Mathathon Sample Question

Since ,angle ABC = 90 degree.

Therefore,AC must be diameter. (angle made by diameter on circumference is 90 degree)

Now, $side^{2}+side^{2}={AC}^{2}$

$side^{2}=12.5$

$area=12.5$

Area Of square $= \frac{diagonal1 × diagonal 2}{2}$ (Square is a rhombus)

$\text{diagonal1} = \text{diagonal2} = \text{Diameter of circle} = 5$

Area = $\frac{5×5}{2}$

$\boxed{Area = \color{#3D99F6}{12.5}}$

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$\text{Diagonal }= 5$

$\text{Pythagoras Theorem} \rightarrow x^2+x^2=5^2 (x\text{ is side length of square)}$

$\boxed{x^2=12.5}$

$\text{The square can be divided into 2 45-45-90 degree triangles.}$

The hypotenuse of a 45-45-90 triangle is $\sqrt{2}$ times more than the sides.

So the side length is $\frac{5}{\sqrt{2}}$ .

The area of a square is $s^2$

$\frac{5^2}{\sqrt{2}^2}$ = $\frac{25}{2}$ = $\boxed{12.5}$

As $ABCD$ is a square $\therefore\angle BCD=90^\degree$

$\Rightarrow \triangle BCD$ is a right angled triangle. Applying Pythagorus theorem $\overline{BC}^2+\overline{CD}^2=\overline{BD}^2$ As $\overline{BC}=\overline{CD}$ because $ABCD$ is a square $\therefore \overline{BC}^2+\overline{BC}^2=\overline{BD}^2$ $2\overline{BC}^2=\overline{BD}^2$ $\overline{BC}^2=\frac{1}{2}\overline{BD}^2$ As $\overline{BD}=diameter=5cm$ $\therefore \overline{BC}^2=\frac{1}{2}\times(5cm)^2=\frac{1}{2}\times25cm^2=12.5cm^2$ Area of $ABCD=BC^2=\boxed{12.5cm^2}$

Pythagoras' Theorem

$a^2 + b^2 = c^2$

$\boxed{\text{As } a = b}$

$2a^2 = c^2$

$2a^2 = 5^2$

$\frac{2a^2}{2} = \frac{5^2}{2}$

$a^2 = 12.5$

As the area of the square is $= a^2$ (where $a$ is the side of the square)

$\boxed{\text{The area of the square is = 12.5}}$

• x^2+x^2=5^2 (pythagorean theorem)
• x=side length of square
• side length = 3.53
• area=3.53^2
• 12.5

There are 3 main ways:
1.
Since the quadrilateral is a square, it’s side length is:
$2s^2=25\Rightarrow s^2=12.5$ Where $s$ is it’s side length. Note that we don’t even need to find $s$ .

2. .
Split the square into two right angled isosceles triangles.
The size of two triangles is: $5\times\frac{5}{2}\times 2=12.5.$

3.
The size of the square is half of a square with side length equal to its diagonal, so $S_\square=$
$\frac{5^2}{2}=12.5.$

Learn more in RadMaths !

1. $A^{2} + B^{2} = C^{2}$

2. $A^{2} = B^{2} = x$

3. $C^{2} = 5^{2} = 25$

4. $2x = 25$

$\red{Q.E.D.: x = 25/2 = 12.5 }$

Although there are many ways to solve this problem, the easiest and quickest way is to rearrange the square into half of a new square, like this:

Since we bisected the square, the $90$ degree angle was split into two $45$ degree angles. Thus, when we rearrange the parts, combining the two $45$ degree angles will give us back a $90$ degree angle, which proves that the resulting figure is half of a new square.

This makes it so much easier! The new square's area is $5 \times 5=25$ units squared. Half of the area of this square is $\frac{1}{2} \times 25=\boxed{12.5}$ units squared, our answer.

$x^2$ = area Pythagorean theorem:

$x^2+x^2 = 5^2$

$2x^2 = 25$

$x^2 = 12.5$

$\boxed{area = 12.5}$