Question

In a circle with radius 4 cm the diameters AC and BD are perpendicular to each other. Prove that ABCD is a square. Find the area of ABCD.

Answer 1

This can be proven by using the Pythagorean theorem.    Draw a circle and label the radius 4 cm.  Then draw the diameters in to create a perpendicular intersection at the center.  If you then label the points that make the diameter, you will see that when connected they create right isosceles triangles.  a^2 + b^2 = c ^2, so 4 x 4 + 4 x 4 = c^2.  32 = c^2.  Because you are going to be multiplying it by itself to calculate the area, you have you answer of 32 square centimeters.  NO need to find the square root and them square it again!

Answer 2

To prove that ABCD is a square we will use the Pythagoras theorem. and we must know the Property of the square.

In ABCD all the sides are equal and diagonals are perpendicular and equal to each other, therefore, ABCD is a square.

Given to us

• circle 'O' with the radius of 4 cm,
• AC and BD are perpendicular to each other.

Property of square

• We know that for a square all the sides are of equal length.
• The diagonals of a square are equal to each other
• the diagonals of the square are perpendicular to each other.

Diagonals of the square

• the diagonals of the square ABCD is the diameter of the circle, therefore, AC and BD. therefore, the diagonals of the square are equal.
• The diagonals are perpendicular as is already mentioned that the diameter AC and BD are perpendicular to each other.

Sides of the square using Pythagoras theorem

1. In ΔAOB

AO = BO = r, radius of the circle O,

$AB = \sqrt{(AO)^2+(BO)^2}\\\\ AB = \sqrt{(r)^2+(r)^2}\\\\ AB = \sqrt{2(r)^2}\\\\ AB = r\sqrt{2}$

2. In ΔBOC

BO = CO = r, radius of the circle O,

$BC = \sqrt{(BO)^2+(CO)^2}\\\\ BC = \sqrt{(r)^2+(r)^2}\\\\ BC = \sqrt{2(r)^2}\\\\ BC = r\sqrt{2}$

3. In ΔCOD

CO = DO = r, radius of the circle O,

$CD = \sqrt{(CO)^2+(DO)^2}\\\\ CD= \sqrt{(r)^2+(r)^2}\\\\ CD= \sqrt{2(r)^2}\\\\ CD= r\sqrt{2}$

4. In ΔAOD

AO = DO = r, radius of the circle O,

$AD= \sqrt{(AO)^2+(DO)^2}\\\\ AD= \sqrt{(r)^2+(r)^2}\\\\ AD= \sqrt{2(r)^2}\\\\ AD = r\sqrt{2}$

Thus, AB = BC = CD =DA = r√2.

Hence, In ABCD all the sides are equal and diagonals are perpendicular and equal to each other, therefore, ABCD is a square.

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