The length of the longer base of an isosceles trapezoid is 12 cm. The length of each side is 5 cm. The measure of each acute angle is 37 degrees. Find the length of the shorter base.

Answer 1

To find the length of the shorter base of the isosceles trapezoid, we can first find the height of the trapezoid using trigonometry.

Since the measure of each acute angle is 37 degrees, we can split the trapezoid into two right triangles by drawing a perpendicular line from one of the vertices to the longer base. The acute angle at the top of each right triangle is 37 degrees.

Using trigonometric ratios, we can write:

\[\tan(37^{\circ}) = \frac{opposite side}{adjacent side}\]
\[\tan(37^{\circ}) = \frac{height}{\frac{12-5}{2}}\]
\[\tan(37^{\circ}) = \frac{height}{3.5}\]

Solving for the height, we get:

\[height = 3.5 \times \tan(37^{\circ})\]
\[height \approx 2.23 \text{ cm}\]

Now that we have the height of the trapezoid, we can find the length of the shorter base by using Pythagoras' theorem in one of the right triangles:

\[a^2 + b^2 = c^2\]
\[(5)^2 + (2.23)^2 = short\ base^2\]
\[25 + 4.97 \approx short\ base^2\]
\[short\ base^2 \approx 29.97\]

Therefore, the length of the shorter base is approximately 5.48 cm.