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Chapter 8: Problem 15

A triangle with sides of lengths 3 in., 4 in., and 5 in. has an area of 6 in \(^{2}\). What is the length of the radius of the inscribed circle?

Short Answer

Expert verified
The radius of the inscribed circle is 1 in.

Step by step solution

01

Understand the Problem

We are given a triangle with sides 3 in., 4 in., and 5 in., which indicates that it is a right-angled triangle. We need to find the radius of the circle inscribed within this triangle.
02

Recall the Formula for the Area of a Triangle

The area of a triangle can be determined using the formula containing the inradius (inscribed circle's radius): Area = \( r \cdot s \), where \( r \) is the radius and \( s \) is the semi-perimeter of the triangle.
03

Calculate the Semi-perimeter

Find the semi-perimeter \( s \) of the triangle. \[ s = \frac{3 + 4 + 5}{2} = 6 \text{ in.} \]
04

Substitute Values into the Formula

Substitute the known values into the formula for the area. \[ 6 = r \times 6 \]
05

Solve for the Radius

Solve the equation for \( r \). \[ r = \frac{6}{6} = 1 \text{ in.} \] In this equation, we simply divide the area by the semi-perimeter.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right-Angled Triangle
A right-angled triangle is a special type of triangle that contains one angle measuring exactly 90 degrees. This is also called a right angle. In the context of the problem, the triangle with sides 3 in., 4 in., and 5 in. is a classic example known as a Pythagorean triple. This specific six unit triangle is very famous in geometry.
  • One side, which measures 5 in., is the hypotenuse. The hypotenuse is always the longest side in a right-angled triangle and is opposite the right angle.
  • The other two sides, 3 in. and 4 in., are called the legs. They form the right angle and can often be calculated using the Pythagorean theorem if the hypotenuse is known.
A right-angled triangle is often used to teach various geometric principles, including trigonometry and the relationships between angles and sides. It's a fundamental shape in math with many applications in science and engineering.
Inradius
The inradius of a triangle is the radius of the circle that can fit perfectly inside the triangle, touching all three sides. This circle is called the incircle. Inradius can be a bit tricky to understand at first, but it's an essential concept in geometry, especially when calculating areas.
  • The formula for finding the area of a triangle in terms of its inradius is: \( \text{Area} = r \cdot s \), where \( r \) is the inradius and \( s \) is the semi-perimeter of the triangle.
  • For right-angled triangles, this relationship often makes it easier to determine the inradius because the area and semi-perimeter are more straightforward to calculate.
Knowing the inradius helps to find the smallest circle that fits inside the triangle without crossing its borders. This geometric feature can be useful in design and structural engineering, where precise fits are necessary.
Semi-perimeter
The semi-perimeter of a triangle is half of its perimeter. It's used in various geometric calculations, especially those involving area or radius sizes within triangles.
  • The semi-perimeter \( s \) is calculated by adding all the side lengths of the triangle together and then dividing by 2. For example, with sides of 3 in., 4 in., and 5 in., the semi-perimeter is \( s = \frac{3 + 4 + 5}{2} = 6 \text{ in.} \).
  • This value is crucial when using the formula that relates the area of a triangle and its inradius, as it simplifies the equation significantly when you know both the area and the semi-perimeter.
Understanding the semi-perimeter can help solve a variety of problems not only related to inscribed circles but also those involving different metrics of triangles, like Heron's formula for area.

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Most popular questions from this chapter

Given a circle with diameter of length \(d\), explain why \(A_{\text {circle }}=\frac{1}{4} \pi d^{2}\)

Find the area of a trapezoid with an altitude of length \(4.2 \mathrm{m}\) and a median of length \(6.5 \mathrm{m}\) Find the area of a trapezoid with an altitude of length \(5 \frac{1}{3} \mathrm{ft}\) and a median of length \(2 \frac{1}{4} \mathrm{ft}\).

Find the ratio \(\frac{A_{1}}{A_{2}}\) of the areas of two similar triangles if a) the ratio of the lengths of the corresponding sides is \(\frac{s_{1}}{s_{2}}=\frac{3}{2}\) b) the lengths of the sides of the first triangle are 6 in., 8 in., and 10 in., and those of the second triangle are 3 in., 4 in., and 5 in.

Two sides of a triangle measure 5 in. and 7 in. What are the limits of the length of the third side?

Given:\( \triangle A B C,\) whose sides are \(10 \mathrm{cm}, 17 \mathrm{cm},\) and \(21 \mathrm{cm}\) Find: a) \(B D,\) the length of the altitude to the \(21-\mathrm{cm}\) side b) The area of \(\triangle A B C,\) using the result from part (a)
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