Question

17.
45 45 90
triangle with 60°, 45°, right angles, sides x, y, 39, z
x = ______
y = ______
z = ______
18.
triangle with 30°, 45°, right angles, sides x, y, 20, z
x = ______
y = ______
z = ______
19.
3c60 90
triangle with 60°, 45°, right angles, sides x, 6√3, y, z
x = ______
y = ______
z = ______
20.
triangle with 30°, 45°, right angles, sides x, y, z, 10√6
x = ______
y = ______
z = ______

1. find the perimeter of the triangle:

isosceles triangle with 60° angles, height 4√15

1. find the perimeter of the square:

square with diagonal 28

1. steel loading ramps are used to load a lawn mower onto a truckbed 37.5 inches above ground. if the ramps make a 30° angle with the ground, find the length of the ramps in feet.
2. the infield of a baseball field is a square with sides measuring 90 feet. a ball thrown from third to first base is caught in 1.2 seconds. find the speed of the ball in feet per second. round to the nearest tenth.

© gina wilson (all things algebra®, llc), 20

Explanation:

Step1: Analyze Triangle 17 (45-45-90 and 30-60-90)

The right triangle with \(45^\circ\) has leg \(39\), so hypotenuse \(z = 39\sqrt{2}\) (45-45-90: hypotenuse = leg\(\sqrt{2}\)). The other right triangle (30-60-90) has adjacent leg \(y\) and opposite leg equal to the height (same as 39, since 45-45-90 legs are equal). For 30-60-90, \(\tan(60^\circ)=\frac{39}{y}\Rightarrow y = \frac{39}{\sqrt{3}} = 13\sqrt{3}\), hypotenuse \(x = 2y = 26\sqrt{3}\)? Wait, no: Wait, the height is 39? Wait, 45-45-90 triangle: legs are equal, so height = 39. Then 30-60-90 triangle: angle \(60^\circ\), opposite side 39, adjacent \(y\), hypotenuse \(x\). \(\tan(60^\circ)=\frac{39}{y}\Rightarrow y=\frac{39}{\sqrt{3}} = 13\sqrt{3}\). \(\sin(60^\circ)=\frac{39}{x}\Rightarrow x=\frac{39}{\sin(60^\circ)}=\frac{39}{\frac{\sqrt{3}}{2}} = 26\sqrt{3}\). Wait, no, 30-60-90: sides are \(y\) (short leg), \(39\) (long leg, opposite 60°), \(x\) (hypotenuse). So long leg = short leg \(\sqrt{3}\), so \(y=\frac{39}{\sqrt{3}} = 13\sqrt{3}\), hypotenuse \(x = 2y = 26\sqrt{3}\)? Wait, no, 30-60-90: short leg (opposite 30°) is \(y\), long leg (opposite 60°) is \(y\sqrt{3}\). So if long leg is 39, then \(y\sqrt{3}=39\Rightarrow y = 13\sqrt{3}\), hypotenuse \(x = 2y = 26\sqrt{3}\). And \(z\) is hypotenuse of 45-45-90 with leg 39, so \(z = 39\sqrt{2}\).

Step2: Triangle 18 (30-60-90 and 45-45-90)

The 45-45-90 triangle has hypotenuse 20? Wait, no, the triangle with 45°: it's a right triangle with angle 45°, so legs \(y\) and the other leg (same as \(y\)). The hypotenuse of 45-45-90 is \(20\)? Wait, no, the side labeled 20 is hypotenuse of 45-45-90? Wait, no, the triangle with 45°: legs \(y\) and the vertical leg, hypotenuse 20? Wait, 45-45-90: legs \(y\), hypotenuse \(y\sqrt{2}=20\Rightarrow y = \frac{20}{\sqrt{2}} = 10\sqrt{2}\). Then the 30-60-90 triangle: angle 30°, opposite side \(y = 10\sqrt{2}\), hypotenuse \(x\), adjacent \(z\). \(\sin(30^\circ)=\frac{y}{x}\Rightarrow x = 2y = 20\sqrt{2}\). \(\cos(30^\circ)=\frac{z}{x}\Rightarrow z = x\cos(30^\circ)=20\sqrt{2}\cdot\frac{\sqrt{3}}{2}=10\sqrt{6}\). Wait, no, maybe I mixed up. Wait, the 45-45-90 triangle: legs are \(y\) (vertical) and the horizontal leg (same as \(y\)), hypotenuse 20? No, the side labeled 20 is the hypotenuse of the 45-45-90? Wait, no, the triangle with 45°: right angle, 45°, so legs equal, hypotenuse 20? Then \(y = \frac{20}{\sqrt{2}} = 10\sqrt{2}\). Then the 30-60-90 triangle: angle 30°, opposite side \(y = 10\sqrt{2}\), so hypotenuse \(x = 2y = 20\sqrt{2}\), adjacent \(z = y\sqrt{3}=10\sqrt{6}\).

