The base and the altitude of a triangle are (3x - 4y) and (6x + 5y) respectively. Find its area.

Given:

Base of a triangle = (3x - 4y)

Altitude of a triangle = (6x + 5y)

As we know that the area of triangle = $\dfrac{1}{2}$ x base x altitude

$\dfrac{1}{2} \times (3x - 4y) \times (6x + 5y)\\[1em] = \dfrac{1}{2} \times (3x (6x + 5y) - 4y (6x + 5y))\\[1em] = \dfrac{1}{2} \times (18x^2 + 15xy - 24xy - 20y^2)\\[1em] = \dfrac{1}{2} \times (18x^2 - 9xy - 20y^2)$

Hence, the area of tirangle = $\dfrac{1}{2}$ (18x² - 9xy - 20y²) sq. unit