Q.Find the area of the triangle whose vertices are $(-1, 1)$, $(0, 5)$ and $(3, 2)$, using integration.

Showing the 12 most recent of 18 on this concept.

1. CBSE 2026Set 65/3/11 markMCQ
   Q.The area of the shaded region of the circle given below (see figure) is equal to: (A) $\int_{1}^{3} \sqrt{9-y^2}\, dy$ (B) $2 \int_{1}^{3} \sqrt{9-y^2}\, dy$ (C) $\int_{0}^{3} \sqrt{9-x^2}\, dx$ (D) $2 \int_{0}^{3} \sqrt{9-x^2}\, dx$

   ›Reveal solutionSolution

   The problem asks for the integral representing the area of a shaded region of a circle. Assuming the shaded region is the quarter circle in the first quadrant of $x^2+y^2=9$, its area is given by $\boxed{\int_{0}^{3} \sqrt{9-x^2}\, dx}$.

   The core concept here is using definite integrals to calculate the area under a curve. When we have a region bounded by a curve $y=f(x)$, the x-axis, and vertical lines $x=a$ and $x=b$, the area is given by $\int_a^b f(x)\, dx$. Similarly, if the region is bounded by a curve $x=g(y)$, the y-axis, and horizontal lines $y=c$ and $y=d$, the area is $\int_c^d g(y)\, dy$.

   The expressions in the options, $\sqrt{9-x^2}$ and $\sqrt{9-y^2}$, immediately point to the equation of a circle. The general equation of a circle centered at the origin with radius $r$ is $x^2+y^2=r^2$. Comparing this with $9-x^2$ or $9-y^2$, we see that $r^2=9$, which means the radius $r=3$.

   For the upper half of this circle, we can express $y$ as a function of $x$: $y^2 = 9-x^2 \implies y = \sqrt{9-x^2}$ (taking the positive root for the upper half).

   For the right half of this circle, we can express $x$ as a function of $y$: $x^2 = 9-y^2 \implies x = \sqrt{9-y^2}$ (taking the positive root for the right half).

   Since the figure is not provided, we must infer the shaded region from the given options. Options (C) and (D) involve integration from $0$ to $3$, which is the radius of the circle. This strongly suggests that the shaded region is either a quarter circle or a semi-circle.

   1. Identify the circle's equation and radius:

      The terms $\sqrt{9-x^2}$ and $\sqrt{9-y^2}$ indicate that the circle has the equation $x^2+y^2=9$. This is a circle centered at the origin $(0,0)$ with a radius $r=3$.

   2. Interpret the shaded region based on options:

      • Option (C) is $\int_{0}^{3} \sqrt{9-x^2}\, dx$. This integral represents the area under the curve $y=\sqrt{9-x^2}$ (the upper semi-circle) from $x=0$ to $x=3$. This region is precisely the quarter circle located in the first quadrant.
      • Option (D) is $2 \int_{0}^{3} \sqrt{9-x^2}\, dx$. This would be twice the area of the quarter circle, meaning it represents the area of the entire upper semi-circle (from $x=-3$ to $x=3$, or by symmetry, $2 \times$ area from $x=0$ to $x=3$).
      • Option (A) is $\int_{1}^{3} \sqrt{9-y^2}\, dy$. This represents the area under the curve $x=\sqrt{9-y^2}$ (the right semi-circle) from $y=1$ to $y=3$. This is a specific segment of the quarter circle, not the entire quarter circle.
      • Option (B) is $2 \int_{1}^{3} \sqrt{9-y^2}\, dy$. This would be twice the area in (A), representing a horizontal strip of the circle symmetric about the y-axis. …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Write the area of the curve $y=\sin x$ between $x=0$ and $x=\pi$ in sq units.

   (a) 3

   (b) 4

   (c) 2

   (d) 5

   ›Reveal solutionSolution

   The area under one arch of $y=\sin x$ from $x=0$ to $x=\pi$ is $2$ square units.

   Since $\sin x\ge0$ throughout $[0,\pi]$, the area is simply the definite integral:

   $A=\int_0^{\pi}\sin x\,dx=\big[-\cos x\big]_0^{\pi}$

   …

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3. CBSE 2026Set ANNUAL1 mark
   Q.Find the area of the circle $x^2 + y^2 = a^2$.

   ›Reveal solutionSolution

   By symmetry, the area of the full circle is 4 times the area in the first quadrant, which is found by integration.

   …

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4. CBSE 2025Set 65/4/11 markMCQ
   Q.The area of the region enclosed by the curve $y = \sqrt{x}$ and the lines $x = 0$ and $x = 4$ and x-axis is : (A) $\frac{16}{9}$ sq. units (B) $\frac{32}{9}$ sq. units (C) $\frac{16}{3}$ sq. units (D) $\frac{32}{3}$ sq. units

   ›Reveal solutionSolution

   The region is the area under $y = \sqrt{x}$ from $x = 0$ to $x = 4$, which is a standard definite integral. The area equals $\frac{16}{3}$ square units, so the correct option is (C).

