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Find, by Heron's formula, the area of a triangle whose sides are (i) \(13,14,15\); (ii) \(3,14,15\).

Short Answer

Expert verified
The areas of the triangles are 84 sq.units and 6 sq.units respectively.

Step by step solution

01

Calculate semi-perimeter (s)

First, one needs to calculate the semi-perimeter of the triangle. This is given by the formula \[s = \frac{a+b+c}{2}\] where \(a, b, c\) represent the lengths of the sides of the triangle.
02

Compute the area using Heron's Formula (Case: i)

Substitute the given values of \(a=13\), \(b=14\), and \(c=15\) into the semi-perimeter formula to get \(s = \frac{13+14+15}{2} = 21\). Next, compute the area \(A\) by plugging \(s, a, b, c\) into Heron's formula \[A = \sqrt{s(s-a)(s-b)(s-c)} => A = \sqrt{21(21-13)(21-14)(21-15)} => A = 84 \, sq.units\]
03

Compute the area using Heron's Formula (Case: ii)

As above, substitute the values of \(a=3\), \(b=14\), and \(c=15\) into the semi-perimeter formula to get \(s = \frac{3+14+15}{2} = 16\). Compute the area \(A\) by plugging \(s, a, b, c\) into Heron's formula \[ A = \sqrt{s(s-a)(s-b)(s-c)} => A = \sqrt{16(16-3)(16-14)(16-15)} => A = 6 \, sq.units\]

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Key Concepts

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

Triangle Area Calculation
Calculating the area of a triangle can sometimes feel tricky, especially when you don't have a right triangle to work with. However, using Heron's formula simplifies this problem for any kind of triangle as long as you know the lengths of all three sides. This method eliminates the need to know the height, which is often required in other area formulas.
To use Heron's formula, we first need to compute what is called the semi-perimeter of the triangle. Once we have the semi-perimeter, we can plug it, along with the sides of the triangle, into Heron’s formula to find the area.
Heron's formula is a great tool because it works efficiently for all different types of triangles – whether they are acute, obtuse, or even scalene. It is particularly useful when dealing with irregular shapes where traditional base-height measurements are inconvenient or impossible to obtain.
Semi-Perimeter
To calculate the area of a triangle with Heron's formula, the first step involves determining the semi-perimeter of the triangle. This is a straightforward process. You simply add up all three sides of the triangle and then divide by two. For example, for a triangle with sides of lengths 13, 14, and 15, the calculation would be
  • Add the sides: 13 + 14 + 15 = 42
  • Divide by 2 to find the semi-perimeter: \[ s = \frac{42}{2} = 21 \]
The concept of semi-perimeter might seem unfamiliar at first, but it just represents half of the perimeter of the triangle. It effectively helps us in applying Heron's formula by serving as an intermediate value that simplifies the final area calculation.
Although it might seem like an extra step, calculating the semi-perimeter ensures that the subsequent steps in Heron's formula become much more manageable and well-organized.
Mathematical Problem Solving
When tackling a mathematical problem like finding the area of a triangle using Heron's formula, it helps to follow a systematic problem-solving approach.
First, clearly understand the problem statement and the given values. In our exercise, we need to find areas for two sets of triangle side lengths.
Next, apply the appropriate formulas. First, calculate the semi-perimeter, and then use Heron's formula to find the area. Breaking the problem into smaller, manageable steps, such as calculating the semi-perimeter before the final area, prevents confusion and errors.
Finally, double-check your computations. Once you've calculated the area, ensure your results make sense given the context and the numbers. Solutions that seem surprisingly large or small often signal errors in calculation.
This structured approach not only applies to area calculations but is useful across all areas of math. Developing these problem-solving skills will make math more approachable and less intimidating.

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Most popular questions from this chapter

If a quadrangle is inscribed in a circle, the product of the distances of a point on the circle from two opposite sides is equal to the product of the distances of the same point from the other two sides, and also to the product of the distances of the same point from the diagonals.

For a triangle \(A B C\), express the inradius \(r\) in terms of \(s, s-a\), \(s-b, s-c\).

In the notation of Figure \(3.3 \mathrm{~B}\), (i) The lines \(P O_{1}, Q O_{2}, R O_{3}\) all pass through \(O\), the circumcenter of \(\triangle A B C\); (ii) The lines \(\mathrm{AO}_{1}, \mathrm{BO}_{2}, \mathrm{CO}_{3}\) are concurrent; (iii) The segments \(A P, B Q, C R\) all have the same length, all pass through the common point \(F\) of the three circumcircles, and meet one another at angles of \(60^{\circ}\). (This point is named \(F\) for Fermat, who obtained it, when no angle of \(\triangle A B C\) exceeds \(120^{\circ}\), as the point whose distances from \(A, B\), and \(C\) have the smallest sum.)

If squares are erected on two sides of a triangle, their circumcircles intersect on the circle whose diameter is the third side, and the centers of these three circles are the vertices of an isosceles right-angled triangle.

Two non-parallel lines are drawn on a sheet of paper so that their theoretical intersection is somewhere off the paper. Through a point \(P\), selected on the part of the paper between the lines, construct the line that would, when sufficiently extended, pass through the intersection of the given lines. What would the same construction yield if we applied it to two parallel lines?

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