Problem 163
Question
Find the in-radius of the triangle having sides \(13,14,15\).
Step-by-Step Solution
Verified Answer
The in-radius of the triangle with sides 13, 14, and 15 is approximately \(2.18\).
1Step 1: Calculate the semi-perimeter of the triangle
Using the formula \(s = (a + b + c) / 2\), and substituting the given lengths of the sides into this formula, we get \(s = (13 + 14 + 15) / 2 = 21\)
2Step 2: Calculate the in-radius of the triangle
Now using the formula for the in-radius \(r = \sqrt{s(s-a)(s-b)(s-c)} / s\), substituting the calculated semi-perimeter and the lengths of the sides into this formula, we get \(r = \sqrt{21 * (21 - 13) * (21 - 14) * (21 - 15)} / 21\) This simplifies to: \(r = \sqrt{21 * 8 * 7 * 6} / 21\)
3Step 3: Simplify the expression
Simplify the expression \(r = \sqrt{21 * 8 * 7 * 6} / 21\) and get the final value of the in-radius. After simplifying, we found that \(r = \sqrt{12348} / 21 \approx 2.18\)
Key Concepts
Semi-Perimeter FormulaTriangle Side LengthsSimplification of Expressions
Semi-Perimeter Formula
The semi-perimeter formula is a crucial concept in geometry, especially when working with triangles. It provides a way to find half of a triangle's perimeter. By knowing the lengths of the sides of the triangle, we can apply this formula to aid in calculations involving certain triangle properties, such as the inradius.
To calculate the semi-perimeter, you add up all the side lengths of the triangle and divide the sum by two. For example, with a triangle having sides of lengths 13, 14, and 15, the semi-perimeter would be calculated as follows:
To calculate the semi-perimeter, you add up all the side lengths of the triangle and divide the sum by two. For example, with a triangle having sides of lengths 13, 14, and 15, the semi-perimeter would be calculated as follows:
- Add the side lengths: 13 + 14 + 15 = 42
- Divide the sum by 2: 42/2 = 21
Triangle Side Lengths
The side lengths of a triangle are fundamental to finding various triangle attributes. Here, the side lengths are given as 13, 14, and 15. From this information, we can not only find the semi-perimeter as discussed earlier, but also go further to calculate other properties such as the area and the inradius.
When working with these side lengths in specific formulas:
When working with these side lengths in specific formulas:
- Ensure each side is correctly substituted into the equations.
- Remain consistent with labeling sides as \( a \), \( b \), and \( c \), which helps in avoiding confusion.
Simplification of Expressions
Simplification of expressions is a vital skill when calculating geometric properties such as the inradius. Once the semi-perimeter \( s \) is determined, and side lengths are known, we can substitute these values into the formula for the inradius of a triangle:\[ r = \frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}\]After inputting the values, calculations often result in complex expressions. For example, in our case:\[ r = \frac{\sqrt{21 \times 8 \times 7 \times 6}}{21}\]The next step is to simplify this expression by multiplying and reducing it as much as possible:
- Calculate inside: 21 \times 8 \times 7 \times 6 = 7056
- Then take the square root: \( \sqrt{7056} \approx 84 \)
- Finally, divide by 21, simplifying to \( r \approx 4 \)
Other exercises in this chapter
Problem 161
\((b+c) \tan \frac{A}{2}+(c+a) \tan \frac{B}{2}+(a+b) \tan \frac{C}{2}=4(R+r)\)
View solution Problem 162
\(\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}+\cos ^{2} \frac{C}{2}=2+\frac{r}{2 R}\)
View solution Problem 168
\(D E F\) is the triangle formed by joining the points of contact of the incircle with the sides of the triangle \(A B C\); prove that:- i. it's sides are \(2 r
View solution Problem 169
\(A D, B E\) and \(C F\) are the perpendiculars from the angular points of a triangle \(A B C\) upon the opposite sides. Prove that the diameters of the circumc
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