Question

The vertices of A THB are shown. B (-5, - 2) H(3, - 4) T(-3, 6) How can A THB be described? Show all work to support your answer.

Answer 1

A THB is an isosceles triangle with two equal sides of 8 units and one different side of 10 units, without a right angle.

Step-by-step explanation:

To describe the figure A THB with vertices B (-5, -2), H(3, -4), and T(-3, 6), we'll first determine the type of triangle by calculating the distances between the vertices to check for any right angles or equal sides.

• The distance between B and H (BH): √[(-5-3)^2 + (-2+4)^2] = √(64) = 8 units.
• The distance between H and T (HT): √[(3+3)^2 + (-4-6)^2] = √(100) = 10 units.
• The distance between T and B (TB): √[(-3 + 5)^2 + (6 + 2)^2] = √(64) = 8 units.

Since TB and BH are both 8 units, A THB is an isosceles triangle with one uneven side (HT).

To determine if A THB has any right angles, we would use the Pythagorean theorem to check if TB^2 + BH^2 = HT^2. Since (8^2) + (8^2) = 128, which is different from HT^2 (100), A THB does not contain a right angle and is therefore an isosceles triangle without a right angle.