Semicircle inside a rectangle and two areas formed by diagonal

This question was uploaded on 02/02/22 on social media accounts.

Solution:

Area of triangle = (a)(b)sin(C)/2

From this formula, if the angle is common, then:

\[\color{black} {\frac{\color{red}{A_{Red}}}{\color{blue}{A_{Blue}}}=\frac{\color{red}{(k)\times(r)}}{\color{blue}{(4k)\times(r)}}=\frac14}\]

Infinite series with a finite solution

This question was uploaded on 01/02/22 on social media accounts.

Solution:

Each term in f(x) is increasing but it has a finite value given so the absolute value of x must be less than 1.

Given series:

\[\color{blue} {f(x)=\frac{\color{red}{(1-x)}(1+x)}{\color{red}{(1-x)}}(1+x^2)(1+x^4)(1+x^8)......}\]

\[\Rightarrow\color{blue} {f(x)=\frac{\color{red}{(1-x^2)}(1+x^2)}{\color{red}{(1-x)}}(1+x^4)(1+x^8)......}\]

Similarly:

\[\Rightarrow\color{blue} {f(x)=\frac{1-x^{\infty}}{1-x}=\frac1{1-x}}\]

Here, higher power of x will tends to zero.

Now:

\[\color{green} {2022=\frac1{1-x}}\]

\[\Rightarrow\color{green} {x=\frac{2021}{2022}}\]

Find the ratio of the radius of both semicircles

This question was uploaded on 31/01/22 on social media accounts.

Solution:

Here, R=1+x

By using intersecting chord theorem:

\[\color{black}{y^2=\color{red}{(2)(2x)}=\color{blue}{(1-x)(1+x)}}\]

\[\color{black}{\Rightarrow y^2=\color{red}{4x}=\color{blue}{1-x^2}}\]

\[\color{black}{\Rightarrow x=\sqrt5-2}\]

\[\color{red}{R=1+x=\sqrt5-1}\]

Now,

\[\color{black} {\frac{\color{red}{R_{Red}}}{\color{blue}{R_{Blue}}}=\frac{\color{red}{\sqrt5-1}}{\color{blue}{1}}=\sqrt5-1}\]

Fraction of blue area and area of the square

This question was uploaded on 30/01/22 on social media accounts.

Solution:

ABC having half of the area of the square.

3, 6, 12 are assumed values and due to similar triangles, they must be in ratio 1:2:4.

ABF and EGF are similar triangles:

\[\color{green} {\frac{9}{12}=\frac{3}{9-x}}\]

\[\Rightarrow \color{green} {x=5}\]

Now, the Blue fraction is half of the ratio of Areas of triangles BDE and ABC:

\[\color{black} {\Rightarrow A_{Fraction}=\frac{\color{blue}{A(BDE)}}{2\times\color{red}{A(ABC)}}=\frac{\color{blue}{3\times5}}{2\times\color{red}{15\times15}}=\frac{1}{30}}\]

Infinite converging series question from JEE Advanced question paper

This question was uploaded on 29/01/22 on social media accounts.

Question from JEE Advanced question paper.

Solution:

Consider 8 as 2³ in all places

We know that:

\[\color{black} {a^3-b^3=(a-b)(a^2+ab+b^2)}\]

\[\color{black} {a^3+b^3=(a+b)(a^2-ab+b^2)}\]

Now,

\[\color{black} {\frac{n^3-2^2}{n^3+2^3}=\color{blue}{\left(\frac{n-2}{n+2}\right)}\color{red}{\left(\frac{n^2+2n+4}{n^2-2n+4}\right)}}\]

Solving in two parts:

Let's consider the blue part from all terms in the series:

\[\color{blue}{\left(\frac{3-2}{3+2}\right)\left(\frac{4-2}{4+2}\right)\left(\frac{5-2}{5+2}\right)....=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot\cdot\cdot\cdot}{5\cdot6\cdot7\cdot8\cdot9\cdot\cdot\cdot\cdot}=24}\]

Now consider the red part from all terms in the series:

\[\color{red}{\frac{19\cdot28\cdot39\cdot52\cdot67\cdot\cdot\cdot\cdot}{7\cdot12\cdot19\cdot28\cdot39\cdot\cdot\cdot\cdot}=\frac{1}{7\cdot12}}\]

Now by combining both parts:

\[\color{black} {S=\left(\color{blue}{24}\right)\left(\color{red}{\frac{1}{7\cdot12}}\right)=\frac27}\]

Quarter circle in a square

This question was uploaded on 28/01/22 on social media accounts.

Solution:

We know that:

\[\color{blue} {s^2+x^2=a^2+b^2}\]

By using cosine rule:

\[\color{black} {2ab\cos(135)=a^2+b^2-2s^2}\]

\[\Rightarrow\color{red} {-\sqrt2ab=a^2+b^2-2s^2}\]

By combining both equations:

\[\color{purple} {2s^2=a^2+b^2+\sqrt2ab}\]

\[\color{green} {2r^2=a^2+b^2-\sqrt2ab}\]

From these two equations:

\[\color{fuchsia} {s^2-r^2=\sqrt2ab}\]

Three sides of the squares are converted to squares

This question was uploaded on 27/01/22 on social media accounts.

Solution:

By applying cosine rule opposite to side A:

\[\color{black} {B^2+C^2-A^2=2BC\cos{x}}\]

By applying cosine rule opposite to angle X:

\[\color{black} {B^2+C^2-3^2=2BC\cos{(180-x)}}\]

\[\Rightarrow \color{black} {3^2-(B^2+C^2)=2BC\cos{x}}\]

From both above equations:

\[\color{black} {B^2+C^2-A^2=3^2-(B^2+C^2)}\]

\[\Rightarrow \color{Blue} {2(B^2+C^2)-A^2=3^2}\]

Similarly:

\[\Rightarrow \color{Red} {2(A^2+C^2)-B^2=4^2}\]

\[\Rightarrow \color{Green} {2(A^2+B^2)-C^2=5^2}\]

By adding the above three equations:

\[\color{Purple} {4(A^2+B^2+C^2)-(A^2+B^2+C^2)=3^2+4^2+5^2}\]

\[\Rightarrow \color{Purple} {A_{Blue}=A^2+B^2+C^2=\frac{50}3}\]

The hyperbolic equation on two variables

This question was uploaded on 26/01/22 on social media accounts.

Solution:

Given equation:

\[\color{blue} {a^2+2ab-3b^2-41=0}\]

\[\Rightarrow \color{blue} {a^2\color{red}{+a^2-a^2}+2ab-3b^2-41=0}\]

\[\Rightarrow \color{blue} {2a^2-(a^2-2ab+b^2)-2b^2-41=0}\]

\[\Rightarrow \color{blue} {2(a^2-b^2)-(a-b)^2-41=0}\]

\[\Rightarrow \color{blue} {2(a+b)(a-b)-(a-b)^2=41}\]

\[\Rightarrow \color{blue} {(a-b)(a+3b)=1\times41}\]

Given that a and b belong to natural numbers. So, both must be positive integers.

Clearly, ( a - b ) < ( a + 3b )

\[\color{red} {a-b=1}\Rightarrow\color{red} {a=1+b}\]

Now put this value in other equation:

\[\color{red} {a+3b=41}\Rightarrow\color{red} {1+4b=41}\]

\[\Rightarrow\color{red} {b=10}\]

\[\Rightarrow\color{red} {a=11}\]

\[\color{fuchsia} {a^2+b^2=11^2+10^2=221}\]