Question

find the area of the triangle using the formula $mathcal{a}=\frac{1}{2}bh$, and verify that the formula $mathcal{a}=\frac{1}{2}absin c$ gives the same result.
the area, using $mathcal{a}=\frac{1}{2}bh$, is
(simplify your answer. type an integer or a fraction. type an exact answer, using radicals as needed.)
the area, using $mathcal{a}=\frac{1}{2}absin c$, is
(simplify your answer. type an integer or a fraction. type an exact answer, using radicals as needed.)

Explanation:

Step1: Identify base and height

The triangle is right-angled at C, so base \( b = 2 \), height \( h = 2 \).

Step2: Calculate area with \( A = \frac{1}{2}bh \)

\( A = \frac{1}{2} \times 2 \times 2 = 2 \)

Step3: Identify sides and angle for \( A = \frac{1}{2}ab\sin C \)

\( a = 2 \), \( b = 2 \), \( C = 45^\circ \), \( \sin 45^\circ = \frac{\sqrt{2}}{2} \)

Step4: Calculate area with sine formula

\( A = \frac{1}{2} \times 2 \times 2 \times \frac{\sqrt{2}}{2} = \sqrt{2} \)? Wait, no—wait the right angle is at C, so angle between sides AC and BC is 90°, not 45°. Oh, correct: sides AC=2, BC=2, angle C=90°, \( \sin 90^\circ=1 \). So \( A = \frac{1}{2} \times 2 \times 2 \times 1 = 2 \)

Answer:

2
2