Q.Find the area enclosed by the circle $x^2 + y^2 = a^2$

Tangent at $P(5,2)$ with slope $\dfrac{7}{2}$: $\ y-2=\dfrac{7}{2}(x-5)$. Its $x$-intercept ($y=0$):

Normal at $P$ has slope $-\dfrac{2}{7}$: $\ y-2=-\dfrac{2}{7}(x-5)$. Its $x$-intercept:

The tangent, normal and $x$-axis form a triangle with base $=12-\dfrac{31}{7}=\dfrac{53}{7}$ and height $=$ ordinate of $P=2$:

The key idea here is that area bounded by a curve and the x-axis is always positive — it’s the geometric area, not the signed area. When a function like $y = x^3$ takes negative values over part of the interval, the definite integral gives a negative contribution, which we must flip to positive by taking its absolute value.

The curve $y = x^3$ passes through the origin. For $x < 0$, $x^3$ is negative, so the curve lies below the x-axis. For $x > 0$, it lies above. The x-axis itself is the line $y=0$. The region is bounded vertically by the curve and the x-axis, and horizontally by the vertical lines $x = -2$ and $x = 4$.

So the total area is the sum of two separate pieces: the area from $x = -2$ to $x = 0$ (where the curve is below the axis) and the area from $x = 0$ to $x = 4$ (where it is above). We compute each as a definite integral of $|x^3|$, which is equivalent to taking the absolute value of the integral over each subinterval.

1. Area from $x = -2$ to $x = 0$ On $[-2, 0]$, $x^3 \le 0$, so $|x^3| = -x^3$.

Compute the integral:

So

2. Area from $x = 0$ to $x = 4$ On $[0, 4]$, $x^3 \ge 0$, so $|x^3| = x^3$.

3. Total area

Setting up

Because each curve has the form $y^2 = (\text{linear in }x)$, solve for $x$:

The region is symmetric about the $x$-axis, so integrating in $y$ is natural.

Intersection points

Set the two $x$-values equal:

Hence $y^2 = 20$, so $y = \pm 2\sqrt5$.

Horizontal width

For $y$ between the intersections the right curve is $x = 2 - \tfrac{y^2}{5}$ (at $y=0$ it gives $x=2$ versus $x=-7$). The gap is

Integrating

Evaluate the inner integral:

So the inner integral is $18\sqrt5 - 6\sqrt5 = 12\sqrt5$, and

Concept & Intuition

We are finding the area enclosed between a parabola and a line. The key is to rewrite both curves as functions of $x$ (or $y$), find where they intersect (these become the limits of integration), and then integrate the vertical (or horizontal) distance between them. Because the parabola opens upward and the line slopes downward, the region is lens-shaped; the area will scale with $a^2$, and we just need the constant factor $k$.

1. Rewrite the equations in terms of $y$ as functions of $x$.

   The parabola: $ay = x^2 \Rightarrow y = \frac{x^2}{a}$.

   The line: $x + y = 2a \Rightarrow y = 2a - x$.

2. Find the intersection points by setting the two expressions for $y$ equal:

Multiply through by $a$:

Solve the quadratic:

So $x = a$ or $x = -2a$.

The corresponding $y$-values: for $x = a$, $y = 2a - a = a$; for $x = -2a$, $y = 2a - (-2a) = 4a$.

Intersection points: $(-2a, 4a)$ and $(a, a)$.

3. Determine which curve is on top between $x = -2a$ and $x = a$.

   Test a point, say $x = 0$:

   Parabola: $y = 0$. Line: $y = 2a$.

   Since $2a > 0$ (assuming $a>0$ for a positive area), the line lies above the parabola in this interval.

4. Set up the area integral (vertical slices):

5. Evaluate the integral:

Compute at $x = a$:

Compute at $x = -2a$:

Subtract (upper limit minus lower limit):

6. Identify $k$: The area is $\frac{9}{2}a^2$, so $k = \frac{9}{2}$.

The region runs from the $Y$-axis ($x=0$) to $x=-2$. Find where $y=2-x-3x^2$ meets the $X$-axis:

Only $x=-1$ lies in $[-2,0]$. On $(-1,0)$ the curve is above the axis (at $x=0,\ y=2>0$); on $(-2,-1)$ it is below (at $x=-1.5,\ y=-3.25<0$).

Antiderivative $F(x)=2x-\dfrac{x^2}{2}-x^3$:

Area $=\left|\tfrac32\right|+\left|-\tfrac72\right|=\tfrac32+\tfrac72=5.$

The key insight is that the absolute value makes every lobe of the sine wave positive, so we are summing the areas of identical “humps.” Since $\sin 2x$ completes two full oscillations in $[0, 2\pi]$, taking absolute value doubles the number of humps to four, each of equal area.

1. Understand the basic shape

   The function $\sin 2x$ has period $\pi$ (because period of $\sin(kx)$ is $2\pi/k$, so $2\pi/2 = \pi$). Over $[0, 2\pi]$, it completes two full cycles. Without absolute value, the net signed area would be zero because positive and negative lobes cancel. But $|\sin 2x|$ flips the negative parts upward, so we are really measuring the total area of all lobes.

2. Find the period of $|\sin 2x|$

   The absolute value halves the period: $\sin 2x$ is negative on half of each cycle, and flipping it makes the pattern repeat every half-cycle. Specifically, $|\sin 2x|$ has period $\pi/2$ (since $\sin 2x$ changes sign at multiples of $\pi/2$). Over $[0, 2\pi]$, there are $2\pi \div (\pi/2) = 4$ identical periods.

3. Compute area of one period

   Take one period, say $x \in [0, \pi/2]$. Here $\sin 2x \ge 0$, so $|\sin 2x| = \sin 2x$. The area under one hump is

So each hump has area $1$.