Step3: Triangle 19 (30-60-90 and 45-45-90)

The 30-60-90 triangle has base \(6\sqrt{3}\) (long leg, opposite 60°), so short leg \(x\) (opposite 30°): \(x\sqrt{3}=6\sqrt{3}\Rightarrow x = 6\). Hypotenuse of 30-60-90 is \(2x = 12\), which is the leg of 45-45-90 triangle. So 45-45-90 triangle: leg 12, so \(y = 12\) (other leg), hypotenuse \(z = 12\sqrt{2}\).

Step4: Triangle 20 (30-60-90 and 45-45-90)

The 45-45-90 triangle has leg \(10\sqrt{6}\)? Wait, no, the side \(10\sqrt{6}\) is leg of 45-45-90? Wait, 45-45-90: legs equal, so the vertical leg (height) is \(10\sqrt{6}\)? No, the triangle with 45°: it's a right triangle with angle 45°, so legs \(z\) (horizontal) and \(10\sqrt{6}\) (vertical)? Wait, no, the side \(10\sqrt{6}\) is hypotenuse? No, 45-45-90: legs \(z\) and \(z\), hypotenuse \(z\sqrt{2}=10\sqrt{6}\Rightarrow z = \frac{10\sqrt{6}}{\sqrt{2}} = 10\sqrt{3}\). Then the 30-60-90 triangle: angle 30°, opposite side \(z = 10\sqrt{3}\), so hypotenuse \(x = 2z = 20\sqrt{3}\), adjacent \(y = z\sqrt{3}=10\sqrt{3}\cdot\sqrt{3}=30\).

Step5: Triangle 21 (Equilateral Triangle)

The triangle has two 60° angles, so it's equilateral. The height is \(4\sqrt{15}\), which in an equilateral triangle is \(\frac{\sqrt{3}}{2} \times\) side. Let side be \(s\), then \(\frac{\sqrt{3}}{2}s = 4\sqrt{15}\Rightarrow s = \frac{8\sqrt{15}}{\sqrt{3}} = 8\sqrt{5}\). Perimeter = \(3s = 24\sqrt{5}\).

Step6: Square 22 (Diagonal 28)

Diagonal of square \(d = s\sqrt{2}\), so \(s = \frac{28}{\sqrt{2}} = 14\sqrt{2}\). Perimeter = \(4s = 56\sqrt{2}\approx 79.2\).

Step7: Ramp 23 (30-60-90)

Height = 37.5 inches = \(\frac{37.5}{12}\) feet = 3.125 feet. In 30-60-90, \(\sin(30^\circ)=\frac{\text{height}}{\text{ramp length}}\Rightarrow \text{ramp length} = 2 \times 3.125 = 6.25\) feet. Wait, 37.5 inches is 3.125 feet. So \(\sin(30^\circ)=0.5=\frac{3.125}{L}\Rightarrow L = 6.25\) feet.

Step8: Baseball Infield 24 (Square Diagonal)

Square side 90 feet, diagonal \(d = 90\sqrt{2}\) feet. Speed = \(\frac{\text{distance}}{\text{time}}=\frac{90\sqrt{2}}{1.2}\approx \frac{127.279}{1.2}\approx 106.1\) ft/s.

Answer:

1. \(x = 26\sqrt{3}\), \(y = 13\sqrt{3}\), \(z = 39\sqrt{2}\)
2. \(x = 20\sqrt{2}\), \(y = 10\sqrt{2}\), \(z = 10\sqrt{6}\)
3. \(x = 6\), \(y = 12\), \(z = 12\sqrt{2}\)
4. \(x = 20\sqrt{3}\), \(y = 30\), \(z = 10\sqrt{3}\)
5. Perimeter: \(24\sqrt{5}\)
6. Perimeter: \(56\sqrt{2}\) (or ~79.2)
7. Ramp length: \(6.25\) feet
8. Speed: ~\(106.1\) ft/s