 = 0\), or \(x^2 + 2x - 48 = 0\). Factoring gives \((x+8)(x-6)=0\), so \(x=-8\) or \(x=6\). A side length must be positive, so \(x=6\). The side lengths are \(6\,\text{cm}\), \(8\,\text{cm}\), and \(10\,\text{cm}\). 2. b) Set up \(x^2 + 10^2 = (x+2)^2\). Expanding gives \(x^2 + 100 = x^2 + 4x + 4\), so \(96 = 4x\) and \(x=24\). The side lengths are \(10\,\text{cm}\), \(24\,\text{cm}\), and \(26\,\text{cm}\), which form a right triangle.

a) \(x=6\). b) Yes. \(x=24\).

- Find the rectangle’s missing side before finding its area. - Relate a square’s diagonal to two congruent side lengths. - Compare the two areas.

1. For the rectangle, the missing side is \(\sqrt{26^2-10^2}=\sqrt{576}=24\,\text{cm}\), so its area is \(10\cdot 24=240\,\text{cm}^2\). 2. For the square, \(d^2=s^2+s^2=2s^2\). Therefore, its area is \(s^2=\frac{26^2}{2}=338\,\text{cm}^2\). 3. Since \(338>240\), the square has the greater area.

The square has the greater area: \(338\,\text{cm}^2\), compared with \(240\,\text{cm}^2\) for the rectangle.

- Convert the side lengths to the same unit. - The altitude bisects the base. - Use the Pythagorean theorem to find the altitude before finding the area.

1. Convert the congruent side length: \(37\,\text{mm}=3.7\,\text{cm}\). 2. The altitude bisects the base, so half the base is \(1.2\,\text{cm}\). 3. Find the altitude: \(h^2=3.7^2-1.2^2=13.69-1.44=12.25\), so \(h=3.5\,\text{cm}\). 4. The area is \(A=\frac{1}{2}\cdot 2.4\cdot 3.5=4.2\,\text{cm}^2\).

The area is \(4.2\,\text{cm}^2\).

- Into what two smaller triangles does the altitude divide the isosceles triangle? - What are the leg lengths of each smaller triangle? - Use the converse of the Pythagorean theorem on the three sides of the original triangle.

1. In an isosceles triangle, the altitude to the base also bisects the base. Each half of the base is \(6\,\text{cm}\). 2. Each smaller right triangle therefore has two legs of length \(6\,\text{cm}\). If \(s\) is either congruent side of the original triangle, then \(s^2 = 6^2+6^2=72\). 3. For the original triangle, \(s^2+s^2=72+72=144\), and \(c^2=12^2=144\). By the converse of the Pythagorean theorem, the original triangle is right. 4. The side of length \(12\,\text{cm}\) is the hypotenuse, so the right angle is at the vertex opposite the base.

The triangle is right because the two congruent sides satisfy \(s^2+s^2=72+72=144=12^2\). The right angle is at the vertex opposite the \(12\,\text{cm}\) base.

- Relate the square’s diagonal to the circle’s diameter. - Use the Pythagorean theorem to express the area \(s^2\) directly.

1. A diagonal of the square is a diameter of the circle, so \(d=2\cdot 5=10\,\text{cm}\). 2. If the side length is \(s\), then \(d^2=s^2+s^2=2s^2\). 3. Therefore, the square’s area is \(s^2=\frac{10^2}{2}=50\,\text{cm}^2\).

The area is \(50\,\text{cm}^2\).

- An altitude divides each isosceles triangle into two right triangles. - In the first half-triangle, use the \(30^\circ\)-\(60^\circ\)-\(90^\circ\) relationship to get one leg, then use the Pythagorean theorem. - Find the base and altitude of each triangle separately.

1. In the first triangle, the altitude creates a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle with hypotenuse \(10\,\text{cm}\). The short leg, which is the altitude, is \(5\,\text{cm}\). Then the Pythagorean theorem gives half the base as \(\sqrt{10^2-5^2}=5\sqrt{3}\,\text{cm}\), so the base is \(10\sqrt{3}\,\text{cm}\). 2. Its area is \(A_1=\frac{1}{2}\cdot 10\sqrt{3}\cdot 5=25\sqrt{3}\,\text{cm}^2\). 3. The second triangle is equilateral, so its base is \(10\,\text{cm}\). The altitude bisects the base, so the Pythagorean theorem gives \(h=\sqrt{10^2-5^2}=5\sqrt{3}\,\text{cm}\). 4. Its area is \(A_2=\frac{1}{2}\cdot 10\cdot 5\sqrt{3}=25\sqrt{3}\,\text{cm}^2\). 5. Therefore, the two areas are equal.

Both triangles have area \(25\sqrt{3}\,\text{cm}^2\approx 43.30\,\text{cm}^2\).

- Compare the sum of the squares of the congruent sides with the square of the base. - Use that comparison to classify the angle opposite the longest side. - Draw the altitude to the base. What happens to the base of an isosceles triangle?

1. a) Compare \(12^2+12^2=144+144=288\) with \(17^2=289\). Since the values are not equal, the triangle is not right. 2. b) Because \(17^2>12^2+12^2\), the angle opposite the \(17\,\text{cm}\) side is obtuse. 3. c) In an isosceles triangle, the altitude to the base bisects the base. Each half is \(8.5\,\text{cm}\). Using either smaller right triangle, \(h_c=\sqrt{12^2-8.5^2}=\sqrt{144-72.25}=\sqrt{71.75}=\frac{\sqrt{287}}{2}\,\text{cm}\approx8.47\,\text{cm}\).

a) The triangle is not right because \(12^2+12^2=288\ne289=17^2\). b) The vertex angle is obtuse because \(17^2>12^2+12^2\). c) \(h_c=\frac{\sqrt{287}}{2}\,\text{cm}\approx8.47\,\text{cm}\).

- Express the longer leg in terms of the shorter leg. - Substitute that expression into the Pythagorean theorem. - Square all factors in expressions such as \(2a\) and \(5\sqrt{5}\).

1. Let \(b=2a\). Substitute into the Pythagorean theorem: \(a^2+(2a)^2=(5\sqrt{5})^2\). 2. Then \(5a^2=125\), so \(a^2=25\). Since a length is positive, \(a=5\,\text{cm}\). 3. Therefore, \(b=2a=10\,\text{cm}\).

\(a=5\,\text{cm}\) and \(b=10\,\text{cm}\)

- Find the square’s area from its diagonal. - Use the equal area to find the rectangle’s other side. - Then apply the Pythagorean theorem.

