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

Equilateral triangle in an equilateral triangle

This question was uploaded on 24/01/22 on social media accounts.

Question is from @mathematics.azerbaijan from Instagram.

Solution:

Blue and green are equilateral triangles.

Yellow triangles are similar.

\[\color{black} {\Rightarrow \frac y3 = \frac {x+y}5 = \frac7{5+3}}\]

\[\color{black} {\Rightarrow x=\frac{16}7}\]

\[\color{blue} {A_{Blue}=\frac{\sqrt3}4\left(\frac{16}7\right)^2=\frac{64\sqrt3}{49}}\]

\[\color{red} {A_{Total}=\frac{\sqrt3}4\left(8\right)^2=16\sqrt3}\]

Blue Fraction:

\[\color{black} {\frac{\color{blue}{A_{Blue}}}{\color{red}{A_{Total}}}=\frac{\color{blue}{64\sqrt3/49}}{\color{red}{16\sqrt3}}=\frac{4}{49}}\]

Two sides of equilateral triangle are tangents to the circle

This question was uploaded on 21/01/22 on social media accounts.

Solution:

Note:

1) Tangent is always perpendicular to the radius at that point.

2) Two tangents with common points are equal in length.

As you can observe in the figure, the perimeter of that circle is two times the length of the side of the equilateral triangle.

From the triangle inside the circle (With sides 1, 1, S), by cosine rule: S = √3

Perimeter of that triangle = 2√3

Three right angled triangles in a semicircle within two parallel lines

This question was uploaded on 20/01/22 on social media accounts.

Solution:

Let the hypotenuse of the blue triangle be 'x' and the hypotenuse of two red triangles be 'a' and 'b'.

All three triangles are similar due to similar angles so their areas can be compared by the length of their hypotenuse.

We know that area of the right angle triangle is proportional to the square of the hypotenuse.

Red/Blue = x²/(a²+b²)

Comparing 'h' in both right angles triangles:

\[\color{Blue} {h^2=\color{#DF00FE}{\left(\frac{a+b}2\right)^2-\left(\frac X2\right)^2}=\color{#1BD62E}{\left(\frac X2\right)^2-\left(\frac{a-b}2\right)^2}}\]

\[\Rightarrow\color{Blue}{X^2}=\color{Red}{a^2+b^2}\]

Now,

\[\frac{\color{Red}{Red}}{\color{Blue}{Blue}}=\frac{\color{Red}{a^2+b^2}}{\color{Blue}{X^2}}=1\]

Two semicircles having parallel diameters

This question was uploaded on 18/01/22 on social media accounts.

Solution:

Here, r = x/2, R = (2a+b)/2

In red triangle:

\[\color{red} {h^2=R^2-r^2}\]

In blue triangle:

\[\color{blue} {h^2=r^2-(R-a)^2}\]

Comparing both equations:

\[\color{fuchsia} {\color{red}{R^2-r^2}=\color{blue}{r^2-(R-a)^2}}\]

\[\Rightarrow\color{fuchsia} {2r^2=R^2+(R-a)^2}\]

\[\Rightarrow\color{fuchsia} {2\left(\frac x2\right)^2=\left(\frac{2a+b}{2}\right)^2+\left(\frac{2a+b}{2}-a\right)^2}\]

\[\Rightarrow\color{fuchsia} {x^2=2a^2+b^2+2ab}\]

\[\Rightarrow\color{fuchsia} {x=\sqrt{2a^2+b^2+2ab}}\]

or

\[\Rightarrow\color{fuchsia} {x=\sqrt{(a+b)^2-a^2}}\]