A circle has an area equal to 25 pi sq.cm. Its diameter AB coincides with one of the sides of triangle ACB in which the vertex C lies on the circle. If the triangle has an area equal to 11 sq.cm, find its perimeter.
Please include solution. Thanks.
Why did the triangle go to the circus? Because it heard there were circles involved! Let's find the solution to this circus act, shall we?
First, let's find the radius of the circle using the formula for the area of a circle: A = π * r^2. We are given that the area of the circle is 25π square cm, so we can set up the equation:
25π = π * r^2
Dividing both sides by π, we get:
25 = r^2
Taking the square root of both sides, we have:
r = 5 cm
Now, we know that the diameter (AB) coincides with one side of the triangle (AC or BC). Since a diameter is twice the length of the radius, AB = 2r = 2 * 5 = 10 cm.
Let's find the length of the base of the triangle. We know that the area of the triangle is 11 square cm, and the base is equal to AB. The formula for the area of a triangle is A = (1/2) * base * height, so we can set up the equation:
11 = (1/2) * 10 * height
Simplifying, we have:
11 = 5 * height
Dividing both sides by 5, we get:
height = 11/5 = 2.2 cm
Since the height (AC or BC) is perpendicular to the base (AB), the triangle is an isosceles triangle. This means that the remaining two sides (AC and BC) are equal in length.
Let's find the perimeter of the triangle. The perimeter is the sum of all three sides, so the perimeter is:
Perimeter = AB + AC + BC = 10 cm + 2.2 cm + 2.2 cm = 14.4 cm
So, the perimeter of the triangle is 14.4 cm, making it quite the circus performer! Hope this solution was as entertaining as the triangle's journey to the circus!
To find the perimeter of the triangle, we need to find the lengths of its sides.
Let's start by finding the radius of the circle. The area of a circle is given by the formula A = πr^2, where A is the area and r is the radius. From the given information, we have:
A = 25π cm^2
Setting this equal to the formula, we get:
25π = πr^2
Dividing both sides by π, we have:
25 = r^2
Taking the square root of both sides, we get:
r = √25
r = 5 cm
Since the diameter AB coincides with one of the sides of the triangle ACB, the length of AB is equal to the diameter of the circle, which is 2 * radius = 2 * 5 = 10 cm.
Now, let's find the height of the triangle. Since vertex C lies on the circle, the height of the triangle is equal to the radius of the circle, which is 5 cm.
The area of the triangle is given by the formula A = (1/2) * base * height. We can rearrange this formula to solve for the base:
base = (2 * A) / height
Plugging in the values, we get:
base = (2 * 11) / 5
base = 22 / 5
base = 4.4 cm
Now, we can use the Pythagorean theorem to find the length of the remaining side of the triangle. Let's call it side AC. Here's the setup:
AC^2 = AB^2 - BC^2
Substituting the values, we get:
AC^2 = 10^2 - 4.4^2
AC^2 = 100 - 19.36
AC^2 = 80.64
Taking the square root of both sides, we get:
AC = √80.64
AC ≈ 8.971 cm
Finally, we can find the perimeter of the triangle by adding up the lengths of all its sides:
Perimeter = AB + BC + AC
Perimeter = 10 + 4.4 + 8.971
Perimeter ≈ 23.371 cm
Therefore, the perimeter of the triangle is approximately 23.371 cm.
To find the perimeter of the triangle, we need to determine the lengths of its sides.
Let's start by finding the radius of the circle. The formula for the area of a circle is A = πr², where A is the area and r is the radius.
Given that the area of the circle is 25π sq.cm, we can set up the equation as follows:
25π = πr²
Dividing both sides by π, we get:
25 = r²
Taking the square root of both sides, we find:
r = √25
r = 5 cm
Now, let's consider the triangle ACB. Since the diameter AB coincides with one of the sides of the triangle, we can consider AC and CB as radii of the circle. Given that the area of the triangle is 11 sq.cm, we can use the formula for the area of a triangle:
A = ½ * base * height
Since AC and CB are radii, the base of the triangle is equal to the diameter, which is 2 * 5 = 10 cm.
So, we have:
11 = ½ * 10 * height
Simplifying this equation, we find:
11 = 5 * height
Dividing both sides by 5, we get:
height = 11/5
height = 2.2 cm
Now, we can use the Pythagorean theorem to find the length of the side AB:
c² = a² + b²
where c is the hypotenuse (AB) and a and b are the other two sides (AC and CB).
We have:
AB² = AC² + CB²
AB² = 5² + 2.2²
AB² = 25 + 4.84
AB² = 29.84
Taking the square root of both sides, we find:
AB = √29.84
AB ≈ 5.47 cm
Now that we know the lengths of all three sides of the triangle, we can calculate its perimeter by adding them together:
Perimeter = AC + AB + CB
Perimeter = 5 + 5.47 + 2.2
Perimeter ≈ 12.67 cm
Therefore, the perimeter of the triangle ACB is approximately 12.67 cm.
The circle radius is 5 cm. That comes from the area.
The triangle ACB is a right triangle with AB as a diameter. The length of AB is 10 cm, since it is a diameter
(AC)^2 + (BC)^2 = (AB)^2 = 100
Area = (1/2)(AB)*(BC) = 11
(AC)*(BC) = 22
(BC)=22/(AC)
(AC)^2 + 484/(AC)^2 = 100
Solve for AC. The short side is about 2.3. Then get BC and the perimenter.
Then get BC