Question:easy

The parabola $y = -x^2 + 16$ intersects the $x$-axis at points A and B. Further, the parabola intersects the line $y = 7$ at points M and N. Then the area of the quadrilateral with vertices A, B, M and N is

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Plotting or visualizing the points on a Cartesian plane quickly reveals that the quadrilateral is a symmetric trapezoid.
This allows you to bypass complex polygon area formulas and use the simple trapezoid area formula directly.
Updated On: Jun 16, 2026
  • 49
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The Correct Option is A

Solution and Explanation

Step 1: Picture the shape.
The parabola $y = -x^2 + 16$ opens downward with its top at $(0, 16)$. Points A and B are where it crosses the $x$-axis, and M and N are where it crosses the horizontal line $y = 7$.

Step 2: Find A and B on the $x$-axis.
Set $y = 0$: $-x^2 + 16 = 0$, so $x^2 = 16$ and $x = \pm 4$. So A and B are $(-4, 0)$ and $(4, 0)$.

Step 3: Find M and N on the line $y = 7$.
Set $y = 7$: $-x^2 + 16 = 7$, so $x^2 = 9$ and $x = \pm 3$. So M and N are $(-3, 7)$ and $(3, 7)$.

Step 4: See that the quadrilateral is a trapezium.
The bottom side AB lies on $y = 0$ and the top side MN lies on $y = 7$, so the two are parallel. That makes ABMN a trapezium.

Step 5: Read off the two parallel lengths and the height.
Bottom length $AB = 4 - (-4) = 8$. Top length $MN = 3 - (-3) = 6$. The vertical gap (height) between $y = 0$ and $y = 7$ is $7$.

Step 6: Apply the trapezium area formula.
Area $= \frac{1}{2}(\text{sum of parallel sides}) \times \text{height} = \frac{1}{2}(8 + 6)(7) = \frac{1}{2}(14)(7) = 49$.
\[ \boxed{49} \]
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