1. The square’s area is \(A=\frac{d_Q^2}{2}=\frac{10^2}{2}=50\,\text{cm}^2\). 2. The rectangle has the same area, so its other side is \(\frac{50}{4}=12.5\,\text{cm}\). 3. Its diagonal is \(d_R=\sqrt{4^2+12.5^2}=\sqrt{172.25}\,\text{cm}\approx 13.12\,\text{cm}\).

\(d_R\approx 13.12\,\text{cm}\)

- The altitude divides the isosceles triangle into two congruent right triangles. - Find half the base first, then double it. - Use the base and altitude to find the area.

1. The altitude bisects the base. If half the base is \(x\), then \(x^2=17^2-15^2=64\), so \(x=8\,\text{cm}\). 2. Therefore, \(c=2\cdot 8=16\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\cdot 16\cdot 15=120\,\text{cm}^2\).

\(c=16\,\text{cm}\) and \(A=120\,\text{cm}^2\)

- Use the area and base to find the altitude. - The altitude bisects the base. - Use a right triangle to find a sloping side.

1. Use the area to find the altitude: \(48=\frac{1}{2}\cdot 12\cdot h\), so \(h=8\,\text{ft}\). 2. The altitude bisects the base. Each roof beam has length \(s=\sqrt{8^2+6^2}=10\,\text{ft}\). 3. The perimeter is \(12+2\cdot 10=32\,\text{ft}\).

Each roof beam is \(10\,\text{ft}\) long, and the perimeter is \(32\,\text{ft}\).

- What does an \(8\%\) grade tell you about rise compared with horizontal run? - Convert the map length to the actual horizontal distance first. - The road itself is the hypotenuse of the right triangle.

1. Convert the map distance to the horizontal ground distance: \(5.6\,\text{cm}\cdot 25{,}000=140{,}000\,\text{cm}=1400\,\text{m}\). 2. An \(8\%\) grade means \(\frac{\text{rise}}{\text{run}}=0.08\). Therefore, the elevation change is \(h=0.08(1400)=112\,\text{m}\). 3. The road is the hypotenuse of a right triangle: \(L=\sqrt{1400^2+112^2}\approx 1404.47\,\text{m}\).

The elevation change is \(112\,\text{m}\), and the road segment is approximately \(1404.47\,\text{m}\) long.

- Use the Pythagorean theorem to relate a square’s side length and diagonal. - The area of a square is the square of its side length. - Express \(3a^2\) as a sum involving \(a^2\) and \(d^2\). - Think of those two known lengths as the legs of a right triangle.

1. a) The diagonal of the original square satisfies \(d^2=a^2+a^2=2a^2\), so \(d=a\sqrt{2}\). A square with side length \(d\) has area \(d^2=2a^2\). 2. b) A square with three times the original area needs side length \(a\sqrt{3}\), because \((a\sqrt{3})^2=3a^2\). 3. Since \(d^2=2a^2\), a right triangle with legs \(a\) and \(d\) has hypotenuse \(h\) satisfying \(h^2=a^2+d^2=a^2+2a^2=3a^2\). Thus, \(h=a\sqrt{3}\), so this hypotenuse is the required side length.

a) The new area is \(2a^2\). b) Construct a right triangle with legs \(a\) and \(d=a\sqrt{2}\). Its hypotenuse has length \(a\sqrt{3}\), which is the side length of a square with area \(3a^2\).

- Write the areas of \(Q_1\), \(Q_2\), and \(Q_3\) in terms of \(s\). - Relate each square’s side length to its area. - Apply the Pythagorean theorem to the two side lengths used as legs. - Remember that the square of the new side length is also the area of the new square.

1. The areas are \(A_1=s^2\), \(A_2=2s^2\), and \(A_3=3s^2\). Therefore, the side lengths of \(Q_2\) and \(Q_3\) are \(s_2=s\sqrt{2}\) and \(s_3=s\sqrt{3}\). 2. Let \(h\) be the hypotenuse of a right triangle with legs \(s_2\) and \(s_3\). Then \(h^2=s_2^2+s_3^2=2s^2+3s^2=5s^2\). 3. A square with side length \(h\) has area \(h^2=5s^2=5A_1\). Therefore, its area is five times the area of \(Q_1\).

The new square has area \(5s^2\), which is exactly five times the area \(s^2\) of \(Q_1\).

- What type of triangles are formed by a diagonal of a square? - Find the diagonal in terms of \(a\). - Use the right-triangle definitions of sine, cosine, and tangent.

1. By the Pythagorean theorem, the diagonal is \(d=\sqrt{a^2+a^2}=a\sqrt{2}\). 2. The diagonal divides the square into two \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangles. 3. Therefore, \(\sin(45^\circ)=\frac{a}{a\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\). 4. Similarly, \(\cos(45^\circ)=\frac{a}{a\sqrt{2}}=\frac{\sqrt{2}}{2}\). 5. Finally, \(\tan(45^\circ)=\frac{a}{a}=1\).

\(\sin(45^\circ)=\frac{\sqrt{2}}{2}\), \(\cos(45^\circ)=\frac{\sqrt{2}}{2}\), and \(\tan(45^\circ)=1\)

- Consider the altitude, which divides the equilateral triangle into two special right triangles. - Use the Pythagorean theorem to find the altitude. - Apply the tangent definition for each acute angle.

1. In one half of the equilateral triangle, \(h^2+\left(\frac{a}{2}\right)^2=a^2\). Thus, \(h=\frac{\sqrt{3}}{2}a\). 2. For the \(30^\circ\) angle, \(\tan(30^\circ)=\frac{a/2}{(\sqrt{3}/2)a}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\). 3. For the \(60^\circ\) angle, \(\tan(60^\circ)=\frac{(\sqrt{3}/2)a}{a/2}=\sqrt{3}\). 4. Since \(90^\circ-30^\circ=60^\circ\), \(\frac{1}{\tan(30^\circ)}=\frac{1}{1/\sqrt{3}}=\sqrt{3}=\tan(60^\circ)\).

a) \(h=\frac{\sqrt{3}}{2}a\) b) \(\tan(30^\circ)=\frac{\sqrt{3}}{3}\) and \(\tan(60^\circ)=\sqrt{3}\) c) \(\tan(60^\circ)=\frac{1}{\tan(30^\circ)}=\sqrt{3}\)

- Find the diagonal of the rectangular base first. - Then use that diagonal and the height as the legs of a second right triangle.

1. The base diagonal is \(d_G=\sqrt{12^2+9^2}=15\,\text{cm}\). 2. The base diagonal, height, and space diagonal form a right triangle: \(h^2=17^2-15^2=64\). 3. Therefore, \(h=8\,\text{cm}\).

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

- Relate the square’s area directly to the square of its diagonal. - The circle’s diameter equals the square’s side length. - Use \(A=\pi r^2\) for the circle.

