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It is given that; the equation of the ellipse is \[25{x^2} + 4{y^2} = 100\].

We have to find the area of the given ellipse.

It is not a general form of an ellipse equation. Hence we have transform the equation we get,

The equation of the ellipse can be written as, \[\dfrac{{25{x^2}}}{{100}} + \dfrac{{4{y^2}}}{{100}} = 1\].

On simplifying we get,

\[\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{{25}} = 1\] … (1)

It means the ellipse cuts the X-axis and Y-axis at \[(2,0)\,\&\, (0,5)\] respectively.

To find the area of the ellipse, we have to take the limit from \[0\] to \[2\].

Let us solve equation (1) for y we get,

\[y = 5\sqrt {1 - \dfrac{{{x^2}}}{4}} \]

As in ellipse major and minor axis divides it into 4 parts

So, the area of the given ellipse becomes

\[A = \int\limits_0^2 {(4)5\sqrt {1 - \dfrac{{{x^2}}}{4}} } dx\]

Let us consider \[x = 2\sin t\]

Differentiate \[x\] with respect to \[t\], we can get, \[dx = 2\cos tdt\]

By substituting limit of \[x\] in \[x\] we get,

Limit from \[0\] to \[2\] changes from \[0\] to \[\dfrac{\pi }{2}\].

By substituting the above values we get,

\[A = \int\limits_0^{\dfrac{\pi }{2}} {(4)5\sqrt {1 - \dfrac{{4{{\sin }^2}t}}{4}} } 2\cos tdt\]

Let us now simplify the fractions in the above equation we get,

\[A = \int\limits_0^{\dfrac{\pi }{2}} {(2)(4)5\sqrt {1 - {{\sin }^2}t} } \cos tdt\]

We know that \[\cos t = \sqrt {1 - {{\sin }^2}t} \]

\[A = \int\limits_0^{\dfrac{\pi }{2}} {40\cos t\cos tdt} \]

Let us solve it further we get,

\[A = \int\limits_0^{\dfrac{\pi }{2}} {40{{\cos }^2}t} dt\]

We know that \[{\cos ^2}t = \dfrac{{1 + \cos 2t}}{2}\] substituting in the above equation, we get,

\[A = 40\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{1 + \cos 2t}}{2}} dt\]

Simplifying we get,

\[A = 20\int\limits_0^{\dfrac{\pi }{2}} {(1 + \cos 2t)} dt\]

We know integration of $1$ is $t$ and $\cos 2t$ is ${\dfrac{\sin 2t}{2}}$

So,Integrating above equation we get,

\[A = 20{\left[ {t + {\dfrac{\sin 2t}{2}}} \right]^{\dfrac{\pi }{2}}}_0\]

Let substitute the limit we get,

\[A = 20[\dfrac{\pi }{2} + 0]\]

On simplifying we get,

\[A = 10\pi \]

Hence, the area of the given ellipse \[25{x^2} + 4{y^2} = 100\] is \[10\pi \].

The problem can be solved by another method.

Let us consider, \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\], be the general form of an ellipse. So, the area of the ellipse is \[4\int\limits_0^a {b\sqrt {1 - \dfrac{{{x^2}}}{{{a^2}}}} } dx = \pi ab\]

Here, \[a = 2,b = 5\]

Substitute the value in the above formula we get,

\[\pi \times 2 \times 5 = 10\pi \]

Hence, the area of the given ellipse \[25{x^2} + 4{y^2} = 100\] is \[10\pi \] sq.units.