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:

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

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:

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

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:

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

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

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:

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

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 squares in a semicircle dividing radius in half

This question was uploaded on 17/01/22 on social media accounts.

Solution:

Here, y is assumed and 2y is due to intersecting chord theorem.

2y length is divided into two parts of x, (2y-x).

Red triangle is right angled triangle with sides (2x+y), (2y-x), 2r.

By intersecting chord theorem:

\[\color{blue} {(2x)(y)=(x)(2y)=(3)(1)}\]

\[\Rightarrow\color{blue} {y=\frac3{2x}}\]

By right angled triangle (Red Triangle):

\[\color{red} {(2x+y)^2+(2y-x)^2=4^2}\]

\[\Rightarrow\color{red} {5(x^2+y^2)=16}\]

\[\Rightarrow\color{red} {5\left(x^2+\left(\color{blue}{\frac3{2x}}\right)^2\right)=16}\]

\[\Rightarrow\color{red} {20x^4-64x^2+45=0}\]

\[\Rightarrow\color{red} {x^2=\frac{16\pm\sqrt{31}}{10}}\]

Blue Area (A):

\[A=\color{blue} {2x^2=\frac{16\pm\sqrt{31}}5}\]

Blue Area cannot be much small, so we will neglect negative symbol:

\[A=\color{blue} {\frac{16+\sqrt{31}}5}\]

Semicircle and sector of circle in a regular hexagon

This question was uploaded on 11/01/22 on social media accounts.

Solution:

Blue Area = A(⌔ABD) - A(⌔OBC) - A(ΔAOC)

\[\Rightarrow Area=\pi(12)^2\left(\frac{30}{360}\right)-\pi(6)^2\left(\frac{60}{360}\right)-\frac12(6)(6)\sin(120)\]

\[\Rightarrow Area=12\pi-6\pi-9\sqrt3=6\pi-9\sqrt3\]

Triangle in a triangle such that one common angle and one similar angle are formed

This question was uploaded on 08/01/22 on social media accounts.

Solution:

Here, one angle is common and one angle is given similar, so both triangles are similar.

One side of both triangles is in the ratio 1:2 and both triangles are similar so all sides are in the ratio of 1:2 and their areas are in the ratio of the square of the ratio of their sides.

Ratio = (1/2)² = 1/4

Ratio = 1:4

Diagonal of a square and base of an equilateral triangle are on the same line

This question was uploaded on 05/01/22 on social media accounts.

Solution 1:

Blue area = A(ABC) - A(ADE)

\[A(ADE)=\frac12(AD)(ED)=\frac12(4\cdot\sin15)(4\cdot\cos15)\]

\[\Rightarrow A(ADE)=2\]

\[A(ABC)=\frac12(AB)(BC)=\frac12(2)(2\cdot\tan15)\]

\[\Rightarrow A(ABC)=4-2\sqrt3\]

\[\Rightarrow A_{Blue}=2-(4-2\sqrt3)=2\sqrt3-2\]

Area between two isosceles triangles in a square

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

Solution 1:

Triangle ABC is 1/4 of the area of the square,

Let, the area of triangle ADE = 8A

Then, the area of triangle ABC = 15A

Now,

\[Fraction=\frac{2\times8}{4\times15}=\frac4{15}\]

Solution 2:

Let equal sides of the triangle have length 5 units.

\[\tan\alpha=\frac12\]

\[\Rightarrow\tan(45-\alpha)=\frac13\]

\[\Rightarrow\frac x4=\frac13\]

\[\Rightarrow x=\frac43\]

Blue Area:

\[\color{blue} {A=2\times\left(\frac12\times4\times\frac43\right)=\frac{16}3}\]

Square Area:

\[\color{red} {A=s^2=4^2+2^2=20}\]

Blue Fraction:

\[Fraction=\frac{\color{blue}{16/3}}{\color{red}{20}}=\frac4{15}\]