1. If the square side is \(a\), then \(a^2=18\). The diagonal satisfies \(d^2=2a^2=36\), so \(d=6.00\,\text{ft}\). 2. The circle’s diameter is \(a\), so \(r=\frac{a}{2}\). Therefore, \(r^2=\frac{a^2}{4}=\frac{18}{4}=4.5\). 3. The circle’s area is \(\pi r^2=4.5\pi\,\text{ft}^2\approx 14.14\,\text{ft}^2\).

The square’s diagonal is \(6.00\,\text{ft}\), and the circle’s area is approximately \(14.14\,\text{ft}^2\).

- Think about the longest segment inside a rectangular prism. - Relate the three edge lengths to the space diagonal. - Write an equation in which the height is the only unknown.

1. For a rectangular prism with space diagonal \(d\), \(d^2=l^2+w^2+h^2\). 2. Substitute the known values: \(13^2=4^2+3^2+h^2\). 3. Simplify: \(169=16+9+h^2\), so \(h^2=144\). 4. Take the positive square root: \(h=\sqrt{144}=12\,\text{in.}\).

The box must be at least \(12\,\text{in.}\) high.

- What does a square base tell you about its two side lengths? - Write the space-diagonal equation with only one unknown. - Remember that the same base edge appears twice in the equation.

1. Because the base is a square, its length and width are both \(a\). 2. Use the space-diagonal relationship: \(9^2=a^2+a^2+7^2\). 3. Simplify: \(81=2a^2+49\), so \(2a^2=32\) and \(a^2=16\). 4. Take the positive square root: \(a=4\,\text{cm}\).

The side length of the square base is \(4\,\text{cm}\).

- What is special about all the edges of a cube? - Simplify the space-diagonal formula when all three edge lengths are equal. - For part c, work symbolically with \(a\) instead of using a decimal value.

1. For a cube, the space diagonal satisfies \(12^2=3a^2\). Thus, \(a^2=48\) and \(a=4\sqrt{3}\,\text{cm}\approx 6.93\,\text{cm}\). 2. A face diagonal satisfies \(d^2=2a^2\). Since \(a^2=48\), \(d=\sqrt{96}\,\text{cm}=4\sqrt{6}\,\text{cm}\approx 9.80\,\text{cm}\). 3. In general, the space diagonal of a cube is \(a\sqrt{3}\), so the ratio of the space diagonal to the edge length is \(\sqrt{3}:1\approx 1.73:1\).

a) \(a=4\sqrt{3}\,\text{cm}\approx 6.93\,\text{cm}\) b) \(d=4\sqrt{6}\,\text{cm}\approx 9.80\,\text{cm}\) c) The ratio is \(\sqrt{3}:1\approx 1.73:1\).

- What is the longest possible segment inside a rectangular prism? - You can apply the Pythagorean theorem first to a base diagonal and then again with the height. - Sketch the box and the rod's position.

1. The longest segment inside the rectangular prism is its space diagonal. 2. Use the three-dimensional Pythagorean relationship: \(d=\sqrt{20^2+30^2+50^2}\). 3. Simplify: \(d=\sqrt{400+900+2500}=\sqrt{3800}\approx 61.6\,\text{in.}\). 4. Because \(61.6\,\text{in.}<62\,\text{in.}\), the box is too short even along its space diagonal.

No. The box's space diagonal is approximately \(61.6\,\text{in.}\), which is shorter than the \(62\,\text{in.}\) rod.

- Divide the trapezoid into a rectangle and two right triangles. - Find the horizontal distance from one lower vertex to the foot of the opposite altitude. - Identify a right triangle in which the diagonal is the hypotenuse.

1. In an isosceles trapezoid, the difference between the base lengths is split equally on the two sides: \(\frac{12-6}{2}=3\,\text{cm}\). 2. The horizontal leg of the right triangle containing a diagonal is \(6+3=9\,\text{cm}\). 3. Apply the Pythagorean theorem: \(d^2=9^2+4^2=97\). 4. Therefore, \(d=\sqrt{97}\,\text{cm}\approx 9.85\,\text{cm}\).

The diagonal is approximately \(9.85\,\text{cm}\) long.

- Express the area of a square in terms of its side length. - Determine how a ninefold area change affects the side length. - Relate the diagonal to the side length using the Pythagorean theorem. - Use variables to show that the result holds for any square.

1. Let the original side length be \(s\). Its area is \(s^2\), and its diagonal is \(d=s\sqrt{2}\). 2. If the new area is nine times the original area, then \(s_{\text{new}}^2=9s^2\). Because lengths are positive, \(s_{\text{new}}=3s\), so the side length is multiplied by \(3\). 3. The new diagonal is \(d_{\text{new}}=s_{\text{new}}\sqrt{2}=3s\sqrt{2}=3d\). Therefore, the diagonal is also multiplied by \(3\).

Both the side length and the diagonal length are multiplied by \(3\).

- Recall the area formula for a triangle. - In an isosceles right triangle, the two legs have equal lengths. - Write a Pythagorean equation before and after the legs are doubled. - Consider what happens to a squared length when the length is doubled.

1. Let each original leg have length \(a\). Then \(c^2=a^2+a^2=2a^2\), so \(c=a\sqrt{2}\). 2. After both legs are doubled, the new hypotenuse is \(c_{\text{new}}=\sqrt{(2a)^2+(2a)^2}=\sqrt{8a^2}=2a\sqrt{2}=2c\). 3. The original area is \(A=\frac{1}{2}a^2\). The new area is \(A_{\text{new}}=\frac{1}{2}(2a)(2a)=2a^2=4A\).

The hypotenuse length doubles, and the area becomes four times as large.

- Recall the angle measures in an equilateral triangle. - Determine how the altitude divides the base and vertex angle. - Use the Pythagorean theorem, then apply the definitions of sine and cosine.

1. An equilateral triangle has three \(60^\circ\) angles. The altitude is perpendicular to the base and bisects the vertex angle, creating angles of \(30^\circ\), \(60^\circ\), and \(90^\circ\). 2. The altitude \(h\) bisects the base, so \(h^2+\left(\frac{s}{2}\right)^2=s^2\). Thus, \(h=\frac{\sqrt{3}}{2}s\). 3. For the \(60^\circ\) angle, \(\sin(60^\circ)=\frac{h}{s}=\frac{\sqrt{3}}{2}\). 4. Also, \(\cos(60^\circ)=\frac{s/2}{s}=\frac{1}{2}\).

a) Each half has angles \(30^\circ\), \(60^\circ\), and \(90^\circ\). b) \(h=\frac{\sqrt{3}}{2}s\) c) \(\sin(60^\circ)=\frac{\sqrt{3}}{2}\) and \(\cos(60^\circ)=\frac{1}{2}\)

- The diagonal forms a \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle. - Apply the Pythagorean theorem to the two equal legs. - In the trigonometric method, identify the side opposite the \(45^\circ\) angle.

