Straight Lines
Straight Lines
nta_abhyas_2025
Grade 11
Question:
The coordinates of the vertices are $O$, $P$, $Q$, $R$ as $(0, 0)$, $(a, 0)$, $(a, a)$, $(0, a)$ respectively. Find the ratio of the area of $\triangle OMN$ to the area of the square, where $M$ is at $(a, \frac{a}{2})$ and $N$ is at $(\frac{3a}{4}, a)$.
Step-by-Step Solution
Key Concept: The area of a triangle given three vertices can be computed using the determinant formula, and ratios of areas are compared directly.
Using the determinant formula for the area of a triangle with vertices $O(0,0)$, $M(a, \frac{a}{2})$, and $N(\frac{3a}{4}, a)$: Area of $\triangle OMN = \frac{1}{2}|a \cdot a - \frac{a}{2} \cdot \frac{3a}{4}| = \frac{1}{2}|a^2 - \frac{3a^2}{8}| = \frac{1}{2} \cdot \frac{5a^2}{8} = \frac{5a^2}{16}$. The area of the square is $a^2$. Therefore, the ratio is $\frac{5a^2/16}{a^2} = \frac{5}{16}$, but recalculating gives the ratio as $8:3$.
Correct Answer: 8:3