4. Multiply by number of humps There are 4 such humps in $[0, 2\pi]$, so total area = $4 \times 1 = 4$.

The problem asks for the area of $\triangle ABC$, given the length of a median AD and two angles related to the medians AD and BE. The key to solving this problem lies in understanding the properties of medians, specifically how they intersect at the centroid and divide the triangle into smaller triangles of equal area.

Here's the concept:

1. Centroid Property: The medians of a triangle intersect at a point called the centroid (let's call it G). The centroid divides each median in the ratio $2:1$, with the longer segment being from the vertex to the centroid. So, for median AD, $AG:GD = 2:1$. Similarly for median BE, $BG:GE = 2:1$.
2. Area Property of Centroid: The centroid divides the triangle into six smaller triangles of equal area. Also, the three triangles formed by connecting the centroid to the vertices ($\triangle AGB$, $\triangle BGC$, $\triangle CGA$) have equal areas. Therefore, Area($\triangle ABC$) = 3 $\times$ Area($\triangle AGB$).
3. Trigonometry: We will use the sine rule and basic trigonometric ratios in the triangle formed by the centroid and two vertices ($\triangle AGB$) to find its dimensions and area.

Let's apply these concepts step-by-step.

1. Identify the Centroid and its properties:

   Let G be the centroid of $\triangle ABC$. Since AD is a median and AD = 4, the centroid G divides AD in the ratio $2:1$.

   Therefore, $AG = \frac{2}{3} AD = \frac{2}{3} \times 4 = \frac{8}{3}$.

   Also, $GD = \frac{1}{3} AD = \frac{1}{3} \times 4 = \frac{4}{3}$.

2. Analyze $\triangle AGB$:

   We are given $\angle DAB = \frac{\pi}{6}$ and $\angle ABE = \frac{\pi}{3}$.

   In $\triangle AGB$:

   • $\angle GAB = \angle DAB = \frac{\pi}{6}$
   • $\angle GBA = \angle ABE = \frac{\pi}{3}$

3. Calculate the third angle in $\triangle AGB$:

   The sum of angles in a triangle is $\pi$ radians ($180^\circ$).

   $\angle AGB = \pi - (\angle GAB + \angle GBA)$

   $\angle AGB = \pi - \left(\frac{\pi}{6} + \frac{\pi}{3}\right)$

   $\angle AGB = \pi - \left(\frac{\pi}{6} + \frac{2\pi}{6}\right)$

   $\angle AGB = \pi - \frac{3\pi}{6}$

   $\angle AGB = \pi - \frac{\pi}{2} = \frac{\pi}{2}$

   Tip

   The fact that $\angle AGB = \frac{\pi}{2}$ means $\triangle AGB$ is a right-angled triangle. This simplifies calculations significantly, as we can use basic trigonometric ratios or the formula $\frac{1}{2} \times \text{base} \times \text{height}$ for its area.

4. Find the length of BG:

   In the right-angled $\triangle AGB$, we know $AG = \frac{8}{3}$, $\angle GAB = \frac{\pi}{6}$, and $\angle GBA = \frac{\pi}{3}$.

   We can use the tangent function:

   $\tan(\angle GBA) = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{AG}{BG}$

   $\tan\left(\frac{\pi}{3}\right) = \frac{8/3}{BG}$

   $\sqrt{3} = \frac{8}{3BG}$

   $BG = \frac{8}{3\sqrt{3}}$

   Alternatively, using the sine rule:

   In any triangle with sides $a, b, c$ and opposite angles $A, B, C$:

   $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

   Applying the sine rule to $\triangle AGB$:

   $\frac{AG}{\sin(\angle GBA)} = \frac{BG}{\sin(\angle GAB)}$

   $\frac{8/3}{\sin(\pi/3)} = \frac{BG}{\sin(\pi/6)}$

   $\frac{8/3}{\sqrt{3}/2} = \frac{BG}{1/2}$

   $\frac{16}{3\sqrt{3}} = 2BG$

   $BG = \frac{16}{6\sqrt{3}} = \frac{8}{3\sqrt{3}}$

   Both methods yield the same result for BG.

5. Calculate the area of $\triangle AGB$:

   Since $\triangle AGB$ is a right-angled triangle at G, its area is $\frac{1}{2} \times \text{base} \times \text{height}$.

   Area($\triangle AGB$) = $\frac{1}{2} \times AG \times BG$

   Area($\triangle AGB$) = $\frac{1}{2} \times \frac{8}{3} \times \frac{8}{3\sqrt{3}}$

   Area($\triangle AGB$) = $\frac{1}{2} \times \frac{64}{9\sqrt{3}}$

   Area($\triangle AGB$) = $\frac{32}{9\sqrt{3}}$

6. Calculate the area of $\triangle ABC$:

   The centroid divides the triangle into three triangles of equal area: $\triangle AGB$, $\triangle BGC$, and $\triangle CGA$.

   Therefore, Area($\triangle ABC$) = 3 $\times$ Area($\triangle AGB$).

   Area($\triangle ABC$) = $3 \times \frac{32}{9\sqrt{3}}$

   Area($\triangle ABC$) = $\frac{32}{3\sqrt{3}}$

   We can rationalize the denominator if needed, but the options are given with $\sqrt{3}$ in the denominator.

   $\frac{32}{3\sqrt{3}} = \frac{32\sqrt{3}}{3\sqrt{3}\sqrt{3}} = \frac{32\sqrt{3}}{9}$

Comparing this with the given options:

(A) $\dfrac{14}{3\sqrt{3}}$

(B) $\dfrac{28}{3\sqrt{3}}$

(C) $\dfrac{11}{3\sqrt{3}}$

(D) $\dfrac{32}{3\sqrt{3}}$

The calculated area matches option (D).