1. By the Pythagorean theorem, \(s^2+s^2=12^2\), so \(2s^2=144\) and \(s=6\sqrt{2}\,\text{cm}\). 2. In one half of the square, \(\sin(45^\circ)=\frac{s}{12}\). Thus, \(s=12\cdot\frac{\sqrt{2}}{2}=6\sqrt{2}\,\text{cm}\). 3. Numerically, \(s\approx 8.49\,\text{cm}\).

\(s=6\sqrt{2}\,\text{cm}\approx 8.49\,\text{cm}\)

- Find the horizontal and vertical coordinate differences. - Use the Pythagorean theorem for the diagonal distance. - Combine one point’s x-coordinate with the other point’s y-coordinate to locate \(P\).

1. The horizontal change is \(|2 - (-4)| = 6\), and the vertical change is \(|-1 - 3| = 4\). 2. By the Pythagorean theorem, \(MN = \sqrt{6^2 + 4^2} = \sqrt{52} = 2\sqrt{13} \approx 7.21\) miles. 3. One possible right-angle vertex is \(P(2, 3)\). Another is \(P(-4, -1)\). 4. The perimeter is \(6 + 4 + \sqrt{52} \approx 17.21\) miles. 5. The area is \(\frac{1}{2}(6)(4) = 12\) square miles.

a) \(MN = 2\sqrt{13} \approx 7.21\) miles b) \(P(2, 3)\), or \(P(-4, -1)\) c) Perimeter \(\approx 17.21\) miles; area \(= 12\) square miles

- Identify the right triangle formed by a radius, the center-to-chord distance, and part of the chord. - The distance from a point to a line segment is measured perpendicularly. - A perpendicular from the center of a circle bisects a chord. - Apply the Pythagorean theorem to one half of the diagram.

1. The perpendicular segment from the center to a chord bisects the chord. Thus, the radius \(r\), distance \(d\), and half-chord \(\frac{s}{2}\) form a right triangle. 2. Apply the Pythagorean theorem: \(13^2=5^2+\left(\frac{s}{2}\right)^2\). 3. Then \(\left(\frac{s}{2}\right)^2=169-25=144\), so \(\frac{s}{2}=12\,\text{cm}\). 4. Therefore, \(s=2\cdot12=24\,\text{cm}\).

The chord length is \(s=24\,\text{cm}\).

- Identify the right angle in the diagram. - A perpendicular from the center of a circle bisects a chord. - The radius is the hypotenuse of the smaller right triangle. - Use half of the chord length in the calculation.

1. The perpendicular from the center to the chord bisects the chord, so half of the chord is \(15\,\text{cm}\). 2. The radius is the hypotenuse of a right triangle whose legs are \(d\) and \(15\,\text{cm}\). 3. Apply the Pythagorean theorem: \(d^2+15^2=17^2\). 4. Then \(d^2=289-225=64\), so \(d=\sqrt{64}=8\,\text{cm}\).

The distance is \(d=8\,\text{cm}\).

- How do the diagonals of a rhombus divide each other? - Substitute \(f=2e\) before applying the Pythagorean theorem. - Use half of each diagonal as a leg of the right triangle.

1. The diagonals of a rhombus are perpendicular bisectors of each other. A side of the rhombus is the hypotenuse of a right triangle with legs \(\frac{e}{2}\) and \(\frac{f}{2}\). 2. Since \(f=2e\), \(\frac{f}{2}=e\). 3. Apply the Pythagorean theorem: \(s^2=\left(\frac{e}{2}\right)^2+e^2=\frac{e^2}{4}+\frac{4e^2}{4}=\frac{5e^2}{4}\). 4. Because \(s>0\), \(s=\frac{\sqrt{5}}{2}e\).

\(s=\frac{\sqrt{5}}{2}e\)

- What additional length is needed to use the rhombus area formula? - Find the missing diagonal from the side length and half of the known diagonal. - Use the formula \(A=\frac{ef}{2}\).

1. The diagonals of a rhombus are perpendicular bisectors. One half of diagonal \(f\) is \(6\,\text{m}\). 2. Let \(\frac{e}{2}\) be half of the other diagonal. In the right triangle, \(\left(\frac{e}{2}\right)^2+6^2=10^2\). 3. Thus, \(\left(\frac{e}{2}\right)^2=64\), so \(\frac{e}{2}=8\,\text{m}\) and \(e=16\,\text{m}\). 4. The area of a rhombus is \(A=\frac{ef}{2}\). Therefore, \(A=\frac{16\cdot12}{2}=96\,\text{m}^2\).

The area is \(96\,\text{m}^2\).

- Divide the trapezoid into a rectangle and a right triangle. - Relate the two base lengths to the horizontal leg of the triangle. - Use the Pythagorean theorem to find that horizontal leg.

1. The slanted side is the hypotenuse of a right triangle with vertical leg \(12\,\text{cm}\). Let \(x\) be its horizontal leg. 2. Then \(x^2=13^2-12^2=169-144=25\), so \(x=5\,\text{cm}\). 3. The longer base is \(a=c+x=7+5=12\,\text{cm}\).

The longer base is \(a=12\,\text{cm}\).

- Draw perpendiculars from the upper base to the lower base. - Use the Pythagorean theorem to find one horizontal section. - Remember that an isosceles trapezoid has one matching section on each side.

1. Dropping perpendiculars from the endpoints of the upper base creates two congruent right triangles. In each triangle, the hypotenuse is \(10\,\text{m}\) and the vertical leg is \(8\,\text{m}\). 2. Let \(x\) be the horizontal leg. Then \(x^2=10^2-8^2=100-64=36\), so \(x=6\,\text{m}\). 3. The lower base contains the upper base and two horizontal legs: \(a=5+2\cdot 6=17\,\text{m}\).

The lower base has length \(a=17\,\text{m}\).

- Use the Pythagorean theorem in the two smaller right triangles. - The full length \(c\) is the sum of \(p\) and \(q\). - Compare \(a^2+b^2\) with \(c^2\). - Use the converse of the Pythagorean theorem to decide whether the angle at \(C\) is right.

1. In right triangle \(BCD\), \(a^2=4^2+8^2=80\), so \(a=\sqrt{80}\,\text{cm}=4\sqrt{5}\,\text{cm}\approx 8.94\,\text{cm}\). 2. The full base is \(c=p+q=2+8=10\,\text{cm}\). Therefore, \(qc=8\cdot 10=80=a^2\), so the relationship holds. 3. In right triangle \(ACD\), \(b^2=4^2+2^2=20\). 4. Since \(a^2+b^2=80+20=100=c^2\), the converse of the Pythagorean theorem shows that triangle \(ABC\) is right at \(C\).

a) \(a=4\sqrt{5}\,\text{cm}\approx 8.94\,\text{cm}\) b) Yes. \(a^2=80\) and \(qc=8\cdot 10=80\). c) Yes. Since \(a^2+b^2=80+20=100=c^2\), triangle \(ABC\) is right at \(C\).

