Question

Can you help me with this exercise?

Answer 1

Answer:

2√3

Step-by-step explanation:

Suppose ∠B is right angle.

Point D is a midpoint of line AC.

If AC is the hypotenuse, point D is a center of circumscribed circle.

But we know that AD = CD = 2 ≠ BD = 3.

Point D is not a center.

Hence AC is not a hypotenuse.

Hence, such ΔABC doesn't exist.

Suppose ∠A is right angle.

Then, ΔABD with ∠A = 90° forms.

If we apply Pythagorean theorem in the right triangle ΔABD,

→ AB² + AD² (Known) = BD² (Known)

→ AB² = 9 - 1

∴ AB² = 8

Back to ΔABC, apply Pythagorean theorem again.

In the right triangle ΔABC,

→ AB² (Known) + CA² (Known) = BC²

→ BC² = 8 + 4, BC > 0

∴ BC = 2√3

Since BC is the hypotenuse of ΔABC, the answer is 2√3.

Suppose ∠C is right angle.

Then, ΔBCD with ∠C = 90° forms.

If we apply Pythagorean theorem in the right triangle ΔBCD,

→ BC² + CD² (Known) = BD² (Known)

→ BC² = 9 - 1

∴ BC² = 8

Back to ΔABC, apply Pythagorean theorem again.

In the right triangle ΔABC,

→ BC² (Known) + CA² (Known) = AB²

→ AB² = 8 + 4, AB > 0

∴ AB = 2√3

Since AB is the hypotenuse of ΔABC, the answer is 2√3.

In both cases, hypotenuse is 2√3.

Hence, the answer is 2√3.

Please use papers. Understanding will be so hard.

Try to follow the steps in big letters.