An isosceles triangle has it's vertex on a circle. If /PQ/=26cm, /QR/=26cm and /PR/=20cm. Calculate the height of the triangle and the radius of the circle to the nearest whole number
Do you mean it has all of its vertices on the circle? If so, then
Let M be the midpoint of PR.
Now you have two 2*(5-12-13) right triangles.
The center of the circle lies 2/3 of the way from Q to M, where the medians meet.
Yes
I need the correct answer of the questions
To calculate the height of the triangle, we need to use the Pythagorean theorem. Since the triangle is isosceles, we can use one of the congruent sides as the base of the triangle, and the other side would be the height.
Let's label the isosceles triangle as ABC, with the vertex at point B. The congruent sides are AB and AC, with length PQ = QR = 26 cm. The base is BC, with length PR = 20 cm.
To find the height, we can use the formula:
Height^2 + (1/2 * PR)^2 = (PR)^2
Height^2 + (1/2 * 20)^2 = 20^2
Height^2 + 10^2 = 20^2
Height^2 + 100 = 400
Height^2 = 400 - 100
Height^2 = 300
Height ≈ √300
Height ≈ 17.32 cm (rounded to two decimal places)
To calculate the radius of the circle, we can use the formula:
Radius^2 = (1/2 * PR)^2 + Height^2
Radius^2 = (1/2 * 20)^2 + 17.32^2
Radius^2 = 10^2 + 17.32^2
Radius^2 = 100 + 300
Radius^2 = 400
Radius ≈ √400
Radius ≈ 20 cm (rounded to the nearest whole number)
So, the height of the triangle is approximately 17 cm and the radius of the circle is approximately 20 cm.