Imagine we have a circle, and a special shape called an isosceles trapezoid is drawn around it. This means the circle is “inscribed” within the trapezoid. We know the diameter of the circle is 15 cm, and each of the non-parallel sides (called legs) of the trapezoid is 17 cm. Our goal is to find the lengths of the two parallel sides (called bases) of the trapezoid.
Solution:
1. Understanding the Special Property:
When a trapezoid can have a circle inscribed inside it, there’s a very important property: the sum of the lengths of its parallel bases is equal to the sum of the lengths of its non-parallel sides (the legs).
In our problem, let’s call the non-parallel sides AB and CD. We are given that AB = CD = 17 cm.
Let the bases be BC (the shorter base) and AD (the longer base).
So, according to this property: BC + AD = AB + CD.
Plugging in the given values: BC + AD = 17 + 17 = 34 cm.
2. Finding the Height of the Trapezoid:
The diameter of the inscribed circle is equal to the height of the trapezoid.
We are given the diameter is 15 cm.
So, the height of the trapezoid, let’s call it CE (where E is a point on the base AD such that CE is perpendicular to AD), is CE = 15 cm. (The problem uses KQ = 15 cm for the diameter, which is correct).
3. Using the Pythagorean Theorem:
Since the trapezoid is isosceles, if we draw heights from both top vertices to the bottom base, we form two right-angled triangles at the ends.
Let’s consider the right-angled triangle \triangle CED, where CD is the hypotenuse (the leg of the trapezoid), CE is the height, and DE is part of the base.
We know CD = 17 cm and CE = 15 cm.
Using the Pythagorean Theorem (a^2 + b^2 = c^2 or c^2 - b^2 = a^2):
DE^2 = CD^2 - CE^2
DE^2 = 17^2 - 15^2
DE^2 = 289 - 225
DE^2 = 64
DE = \sqrt{64}
DE = 8 cm.
4. Relating the Bases:
In an isosceles trapezoid, when we drop perpendiculars (heights) from the endpoints of the shorter base to the longer base, the longer base AD can be expressed as the sum of the shorter base BC and two segments like DE.
So, AD = BC + 2 \cdot DE.
We found DE = 8 cm, so 2 \cdot DE = 16 cm.
Therefore, AD = BC + 16 cm.
5. Solving for the Bases:
Now we have a system of two equations with two unknowns (BC and AD):
Equation 1: BC + AD = 34
Equation 2: AD = BC + 16
Let’s substitute Equation 2 into Equation 1:
BC + (BC + 16) = 34
2 \cdot BC + 16 = 34
Subtract 16 from both sides:
2 \cdot BC = 34 - 16
2 \cdot BC = 18
Divide by 2:
BC = 9 cm.
Now that we have BC, we can find AD using Equation 2:
AD = BC + 16
AD = 9 + 16
AD = 25 cm.
Answer:
The lengths of the bases of the trapezoid are 9 cm and 25 cm.