- Draw the altitude from the vertex to create two right triangles. - Use the Pythagorean theorem to express the altitude in terms of \(x\). - Use the triangle area formula. - For part b, set your area expression equal to the given area.

1. Draw the altitude to the base. In an isosceles triangle, it bisects the \(10x\) base into two segments of length \(5x\). 2. Use the Pythagorean theorem to find the altitude \(h\): \(h^2=(13x)^2-(5x)^2=144x^2\). Since lengths are positive, \(h=12x\). 3. The area is \(A=\frac{1}{2}\cdot 10x\cdot 12x=60x^2\). 4. For an area of \(240\,\text{cm}^2\), solve \(60x^2=240\). Then \(x^2=4\), so \(x=2\,\text{cm}\).

a) \(A=60x^2\) b) \(x=2\,\text{cm}\)

- Draw a height that creates a right triangle. - Find the short horizontal leg from the difference of the parallel side lengths. - Use the trapezoid area formula after finding the height. - For part b, set your area expression equal to the given area.

1. Draw the height to form a right triangle. Its horizontal leg is \(12k-7k=5k\), and its hypotenuse is \(13k\). 2. By the Pythagorean theorem, \(h^2=(13k)^2-(5k)^2=144k^2\). Since lengths are positive, \(h=12k\). 3. The trapezoid area is \(A=\frac{12k+7k}{2}\cdot 12k=114k^2\). 4. For an area of \(456\,\text{cm}^2\), solve \(114k^2=456\). Then \(k^2=4\), so \(k=2\,\text{cm}\).

a) \(A=114k^2\) b) \(k=2\,\text{cm}\)

- Draw heights from both endpoints of the shorter base. - Find the length of one small horizontal section. - Use the Pythagorean theorem to find the height. - Then apply the trapezoid area formula.

1. Drawing the two heights creates two congruent right triangles. Each horizontal leg has length \(\frac{15-9}{2}=3\,\text{m}\). 2. In either right triangle, \(h^2+3^2=5^2\). Thus \(h^2=16\), so \(h=4\,\text{m}\). 3. Use the trapezoid area formula: \(A=\frac{15+9}{2}\cdot 4=48\,\text{m}^2\).

1. \(h=4\,\text{m}\) 2. \(A=48\,\text{m}^2\)

- Draw the height to create a right triangle. - Find the triangle’s horizontal leg from the parallel side lengths. - Use the Pythagorean theorem to find the height. - Then use the trapezoid area formula.

1. Drawing the height creates a right triangle. Its horizontal leg is the difference of the base lengths: \(14-8=6\,\text{m}\). 2. The slanted side is the hypotenuse, so \(h^2+6^2=10^2\). Thus \(h^2=64\), and \(h=8\,\text{m}\). 3. The area is \(A=\frac{14+8}{2}\cdot 8=88\,\text{m}^2\).

1. \(h=8\,\text{m}\) 2. \(A=88\,\text{m}^2\)

- Find the hypotenuse in terms of \(a\). - Apply the right-triangle definitions of sine and cosine. - Rationalize the denominator.

1. By the Pythagorean theorem, \(d^2=a^2+a^2=2a^2\), so \(d=a\sqrt{2}\). 2. Therefore, \(\sin(45^\circ)=\frac{a}{a\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\). 3. Because the legs are equal, the cosine ratio is the same: \(\cos(45^\circ)=\frac{a}{a\sqrt{2}}=\frac{\sqrt{2}}{2}\).

\(\sin(45^\circ)=\frac{\sqrt{2}}{2}\) and \(\cos(45^\circ)=\frac{\sqrt{2}}{2}\)

- First find the altitude in terms of \(s\). - Identify the opposite and adjacent legs relative to \(60^\circ\). - Substitute the lengths into the tangent ratio and simplify.

1. The altitude bisects the base, so one right triangle has hypotenuse \(s\) and one leg \(\frac{s}{2}\). 2. By the Pythagorean theorem, \(h=\sqrt{s^2-\left(\frac{s}{2}\right)^2}=\frac{\sqrt{3}}{2}s\). 3. Relative to the \(60^\circ\) angle, \(h\) is opposite and \(\frac{s}{2}\) is adjacent. 4. Thus, \(\tan(60^\circ)=\frac{(\sqrt{3}/2)s}{s/2}=\sqrt{3}\).

\(\tan(60^\circ)=\sqrt{3}\)

- View the figure as an inner square surrounded by four congruent right triangles. - Determine the two leg lengths of a corner triangle. - Use the Pythagorean theorem to relate those legs to a side of the inner square.

1. The inner square leaves four congruent right triangles at the corners of the outer square. 2. Each right triangle has legs \(2\,\text{cm}\) and \(8-2=6\,\text{cm}\). 3. A side \(s\) of the inner square is the hypotenuse of one corner triangle, so \(s^2=2^2+6^2=40\). 4. Because the area of a square is the square of its side length, the area of \(PQRS\) is \(40\,\text{cm}^2\).

The area of square \(PQRS\) is \(40\,\text{cm}^2\).

- First find the hypotenuse. - Write two area formulas for the same triangle, using each possible base-height pair.

1. Find the hypotenuse: \(c=\sqrt{8^2+6^2}=10\,\text{cm}\). 2. Compute the area in two ways: \(A=\frac{1}{2}ab\) and \(A=\frac{1}{2}ch\). 3. Set the expressions equal: \(\frac{1}{2}\cdot 8\cdot 6=\frac{1}{2}\cdot 10\cdot h\). 4. Thus, \(24=5h\), so \(h=4.8\,\text{cm}\).

\(h=4.8\,\text{cm}\)

- In an equilateral triangle, the centroid, incenter, and circumcenter are the same point. - That point divides each median in a \(2:1\) ratio.

1. The triangle's height is \(h=\frac{12\sqrt{3}}{2}=6\sqrt{3}\,\text{cm}\). 2. In an equilateral triangle, the common center divides each median in a \(2:1\) ratio. 3. Thus, \(R=\frac{2}{3}h=4\sqrt{3}\,\text{cm}\approx 6.93\,\text{cm}\). 4. Also, \(r=\frac{1}{3}h=2\sqrt{3}\,\text{cm}\approx 3.46\,\text{cm}\).

\(R\approx 6.93\,\text{cm}\) and \(r\approx 3.46\,\text{cm}\)

- Divide the regular hexagon into six congruent equilateral triangles. - Use the side relationships in an equilateral triangle.

