Question : What is the area (in sq. units) of the triangle formed by the graphs of the equations $2x + 5y - 12=0, x + y = 3,$ and $y = 0$?
Option 1: 3
Option 2: 2
Option 3: 5
Option 4: 6
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Correct Answer: 3
Solution : According to the question Intersection point (P) of line $2x + 5y - 12 = 0$ and $y = 0$ ⇒ $2x + 5 × 0 - 12 = 0$ ⇒ $x = 6$ The point of intersection is P (6, 0). Intersection point (Q) of line $x + y = 3$ and $y = 0$ ⇒ $x + 0 = 3$ ⇒ $x = 3$ The point of intersection is Q (3, 0). Intersection point (R) of line $2x + 5y - 12 = 0$ and $x + y -3 = 0$ ⇒ $2(3 -y) + 5y - 12 = 0$ ⇒ $y = 2$ and $x = 1$ The point of intersection is R (1, 2). Length of the base = PQ = 6 – 3 = 3 units Height of point R from base = y coordinate of point R = 2 units Area of triangle PQR = $\frac{1}{2}$ × 3 × 2 = 3 units Hence, the correct answer is 3.
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Question : The area of the triangle formed by the graph of the straight lines $x-y=0$, $x+y=2$, and the $x$-axis is:
Option 1: 1 sq. unit
Option 2: 2 sq. units
Option 3: 4 sq. units
Option 4: 5 sq. units
Question : The area of triangle with vertices A(0, 8), O(0, 0), and B(5, 0) is:
Option 1: 8 sq. units
Option 2: 13 sq. units
Option 3: 20 sq. units
Option 4: 40 sq. units
Question : What is the area (in unit squares) of the region enclosed by the graphs of the equations $2x - 3y + 6 = 0, 4x + y = 16$ and $y = 0$?
Option 1: 12
Option 2: 10.5
Option 3: 14
Option 4: 11.5
Question : What is the area (in unit squares) of the triangle enclosed by the graphs of $2 x+5 y=12, x+y=3$ and the x-axis?
Option 1: 2.5
Option 2: 3.5
Option 3: 3
Option 4: 4
Question : The graphs of the equations $4 x+\frac{1}{3} y=\frac{8}{3}$ and $\frac{1}{2} x+\frac{3}{4} y+\frac{5}{2}=0$ intersect at a point P. The point P also lies on the graph of the equation:
Option 1: $x + 2y - 5 = 0$
Option 2: $3x - y - 7 = 0$
Option 3: $x - 3y - 12= 0$
Option 4: $4x - y + 7= 0$
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