Using integration, the area (in sq. units) of the triangle with vertices (1, 1), (3, 5) and (5, 2) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 7
Let A(1, 1), B(3, 5) and C(5, 2). The equations of the sides are
AB: \(y = 2x-1\)
\( \displaystyle \text{BC: } y = \frac{19-3x}{2} \)
\( \displaystyle \text{AC: } y = \frac{x+3}{4} \)
The side AC is the bottom of the triangle for all x from 1 to 5. The top is AB up to x = 3 and BC after it, so the integral is split at x = 3.
\( \displaystyle \int_{1}^{3}\left(2x-1-\frac{x+3}{4}\right)dx = \int_{1}^{3}\frac{7(x-1)}{4}\,dx \)
\( \displaystyle = \frac{7}{4}\left[\frac{(x-1)^{2}}{2}\right]_{1}^{3} = \frac{7}{4}\times2 = \frac{7}{2} \)
\( \displaystyle \int_{3}^{5}\left(\frac{19-3x}{2}-\frac{x+3}{4}\right)dx = \int_{3}^{5}\frac{7(5-x)}{4}\,dx \)
\( \displaystyle = \frac{7}{4}\left[-\frac{(5-x)^{2}}{2}\right]_{3}^{5} = \frac{7}{4}\times2 = \frac{7}{2} \)
\( \displaystyle A = \frac{7}{2}+\frac{7}{2} = 7 \)