1. A regular hexagon can be divided into six equilateral triangles, so the circumradius equals the side length: \(R=6\,\text{cm}\). 2. The inradius is the altitude of one equilateral triangle: \(r=\frac{6\sqrt{3}}{2}=3\sqrt{3}\,\text{cm}\approx 5.20\,\text{cm}\).

\(R=6.00\,\text{cm}\) and \(r\approx 5.20\,\text{cm}\)

- Draw an altitude to one side of the equilateral triangle. - What are the side lengths of either resulting right triangle? - Find the altitude first, then use the triangle area formula.

1. Drawing an altitude creates two right triangles with hypotenuse \(4\,\text{cm}\) and one leg \(2\,\text{cm}\). 2. The altitude is \(h=\sqrt{4^2-2^2}=\sqrt{12}=2\sqrt{3}\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\cdot 4\cdot 2\sqrt{3}=4\sqrt{3}\,\text{cm}^2\approx 6.93\,\text{cm}^2\).

\(A=4\sqrt{3}\,\text{cm}^2\approx 6.93\,\text{cm}^2\)

- The altitude of an isosceles triangle bisects the base. - Use the Pythagorean theorem to find the altitude.

1. The altitude bisects the base, so each half is \(6\,\text{cm}\). 2. Use the Pythagorean theorem to find the altitude: \(h^2=10^2-6^2=64\), so \(h=8\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\cdot 12\cdot 8=48\,\text{cm}^2\).

The area is \(48\,\text{cm}^2\).

- Recall the side relationships in a \(30^\circ\text{-}60^\circ\text{-}90^\circ\) triangle. - Use the triangle angle sum. - Use the Pythagorean theorem for the remaining side.

1. The wall and ground form a right angle, so the angle between the ladder and wall is \(180^\circ-90^\circ-60^\circ=30^\circ\). 2. In a \(30^\circ\text{-}60^\circ\text{-}90^\circ\) triangle, the side opposite \(30^\circ\) is half the hypotenuse. Thus, the ground distance is \(\frac{12}{2}=6\,\text{ft}\). 3. The height is \(h=\sqrt{12^2-6^2}=\sqrt{108}=6\sqrt{3}\,\text{ft}\approx 10.39\,\text{ft}\).

a) \(30^\circ\) b) \(6\,\text{ft}\), because the shorter leg is half the hypotenuse. c) \(6\sqrt{3}\,\text{ft}\approx 10.39\,\text{ft}\)

- Test whether either triangle is right. - For the isosceles triangle, draw an altitude to the base. - Find both areas before comparing them.

1. For Triangle 1, \(15^2+36^2=1521=39^2\), so it is a right triangle. Its area is \(A_1=\frac{1}{2}\cdot 15\cdot 36=270\,\text{cm}^2\). 2. Triangle 2 is isosceles. Its altitude bisects the \(14\,\text{cm}\) base, so \(h^2=25^2-7^2=576\), giving \(h=24\,\text{cm}\). 3. Its area is \(A_2=\frac{1}{2}\cdot 14\cdot 24=168\,\text{cm}^2\). 4. Since \(270>168\), Triangle 1 has the greater area.

Triangle 1 has area \(270\,\text{cm}^2\); Triangle 2 has area \(168\,\text{cm}^2\). Triangle 1 has the greater area.

- First determine when two sides of length \(s\) can form a triangle with a \(10\,\text{cm}\) base. - Compare the square of the base with the sum of the squares of the congruent sides. - For an obtuse vertex angle, decide which side of that comparison must be larger.

1. The triangle inequality requires \(s+s>10\), so \(s>5\,\text{cm}\). 2. The vertex angle is opposite the base. For that angle to be obtuse, the square of the base must be greater than the sum of the squares of the congruent sides: \(10^2>s^2+s^2\). 3. Simplifying gives \(100>2s^2\), so \(s^2<50\). Because \(s\) is positive, \(s<\sqrt{50}=5\sqrt{2}\,\text{cm}\). 4. Combining the two conditions gives \(5\,\text{cm}<s<5\sqrt{2}\,\text{cm}\), or approximately \(5\,\text{cm}<s<7.07\,\text{cm}\).

The required range is \(5\,\text{cm}<s<5\sqrt{2}\,\text{cm}\), approximately \(5\,\text{cm}<s<7.07\,\text{cm}\).

- Use the triangle inequality to determine the possible base lengths. - For a right vertex angle, the base would be the hypotenuse. - Compare the square of the base with the sum of the squares of the congruent sides.

1. a) The base must be positive, and the triangle inequality requires \(b<10+10\). Therefore, \(0\,\text{cm}<b<20\,\text{cm}\). 2. b) For a right vertex angle, the base is the hypotenuse, so \(10^2+10^2=b^2\). Thus, \(b=\sqrt{200}=10\sqrt{2}\,\text{cm}\). 3. c) Compare \(10^2+10^2=200\) with \(14.1^2=198.81\). Since \(200>198.81\), the vertex angle is acute. Because the base is the longest side, the vertex angle is the largest angle, so the entire triangle is acute.

a) \(0\,\text{cm}<b<20\,\text{cm}\). b) \(b=10\sqrt{2}\,\text{cm}\). c) The triangle is acute because \(10^2+10^2=200>198.81=14.1^2\).

- Use the perimeter to find each congruent side. - Find the altitude with the Pythagorean theorem. - Calculate the area before comparing.

1. Let each congruent side have length \(s\). Then \(10+2s=36\), so \(s=13\,\text{cm}\). 2. The altitude bisects the base, so \(h^2=13^2-5^2=144\), giving \(h=12\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\cdot 10\cdot 12=60\,\text{cm}^2\). 4. Since \(60>50\), the area is greater than \(50\,\text{cm}^2\).

The area is \(60\,\text{cm}^2\), which is greater than \(50\,\text{cm}^2\).

- Represent each screen’s width and height using its aspect ratio. - Use the diagonal and the Pythagorean theorem to find the scale factor. - Multiply width by height, then compare the areas.

1. For the \(4:3\) screen, let the dimensions be \(4x\) and \(3x\). Then \(25x^2=40^2\), so \(x=8\). The dimensions are \(32\,\text{in.}\) by \(24\,\text{in.}\), and the area is \(768\,\text{in.}^2\). 2. For the \(16:9\) screen, let the dimensions be \(16k\) and \(9k\). Then \(337k^2=1600\), so its area is \(144k^2=\frac{230400}{337}\,\text{in.}^2\approx 683.68\,\text{in.}^2\). 3. The \(4:3\) screen has the greater area. The difference is \(768-683.68=84.32\,\text{in.}^2\).

