QUESTION IMAGE
Question
special right triangles
complete the tables for the side lengths based on the given leg length(s).
45°–45°–90° △
| leg | leg | hypotenuse |
|---|
30°–60°–90° △
| short leg | long leg | hypotenuse |
|---|
(then two triangle diagrams: one 45-45-90 with labels x, x√2, and one 30-60-90 with labels y, y√3)
Step1: Solve 45°-45°-90° Triangle Table
In a 45°-45°-90° triangle, the hypotenuse \( c \) is related to the legs \( a = b = 1 \) by the Pythagorean theorem \( c=\sqrt{a^{2}+b^{2}} \). Substituting \( a = 1 \), \( b = 1 \), we get \( c=\sqrt{1^{2}+1^{2}}=\sqrt{2} \). So the hypotenuse is \( \sqrt{2} \), filling the blank with 2.
Step2: Solve 30°-60°-90° Triangle Table (Long Leg)
In a 30°-60°-90° triangle, the long leg (opposite 60°) is \( \sqrt{3} \) times the short leg (opposite 30°). Given short leg \( = 1 \), long leg \( = 1\times\sqrt{3}=\sqrt{3} \), so the blank under long leg is 3.
Step3: Solve 30°-60°-90° Triangle Table (Hypotenuse)
In a 30°-60°-90° triangle, the hypotenuse is twice the short leg. Given short leg \( = 1 \), hypotenuse \( = 2\times1 = 2 \), so the hypotenuse blank is 2.
Step4: 45°-45°-90° Triangle (Side AC)
In \( \triangle ABC \), it's a 45°-45°-90° triangle, so legs \( BC = AC = x \) (since angles at B and A are 45°, legs are equal).
Step5: 30°-60°-90° Triangle (Side GE)
In \( \triangle GFE \), it's a 30°-60°-90° triangle. The hypotenuse (opposite right angle) is twice the short leg. Short leg is \( y \) (opposite 30° at E), so hypotenuse \( GE = 2y \).
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- 45°-45°-90° Table Hypotenuse: \( \sqrt{2} \) (blank filled with 2)
- 30°-60°-90° Table Long Leg: \( \sqrt{3} \) (blank filled with 3)
- 30°-60°-90° Table Hypotenuse: \( 2 \)
- \( \triangle ABC \) side \( AC \): \( x \)
- \( \triangle GFE \) side \( GE \): \( 2y \)