   The problem asks for the area enclosed by the curve $y = \sqrt{x}$, the vertical lines $x = 0$ and $x = 4$, and the x-axis. This is a classic "area under a curve" problem — the region is bounded above by the curve, below by the x-axis, and on the sides by two vertical lines. The key idea is that the area between a curve $y = f(x)$ and the x-axis from $x = a$ to $x = b$ is given by the definite integral $\int_a^b f(x) \, dx$, provided $f(x) \ge 0$ on that interval. Here, $\sqrt{x}$ is non-negative for $x \ge 0$, so we can directly integrate.

   Watch out

   A common mistake is to confuse the area under $y = \sqrt{x}$ with the area under $y = x^2$ or to misapply the power rule. Always check the exponent: $\sqrt{x} = x^{1/2}$, not $x^2$.

   Let’s work through the calculation step by step.

   1. Set up the integral. The region is bounded by $x = 0$ on the left and $x = 4$ on the right. The curve is $y = \sqrt{x}$, and the lower boundary is the x-axis ($y = 0$). So the area $A$ is:

   $A = \int_{0}^{4} \sqrt{x} \, dx$

   2. Rewrite the integrand. Recall that $\sqrt{x} = x^{1/2}$. This makes the power rule for integration straightforward:

   $A = \int_{0}^{4} x^{1/2} \, dx$

   3. Apply the power rule. …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.What is the area of the region bounded by $y=e^{x}$, $X$-axis, $x=1$ and $x=3$ in square unit?

   (i) $e(e^{2}-1)$

   (ii) $e^{3}-1$

   (iii) $e(1-e^{2})$

   (iv) $e^{2}(1-e)$

   ›Reveal solutionSolution

   Area under $y=e^x$ from $x=1$ to $x=3$ is $\displaystyle\int_1^3 e^x\,dx$.

   Since $y=e^x > 0$ throughout $[1,3]$, the region between the curve and the $X$-axis has area: …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.The area bounded by $x$-axis, $y$-axis, $y=\cos x$, $0\le x\le \dfrac{\pi}{2}$ will be -

   (a) $1$

   (b) $0$

   (c) $-1$

   (d) $2$

   ›Reveal solutionSolution

   The required area is $\int_0^{\pi/2}\cos x\,dx$.

   …

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7. CBSE 2025Set ANNUAL1 mark
   Q.Find the area lying in the first quadrant and bounded by the circle $x^2+y^2=4$.

   ›Reveal solutionSolution

   The full circle $x^2+y^2=4$ has radius $2$; the first-quadrant portion is one quarter of the full circle.

   Full circle area $=\pi r^2=\pi(2)^2=4\pi$.

   Area in the first quadrant $=\dfrac{1}{4}\times4\pi=\pi$.

   …

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8. CBSE 2025Set ANNUAL1 mark
   Q.If area of triangle is 35 sq. units with vertices $(2, -6)$, $(5, 4)$ and $(k, 4)$, then $k$ is ______ .

   ›Reveal solutionSolution

   Using the determinant formula for the area of a triangle with the given vertices and setting it to 35 gives two valid values of $k$.

   For vertices $(x_1,y_1)=(2,-6)$, $(x_2,y_2)=(5,4)$, $(x_3,y_3)=(k,4)$, the area is

   $\text{Area} = \frac12\big|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big|$

   $= \frac12\big|2(4-4)+5(4-(-6))+k(-6-4)\big| = \frac12|0+50-10k| = \frac12|50-10k|$ …

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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The area enclosed by circle $x^2 + y^2 = 2$ is equal to:

   (a) $4\pi$ sq. units

   (b) $2\sqrt{2}\pi$ sq. units

   (c) $4\pi^2$ sq. units

   (d) $2\pi$ sq. units

   ›Reveal solutionSolution

   $2\pi$ sq. units — option (d).

   The circle $x^2+y^2=2$ has radius $R=\sqrt2$ (comparing with $x^2+y^2=R^2$).

   …

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10. CBSE 2023Set 65/2/11 markMCQ
    Q.If $(a, b)$, $(c, d)$ and $(e, f)$ are the vertices of $\triangle ABC$ and $\Delta$ denotes the area of $\triangle ABC$, then $\begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}^2$ is equal to:

    (a) $2\Delta^2$

    (b) $4\Delta^2$

    (c) $2\Delta$

    (d) $4\Delta$

    ›Reveal solutionSolution

    The area of a triangle can be expressed using a determinant of its vertices. The given expression is the square of a determinant which is the transpose of the one used in the area formula, leading to a result of $\mathbf{4\Delta^2}$.

    Concept and Intuition

    The area of a triangle whose vertices are given by coordinates is a fundamental concept in coordinate geometry. While you might be familiar with the base-height formula or Heron's formula, when coordinates are involved, a powerful tool is the determinant.

    The determinant method for calculating the area of a triangle arises from vector geometry. If we consider two vectors forming two sides of a triangle, say $\vec{AB}$ and $\vec{AC}$, then the area of the triangle is half the magnitude of their cross product, i.e., $\frac{1}{2} |\vec{AB} \times \vec{AC}|$. When these vectors are expressed in coordinates, this cross product magnitude simplifies to a determinant.

