Square inside a semicircle such that one edge is on the semicircle and one side on the diameter

This question was uploaded on 06/12/21 on social media accounts.

Solution:

There are two right angled triangles:

\[a = \sqrt{4^2-3^2} = \sqrt7\]

By similar triangles:

\[\frac a4 = \frac x3\]

\[\Rightarrow x = \frac {3\sqrt7}4\]

\[\Rightarrow Area = x^2 = \frac {63}{16}\]

Dare2solve | Two squares hanging on one common vertex

This question was uploaded on 13/11/21 on social media accounts.

Solutions:

1) Solution from @JaihindSigh3 from Twitter

2) Solution from @JohnMic58788564 from Twitter

Dare2solve | Farthest vertex of two squares are connected by line - Angle at midpoint of line and vertex of square

This question was uploaded on 10/11/21 on social media accounts.

Solution:

Three blue lines are equal and the orange line is starting from the ends of blue lines:

\[\Rightarrow Angle=90°\]

Dare2solve | Circle in a square and two tangents from a vertex

This question was uploaded on 09/11/21 on social media accounts.

Solution:

Area:

\[A = \frac12\times2\times2\times\sin(120)=\sqrt3\]

Dare2solve | Ratio of area of circle and sum of areas of 4 squares

This question was uploaded on 07/11/21 on social media accounts.

Solution:

We know that:

\[a^2+b^2+c^2+d^2=4r^2\]

Click here to read the proof of this identity.

\[Orange = \pi r^2\]

\[Green = a^2+b^2+c^2+d^2=4r^2\]

\[\frac{Orange}{Green}=\frac{\pi r^2}{4r^2}=\frac\pi4\]

Dare2solve | Area between two lines drawn in a square

This question was uploaded on 31/10/21 on social media accounts.

Two lines in a square from a vertex inclined with equal angle. Find the area under both lines and sides of the square.

Solution:

Triangle with two red sides is an equilateral triangle.

One triangle is 6-8-10 triangle.

The side of the square is 8.

\[Area=\frac{(4)(8)}2+\frac{(2)(8)}2=24\]

Dare2solve | Square in a semicircle such that one end of diameter and two opposite edges of square are collinear

This question was uploaded on 29/10/21 on social media accounts.

Square in a semicircle such that one line can pass through one end of diameter and two opposite edges of the square. One edge of the square is on the origin and one on the semicircle.

Solution:

Purple triangle is an equilateral triangle.

By inscribed angle theorem:

\[x=\frac{60+45}2=\frac{105}2=72.5\]

Puzzle related to Geometry, Circle, Square, Angle.

Dare2solve | Square and triangle - One vertex in common

This question was uploaded on 26/10/21 on social media accounts.

Square and an equilateral triangle such that one side of the square is on one side of the triangle and also one vertex for both figures is the same.

Solution:

Green Length (By cosine rule):

\[\cos30 = \frac{(\sqrt3)^2+(1+\sqrt3)^2-x^2}{2(\sqrt3)(1+\sqrt3)}\]

\[\Rightarrow x^2 = 4-\sqrt3\]

\[\Rightarrow x = \sqrt{4-\sqrt3}\]

Puzzle related to Geometry, Length, Square, Triangle.

Dare2solve | Area between square and triangle - Ratio with complete figure

This question was uploaded on 25/10/21 on social media accounts.

Find the ratio of the area between the equilateral triangle and the square to the complete area of the figure. Square is placed on the base of the triangle as shown in the figure.

Solution 1:

Now you can see 1 small equilateral triangle (1), one rectangle (3), and three 30-60-90 triangles (2, 4, 5).

Areas:

\[A_1=\frac{\sqrt3}4(\sqrt3-1)^2=\frac{\sqrt3}2(2-\sqrt3)\]

\[A_2=A_4=A_5=\frac{1\times\sqrt3}2=\frac{\sqrt3}2\]

\[A_3=\sqrt3(\sqrt3-1)\]

Blue fraction:

\[Fraction=\frac{A_3+A_4}{A_1+A_2+A_3+A_4+A_5}\]

\[\Rightarrow Fraction=\frac{2\sqrt3-1}{3+\sqrt3}=\frac{7\sqrt3-9}6\]

Puzzle related to Geometry, Square, Triangle, Ratio.

Dare2solve | Square hanging on the vertex of the other square

This question was uploaded on 20/10/21 on social media accounts.

Two squares with one same vertex and one side is passing through the vertex of another square.

Solution:

Angle 45 is the same due to the angle inscribed in the same arc.

\[\alpha+45=90\]

\[\Rightarrow\alpha=45\]

Puzzle related to geometry, square, angle.