Question

A​ large, triangular-shaped table is supported by a single pole at the center of gravity. how far is vertex c from the center of​ gravity?

Answer 1

The distance of vertex (C) from the center of​ gravity (G) is equal to 2 feet.

How to calculate the distance of vertex (C).

Let center of gravity of the given triangular-shaped table be CD, which is also its median or centroid of triangle. Thus, we would find side CD by applying Pythagorean's Theorem.

Mathematically, Pythagorean's Theorem for a right-angled triangle is given by this formula:

$c^2 = a^2 + b^2$

Where:

• c is the hypotenuse.

• a is the adjacent side.

• b is the opposite side.

For this triangular-shaped table, we have;

$AC^2=CD^2+AD^2\\\\5^2=CD^2+4^2\\\\25=CD^2+16\\\\CD^2=25-16\\\\CD^2=9\\\\CD=\sqrt{9}$

CD = 3 feet.

Based on the concurrency of median theorem, we have:

$CG=\frac{2}{3} CD\\\\CG=\frac{2}{3} \times 3$

CG = 2 feet.

Read more on Pythagorean Theorem here: https://brainly.com/question/16176867