    Alternatively, you can think of it as a generalization of the "shoelace formula" for polygon areas. The determinant essentially calculates a signed area, where the sign depends on the order of vertices (clockwise or counter-clockwise). Since area is always positive, we take the absolute value of the determinant.

    The area $\Delta$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

    $\Delta = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$

    The absolute value bars are crucial because the determinant itself can be negative, but area must be positive.

    Step-by-Step Solution

    1. Identify the vertices and the standard area formula: The vertices of $\triangle ABC$ are given as $(a, b)$, $(c, d)$, and $(e, f)$. Using the determinant formula for the area of a triangle, we can write:

    $\Delta = \frac{1}{2} \left| \begin{vmatrix} a & b & 1 \\ c & d & 1 \\ e & f & 1 \end{vmatrix} \right|$

    2. Isolate the determinant from the area formula: From the formula above, we can multiply both sides by 2:

    $2\Delta = \left| \begin{vmatrix} a & b & 1 \\ c & d & 1 \\ e & f & 1 \end{vmatrix} \right|$

    Let's denote the determinant inside the absolute value as $D$:
    

    $D = \begin{vmatrix} a & b & 1 \\ c & d & 1 \\ e & f & 1 \end{vmatrix}$

    So, we have $2\Delta = |D|$.
    

    3. Consider the given expression:

    We need to evaluate $\begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}^2$.

    Let's call the determinant in this expression $D'$.

    $D' = \begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}$

    4. Relate $D'$ to $D$ using determinant properties: A fundamental property of determinants states that the determinant of a matrix is equal to the determinant of its transpose. That is, $\det(A) = \det(A^T)$. If we compare $D$ and $D'$, we can see that $D'$ is the transpose of $D$. …

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11. CBSE 2023Set 65/3/11 markMCQ
    Q.Let $A$ be the area of a triangle having vertices $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$. Which of the following is correct ?

    (a) $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm A$

    (b) $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm 2A$

    (c) $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm \frac{A}{2}$

    (d) $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2 = A^2$

    ›Reveal solutionSolution

    The area of a triangle $A$ with given vertices is half the absolute value of a specific $3 \times 3$ determinant. This means the determinant itself is equal to $\pm 2A$.

    The area of a triangle in coordinate geometry is a fundamental concept. While you might be familiar with the base-height formula, when the vertices are given as coordinates, a more direct formula exists. This formula can be elegantly expressed using a determinant, which is what this question explores.

    The core idea is that a determinant involving the coordinates of the vertices provides a value that is directly proportional to the area of the triangle. The sign of this determinant tells us about the orientation of the vertices (whether they are listed in a clockwise or counter-clockwise order), while its absolute value gives twice the area. Since area is always a positive quantity, we take the absolute value of the determinant expression.

    1. Recall the Area Formula for a Triangle with Given Vertices The area $A$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

    $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

    The absolute value is crucial here because area must be non-negative. The expression inside the absolute value can be positive or negative depending on the order in which the vertices are taken.
    
    > [!FORMULA]
    > The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is:
    > $$A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$
    

    2. Define the Determinant in Question

    Let's consider the determinant given in the options:

    $D = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$

    3. Expand the Determinant We expand this $3 \times 3$ determinant along the first row:

    $D = x_1 \begin{vmatrix} y_2 & 1 \\ y_3 & 1 \end{vmatrix} - y_1 \begin{vmatrix} x_2 & 1 \\ x_3 & 1 \end{vmatrix} + 1 \begin{vmatrix} x_2 & y_2 \\ x_3 & y_3 \end{vmatrix}$

    Now, evaluate the $2 \times 2$ determinants:
    

    $D = x_1(y_2 \cdot 1 - 1 \cdot y_3) - y_1(x_2 \cdot 1 - 1 \cdot x_3) + 1(x_2 y_3 - y_2 x_3)$

    $D = x_1(y_2 - y_3) - y_1(x_2 - x_3) + (x_2 y_3 - x_3 y_2)$

    Rearranging the terms to match the area formula's structure:
    

    $D = x_1(y_2 - y_3) + x_2 y_3 - x_2 y_1 + x_3 y_1 - x_3 y_2$

    This can be rewritten as:
    

    $D = x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$

    Notice that this is exactly the expression inside the absolute value in the area formula from Step 1.
    

    4. Relate the Determinant to the Area

    From Step 1, we have $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.

    From Step 3, we found that $D = x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$.

    Therefore, we can write:

    $A = \frac{1}{2} |D|$

    Multiplying both sides by 2, we get: …
    

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12. CBSE 2022Set ANNUAL1 markMCQ
    Q.Write the area of the region bounded by $y=x$, $X$-axis, $x=1$ and $x=3$.

    (a) 8 sq. units

    (b) 4 sq. units

    (c) 2 sq. units

    (d) 1 sq. unit

    ›Reveal solutionSolution

    The area under a straight line $y=x$ between two vertical lines is a definite integral, here it also equals the area of a trapezium.

    …

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