SOLUTION: Two chords and a diameter form a triangle inside a circle. The radius is 5cm and one chord is 2cm longer than the other. Find the perimeter and area of the triangle

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Question 1206348: Two chords and a diameter form a triangle inside a circle. The radius is 5cm and one chord is 2cm longer than the other. Find the perimeter and area of the triangle
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52797)   (Show Source): You can put this solution on YOUR website!
.

The side which is the diameter is 10 = 5*2 cm long.


Two other sides form a right angle, since they lean on the diameter.


So, we have a right-angled triangle with the hypotenuse of 10 cm
and one leg is 2 cm longer than another.


Hence, as probably you just can to guess, this triangle is (3,4,5)-triangle 
with the sides 6 cm, 8 cm and 10 cm.


Its perimeter is  6+8+10 = 24 cm.

Its area is  = 3*8 = 24 cm^2.

Solved.



Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Apply Thales Theorem to determine we have a right triangle.
Thales Theorem is a special case of the Inscribed Angle Theorem.
This is what your diagram would look like (imagine adding the proper labels)

Image credit:
https://en.wikipedia.org/wiki/Thales%27s_theorem

radius = 5
diameter = 2*radius = 2*5 = 10
The diameter is the hypotenuse of the right triangle.
A diameter is a special type of chord that passes through the center. It's the longest chord of the circle.

x = shorter leg of the right triangle
x+2 = longer leg of the right triangle
x > 0 since a negative side length makes no sense.

Due to the Pythagorean Theorem , we can say

Let's expand things out to solve for x.








or
or
The quadratic formula is an alternative approach.
Ignore the negative result because we stated x > 0 earlier.

leads to

We have a 6-8-10 right triangle.

perimeter = add the sides = 6+8+10 = 24 cm
area = 0.5*base*height = 0.5*6*8 = 24 square cm
This is one of the fairly rare moments when the perimeter and area are the same value (except for the differing unit types of course).
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