Mathematics, 31.07.2019 08:00 zurfluhe

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.

Answers: 1

Show answers

Answers

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 from: thicklooney

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 uscircle 'O' with the radius of 4 cm,AC and BD are perpendicular to each other.Property of squareWe know that for a square all the sides are of equal length. The diagonals of a square are equal to each otherthe diagonals of the square are perpendicular to each other.Diagonals of the squarethe 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.

Learn more about Property of square:

link

In a circle with radius 4 cm the diameters ac and bd are perpendicular to each other. prove that abc

Answer from: Quest

1.0

step-by-step explanation:

Answer from: Quest

W-14=t. that i think is the answer idk

Another question on Mathematics

Mathematics, 21.06.2019 14:00

Afunction is given: f(x)=3x+12 a. determine the inverse of this function and name it g(x) b. use composite functions to show that these functions are inverses. c. f(g(– explain: what is the domain?

Answers: 1

Answer

Mathematics, 21.06.2019 18:00

Simplify 15(x - 4)(x -1)-3(x + 4)(x - 1) with steps

Answers: 1

Answer

Mathematics, 21.06.2019 19:00

Draw a scaled copy of the circle using a scale factor of 2

Answers: 1

Answer

Mathematics, 21.06.2019 21:00

These box plots show the number of electoral votes one by democratic and republican presidential candidates for the elections from 1984 through 2012. which statement best compares the spread of the data sets

Answers: 2

Answer

You know the right answer?

In a circle with radius 4 cm the diameters ac and bd are perpendicular to each other. prove that abc...

Questions