The \(4:3\) screen has area \(768\,\text{in.}^2\); the \(16:9\) screen has area approximately \(683.68\,\text{in.}^2\). The difference is approximately \(84.32\,\text{in.}^2\).

- Sketch each figure and draw an altitude to create right triangles. - In the trapezoid, split the difference between the base lengths equally. - Identify the horizontal distance beneath a trapezoid diagonal. - In an equilateral triangle, the altitude bisects the opposite side.

1. The altitude of an equilateral triangle bisects its base. The resulting right triangle gives \(a^2=9^2+\left(\frac{a}{2}\right)^2\). 2. Solve: \(\frac{3}{4}a^2=81\), so \(a=\frac{18}{\sqrt{3}}=6\sqrt{3}\,\text{cm}\approx 10.4\,\text{cm}\). 3. The area is \(A=\frac{1}{2}ah=\frac{1}{2}\cdot 6\sqrt{3}\cdot 9=27\sqrt{3}\,\text{cm}^2\approx 46.8\,\text{cm}^2\). 4. For the trapezoid, the horizontal leg of each outer right triangle is \(\frac{20-8}{2}=6\,\text{cm}\). 5. The height is \(h=\sqrt{10^2-6^2}=8\,\text{cm}\). 6. A diagonal is the hypotenuse of a right triangle with legs \(8\,\text{cm}\) and \(8+6=14\,\text{cm}\). Thus, \(d=\sqrt{8^2+14^2}=\sqrt{260}\,\text{cm}\approx 16.1\,\text{cm}\).

1) \(a\approx 10.4\,\text{cm}\); \(A\approx 46.8\,\text{cm}^2\) 2) \(h=8\,\text{cm}\); \(d\approx 16.1\,\text{cm}\)

- Divide the isosceles trapezoid into right triangles. - Use the difference of the base lengths to find each horizontal offset. - Identify the horizontal segment that forms a right triangle with \(h\) and diagonal \(e\). - Use the converse of the Pythagorean theorem to test \(\triangle ABC\).

1. a) Because the trapezoid is isosceles, each horizontal offset is \(x=\frac{14-4}{2}=5\,\text{cm}\). 2. A leg is the hypotenuse of a right triangle with legs \(x\) and \(h\). Thus, \(h=\sqrt{13^2-5^2}=\sqrt{144}=12\,\text{cm}\). 3. For diagonal \(AC\), the horizontal distance is \(14-5=9\,\text{cm}\). Therefore, \(e=\sqrt{9^2+12^2}=\sqrt{225}=15\,\text{cm}\). 4. b) Using base \(AB=14\,\text{cm}\) and height \(12\,\text{cm}\), the area is \(\frac{1}{2}\cdot14\cdot12=84\,\text{cm}^2\). 5. c) The side lengths of \(\triangle ABC\) are \(13\), \(14\), and \(15\) centimeters. Since \(13^2+14^2=365\ne225=15^2\), the triangle is not right.

a) \(h=12\,\text{cm}\) and \(e=15\,\text{cm}\). b) \(84\,\text{cm}^2\). c) No. \(\triangle ABC\) is not right because \(13^2+14^2\ne15^2\).

- Sketch the trapezoid and label the known lengths. - Use the symmetry of an isosceles trapezoid to split the extra length of the longer base equally. - Identify the right triangles that contain a leg and a diagonal. - Use the converse of the Pythagorean theorem to test triangle \(ABC\).

1. The difference between the base lengths is split equally, so each outer horizontal segment is \(\frac{25-7}{2}=9\,\text{cm}\). 2. Each leg is the hypotenuse of a right triangle with legs \(9\,\text{cm}\) and \(12\,\text{cm}\): \(b=d=\sqrt{9^2+12^2}=15\,\text{cm}\). 3. The diagonal \(e=AC\) is the hypotenuse of a right triangle with legs \(25-9=16\,\text{cm}\) and \(12\,\text{cm}\): \(e=\sqrt{16^2+12^2}=20\,\text{cm}\). 4. Triangle \(ABC\) has side lengths \(15\), \(20\), and \(25\). Because \(15^2+20^2=25^2\), it is a right triangle. 5. The side \(AB\) is the hypotenuse, so the right angle is at \(C\).

a) \(b=d=15\,\text{cm}\); \(e=20\,\text{cm}\) b) Yes. Triangle \(ABC\) is right because \(15^2+20^2=25^2\). The right angle is at \(C\).

- Identify the right triangles formed by the perpendicular diagonals. - First find the part of diagonal \(f\) in the triangle containing side \(a\). - Subtract that length from \(f\), then use the remaining part to find \(b\).

1. The diagonals of a kite are perpendicular. Because \(f\) bisects \(e\), each half of \(e\) is \(8\,\text{cm}\). 2. Let \(f_1\) be the part of diagonal \(f\) in the right triangle with hypotenuse \(10\,\text{cm}\). Then \(f_1^2=10^2-8^2=36\), so \(f_1=6\,\text{cm}\). 3. The remaining part is \(f_2=21-6=15\,\text{cm}\). 4. The longer side is the hypotenuse of a right triangle with legs \(8\,\text{cm}\) and \(15\,\text{cm}\): \(b^2=8^2+15^2=289\). Therefore, \(b=17\,\text{cm}\).

The longer side has length \(b=17\,\text{cm}\).

- Draw both diagonals and identify which one is bisected. - The unknown diagonal is made of one leg from each of two right triangles. - Find those two parts separately with the Pythagorean theorem.

1. The line-of-symmetry diagonal is perpendicular to and bisects the \(24\,\text{cm}\) diagonal, so each half is \(12\,\text{cm}\). 2. Let the two parts of diagonal \(d\) be \(d_1\) and \(d_2\). 3. In the triangle with hypotenuse \(13\,\text{cm}\), \(d_1^2=13^2-12^2=25\), so \(d_1=5\,\text{cm}\). 4. In the triangle with hypotenuse \(20\,\text{cm}\), \(d_2^2=20^2-12^2=256\), so \(d_2=16\,\text{cm}\). 5. Therefore, \(d=d_1+d_2=5+16=21\,\text{cm}\).

The other diagonal has length \(d=21\,\text{cm}\).

- What equation results when you add the two given equations? - Factor the common factor on the right side. - How are \(p\), \(q\), and \(c\) related? - Which converse theorem connects an equation involving three squared side lengths to a right angle?

1. Add the given equations: \(a^2+b^2=qc+pc\). 2. Factor the right side: \(a^2+b^2=c(q+p)\). 3. Because \(D\) lies on \(AB\), \(p+q=c\). 4. Substitute: \(a^2+b^2=c^2\). 5. By the converse of the Pythagorean theorem, triangle \(ABC\) is a right triangle with hypotenuse \(AB\). Therefore, the right angle is at \(C\).

