Question

for the right triangles below, find the exact values of the side lengths d and h. if necessary, write your responses in simplified radical form. d = h =

Explanation:

Step1: Find \( d \) (45-45-90 triangle)

In a 45-45-90 right triangle, the legs are equal, and the hypotenuse \( c = a\sqrt{2} \) (where \( a \) is a leg). Here, hypotenuse is 7, and leg \( d \) satisfies \( \sin(45^\circ)=\frac{d}{7} \) or since it's isoceles, legs are equal. Using \( \sin(45^\circ)=\frac{\sqrt{2}}{2} \), so \( d = 7\sin(45^\circ)=7\times\frac{\sqrt{2}}{2}=\frac{7\sqrt{2}}{2} \)? Wait, no—wait, 45-45-90: legs are equal, hypotenuse \( = \) leg \( \times\sqrt{2} \). So if hypotenuse is 7, then leg \( d = \frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2} \)? Wait, no, wait the triangle: the left triangle has angles 45, 45, 90, hypotenuse 7, and leg \( d \). So in 45-45-90, legs are equal, hypotenuse \( = \) leg \( \times\sqrt{2} \). So \( \text{leg} = \frac{\text{hypotenuse}}{\sqrt{2}} \). So \( d = \frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2} \)? Wait, no, maybe I mixed up. Wait, the left triangle: angle 45, 45, 90, so it's an isoceles right triangle, so the two legs (one is \( d \), the other is the vertical leg) are equal, and hypotenuse is 7. So by Pythagoras: \( d^2 + d^2 = 7^2 \Rightarrow 2d^2 = 49 \Rightarrow d^2 = \frac{49}{2} \Rightarrow d = \frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2} \). Wait, but maybe the hypotenuse is 7, so leg \( d = 7\cos(45^\circ)=7\times\frac{\sqrt{2}}{2}=\frac{7\sqrt{2}}{2} \). Correct.

Step2: Find \( h \) (30-60-90 triangle)

In a 30-60-90 right triangle, the sides are in ratio \( 1 : \sqrt{3} : 2 \) (opposite 30°, 60°, 90°). The side opposite 30° is the shortest leg (here, the horizontal leg is 6, opposite 30°? Wait, the right triangle on the right: angles 30°, 60°, 90°. The horizontal leg is 6 (adjacent to 60° or opposite 30°?). Wait, angle 30° is at the top, 60° at the bottom left, right angle at bottom right. So the side opposite 30° is the horizontal leg (length 6), the side opposite 60° is \( h \) (vertical leg), and hypotenuse is the other side. In 30-60-90, \( \text{opposite 30°} = \frac{1}{2}\text{hypotenuse} \), \( \text{opposite 60°} = \text{opposite 30°} \times\sqrt{3} \). So if the side opposite 30° is 6, then \( h \) (opposite 60°) is \( 6\sqrt{3} \). Wait, let's check: \( \tan(60^\circ)=\frac{h}{6} \), and \( \tan(60^\circ)=\sqrt{3} \), so \( h = 6\tan(60^\circ)=6\sqrt{3} \). Alternatively, \( \cos(30^\circ)=\frac{6}{\text{hypotenuse}} \), but we need \( h \). So \( \tan(60^\circ)=\sqrt{3}=\frac{h}{6} \Rightarrow h = 6\sqrt{3} \).

Wait, but wait the first triangle: 45-45-90, hypotenuse 7, so legs are \( \frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2} \). Wait, but maybe I made a mistake. Wait, the left triangle: angle 45, 45, 90, so it's isoceles, so legs are equal. So if hypotenuse is 7, then leg \( d = 7\sin(45^\circ)=7\times\frac{\sqrt{2}}{2}=\frac{7\sqrt{2}}{2} \). Correct.

Step1 (corrected for \( d \)):

In 45-45-90 triangle, legs are equal, hypotenuse \( = \) leg \( \times\sqrt{2} \). So leg \( d = \frac{\text{hypotenuse}}{\sqrt{2}} = \frac{7}{\sqrt{2}} = \frac{7\sqrt{2}}{2} \). Wait, but maybe the hypotenuse is 7, so leg \( d = 7\cos(45^\circ) = 7\times\frac{\sqrt{2}}{2} = \frac{7\sqrt{2}}{2} \). Correct.

Step2 (for \( h \)):

In 30-60-90 triangle, the side opposite 30° is the shorter leg (length 6), so the side opposite 60° ( \( h \)) is \( 6\sqrt{3} \) (since \( \tan(60^\circ) = \sqrt{3} = \frac{h}{6} \Rightarrow h = 6\sqrt{3} \)).

Wait, but let's recheck:

Left triangle (45-45-90):

- Angles: 45°, 45°, 90°
- Hypotenuse: 7
- Legs: \( d \) (horizontal) and vertical leg (equal to \( d \))
- By Pythagoras: \( d^2 + d^2 = 7^2 \Rightarrow 2d^2 = 49 \Rightarrow d^2 = \frac{49}{2} \Rightarrow d = \frac{7}{\sqrt{2}} = \frac{7\sqrt{2}}{2} \) (rationalized).

Right triangle (30-60-90):

- Angles: 30°, 60°, 90°
- Shorter leg (opposite 30°): 6 (horizontal leg)
- Longer leg (opposite 60°): \( h \)
- In 30-60-90, longer leg = shorter leg \( \times\sqrt{3} \), so \( h = 6\sqrt{3} \).

Wait, but maybe the left triangle's hypotenuse is 7, and \( d \) is a leg, so \( \sin(45^\circ) = \frac{d}{7} \Rightarrow d = 7\sin(45^\circ) = 7\times\frac{\sqrt{2}}{2} = \frac{7\sqrt{2}}{2} \). Correct.

Answer:

\( d = \frac{7\sqrt{2}}{2} \)
\( h = 6\sqrt{3} \)