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:

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

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:

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

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:

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

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

Two angle bisectors and one altitude in a right angled triangle

This question was uploaded on 25/01/22 on social media accounts.

Solution:

Given that, a:b = 3:4

Let AB = 15 and BC = 20 so AC = 25, AE = 9, CE = 16

2x + 2y = 90  (or)  x + y = 45

<BDC = 90 - x = x + 2y

<BFA = 90 - y = 2x + y

By this, Triangles ABF and CBD are isosceles.

So, AF = 15 and CD = 20

Also, AD = 5 and FC = 10

Now, DE = AE - AD = 9 - 5 = 4

and, FE = CE - CF = 16 - 10 = 6

c:d = DE:FE = 4:6 = 2:3