Adding the equations gives \(a^2+b^2=c(p+q)\). Since \(p+q=c\), \(a^2+b^2=c^2\). By the converse of the Pythagorean theorem, triangle \(ABC\) is right at \(C\).

- Apply the Pythagorean theorem in right triangle \(BCD\). - Replace \(c\) with \(p+q\) in the given equation. - After finding \(h^2\), apply the Pythagorean theorem in right triangle \(ACD\). - Combine the resulting expressions for \(a^2\) and \(b^2\).

1. In right triangle \(BCD\), \(a^2=h^2+q^2\). 2. The given condition and \(c=p+q\) give \(a^2=qc=q(p+q)=pq+q^2\). 3. Equate the two expressions for \(a^2\): \(h^2+q^2=pq+q^2\). Therefore, \(h^2=pq\). 4. In right triangle \(ACD\), \(b^2=h^2+p^2=pq+p^2=p(p+q)=pc\). 5. Add the two leg equations: \(a^2+b^2=qc+pc=c(p+q)=c^2\). 6. By the converse of the Pythagorean theorem, triangle \(ABC\) is right at \(C\).

Since \(a^2=h^2+q^2\) and \(a^2=q(p+q)=pq+q^2\), it follows that \(h^2=pq\). Then \(b^2=h^2+p^2=pc\), so \(a^2+b^2=qc+pc=c^2\). Therefore, triangle \(ABC\) is right at \(C\).

- Identify the horizontal, depth, and vertical changes from \(A\) to \(M\). - Apply the Pythagorean theorem twice, or use the three-dimensional distance relationship. - For surface area, add the areas of all six faces.

1. Using \(A\) as the origin, the displacement from \(A\) to \(M\) is \((3, 4, 12)\): half the width, the full depth, and the full height. 2. Apply the Pythagorean theorem in three dimensions: \(AM=\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\,\text{cm}\). 3. The surface area is \(2(ab+bc+ac)=2(6\cdot 4+4\cdot 12+6\cdot 12)=288\,\text{cm}^2\).

a) \(AM=13\,\text{cm}\) b) \(288\,\text{cm}^2\)

- Use the \(30^\circ\) angle in right triangle \(ADC\). - Subtract \(CD\) from the full equal side to find \(BD\). - Find the base angle of the isosceles triangle. - Use right triangle \(ABD\) to form the sine ratio.

1. In right triangle \(ADC\), \(AD=2\sin(30^\circ)=1\) and \(CD=2\cos(30^\circ)=\sqrt{3}\). 2. Since \(BC=2\), \(BD=2-\sqrt{3}\). 3. In right triangle \(ABD\), \(AB^2=AD^2+BD^2=1+(2-\sqrt{3})^2=8-4\sqrt{3}\). 4. Thus, \(AB=2\sqrt{2-\sqrt{3}}=\sqrt{6}-\sqrt{2}\). 5. The base angles of the isosceles triangle are \(75^\circ\). In triangle \(ABD\), \(\sin(75^\circ)=\frac{AD}{AB}=\frac{1}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4}\).

a) \(AD=1\), \(CD=\sqrt{3}\), and \(BD=2-\sqrt{3}\) b) \(AB=\sqrt{6}-\sqrt{2}\) c) \(\sin(75^\circ)=\frac{\sqrt{6}+\sqrt{2}}{4}\)

- Use the base angles of isosceles triangle \(ABD\). - Apply the exterior angle theorem at \(D\). - Use the \(30^\circ\) right triangle to find \(CD\) and \(BD\). - Express \(AC\) as a sum of collinear segments.

1. Since \(AD=BD\), triangle \(ABD\) is isosceles. Exterior angle \(\angle BDC=30^\circ\) equals the sum of the two equal remote interior angles, so each is \(15^\circ\). Therefore, \(\alpha=15^\circ\). 2. In right triangle \(BCD\), \(CD=\frac{1}{\tan(30^\circ)}=\sqrt{3}\) and \(BD=\frac{1}{\sin(30^\circ)}=2\). 3. Since \(AD=2\), \(AC=AD+CD=2+\sqrt{3}\). 4. In right triangle \(ABC\), \(\tan(15^\circ)=\frac{BC}{AC}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\).

a) \(\alpha=15^\circ\) b) \(CD=\sqrt{3}\), \(BD=2\), and \(AC=2+\sqrt{3}\) c) \(\tan(15^\circ)=2-\sqrt{3}\)

- Let \(EC=x\). - Use the side ratios of a \(30^\circ\text{-}60^\circ\text{-}90^\circ\) triangle to express \(BE\). - Find all three angles in triangle \(ABE\) and identify its special property.

1. Let \(EC=x\). Triangle \(BEC\) is a \(30^\circ\text{-}60^\circ\text{-}90^\circ\) triangle, so its hypotenuse is \(BE=2x\). 2. Because \(A\), \(E\), and \(C\) are collinear, \(\angle AEB=180^\circ-60^\circ=120^\circ\). 3. In triangle \(ABE\), \(\angle ABE=180^\circ-120^\circ-30^\circ=30^\circ\). 4. Since \(\angle EAB=\angle ABE=30^\circ\), triangle \(ABE\) is isosceles and \(AE=BE\). 5. Therefore, \(AE=BE=2x=2EC\).

Triangle \(BEC\) gives \(BE=2EC\), and triangle \(ABE\) is isosceles with \(AE=BE\). Therefore, \(AE=2EC\).

- Use the symmetry of the isosceles trapezoid at the intersection of the diagonals. - What special triangles are formed when the diagonals are perpendicular? - Relate the altitude to the hypotenuse in each isosceles right triangle.

1. Let \(S\) be the intersection of the diagonals. By the symmetry of an isosceles trapezoid, \(SA = SB\) and \(SC = SD\). 2. Since the diagonals are perpendicular, \(\triangle ASB\) and \(\triangle CSD\) are isosceles right triangles with hypotenuses \(AB = a\) and \(CD = c\). 3. In an isosceles right triangle, the altitude from the right-angle vertex to the hypotenuse has length one-half the hypotenuse. Therefore, the distance from \(S\) to \(AB\) is \(\frac{a}{2}\), and the distance from \(S\) to \(CD\) is \(\frac{c}{2}\). 4. These distances add to the trapezoid’s height: \(h = \frac{a}{2} + \frac{c}{2} = \frac{a+c}{2}\). 5. Since the trapezoid midsegment has length \(m = \frac{a+c}{2}\), it follows that \(h = m\).

The perpendicular diagonals create two isosceles right triangles whose altitudes are \(\frac{a}{2}\) and \(\frac{c}{2}\). Thus, \(h = \frac{a}{2} + \frac{c}{2} = \frac{a+c}{2} = m\).