Area of a Triangle

From Heron's Formula, we have Area of triangle = \sqrt{(s)(s-a)(s-b)(s-c)}

Where s = (a + b + c)/2 (i.e. semiperimeter) a, b, c are the sides of the triangle (13,14,15 respectively)

Subbing in the values would thus lead to the answer, 84. (can be done without a calculator by prime factorisation)

area of triangle $= (ab \sin x)/2$ where $x$ is the angle between sides $a$ and $b$ , This follows from seeing that with a base of $a$ , the corresponding height will be $b \sin x$ . Now from cosine rule, $\cos x = \frac {a^2 + b^2 - c^2}{2ab}$ , where $c$ is the third side, Now taking $a, b, c$ as 14, 15, 13 respectively, $\cos x = 3/5$ therefore, $\sin x = 4/5$ so area $= (ab \sin x)/2 = (14*15*4/5)/2 = 84$ .

You can use Heron's Formula to find the area of this triangle. First, find the semi-perimeter of the triangle which is (13+14+15)/2=21. Next, you plug in the values into Heron's Formula: Recap of Heron's Formula. s=semi-perimeter, a,b,c=sides of the triangle:
sqrt[s(s-a)(s-b)(s-c)]. So now you have sqrt[21(21-13)(21-14)(21-15)]=sqrt[21(8)(7)(6)]=84

s=a+b+c=42 area of triangle=root of s(s-a)(s-b)(s-c) =42(29)28)(27) by taking away square root =3 7 3*2 =84

To find the area of a scalene triangle when the length of all its three sides are given Heron’s formula is used \sqrt{s(s-a)(s-b)(s-c)} Where s is the Semi perimeter of the triangle that is s=\frac {a+b+c}{2} a,b and c are sides of the triangle .In this case the value of a,b and c is 13,14 and 15 respectively. Now we will find s first s=\frac {13+14+15}{2} s=\frac {42}{2} s=21 Now putting these values in Heron's formula \sqrt{21(21-13)(21-14)(21-15)} \sqrt{21(8)(7)(6)} \sqrt{7056} 84 So the area of the triangle is 84

Solution 1

Label the triangle as $ABC$ with $AB = 13$ , $AC = 14$ and $BC = 15$ . Let $AD$ be the height of the triangle from A to BC. Let $x = \overline{CD}$ . After solving for $x$ by using the Pythagorean Theorem, the height $AD$ can be solved as well. Doing so

$\sqrt{\overline{AB}^2 - \overline{BD}^2} = \sqrt{\overline{AC}^2 - \overline{CD}^2}$ $\Rightarrow\overline{AB}^2 - \overline{BD}^2 = \overline{AC}^2 - \overline{CD}^2$ $\Rightarrow13^2 - (15 - x)^2 = 14^2 - x^2$ $\Rightarrow169 - 225 +30x -x^2 = 196 - x^2$ $\Rightarrow30x = 225 + 196 -169$ $\Rightarrow x = \overline{CD} = \frac{42}{5}$

Hence,

$\overline{AD} = \sqrt{14^2 - (\frac{42}{5})^2}$ $\Rightarrow \overline{AD} = \frac{56}{5}$

And the area of triangle $ABC$ is

$Area = \frac{1}{2} \times15 \times \frac{56}{5}$ $\Rightarrow Area = 84$

Solution 2

Use the Heron's Formula,

$Area = \sqrt{(s)(s - a)(s - b)(s - c)}$

where $a$ , $b$ and $c$ are the sides and $s$ is half of the perimeter. Thus, the area of a $13-14-15$ triangle is

$s = \frac{1}{2}(13 + 14 + 15) = 21$ $Area = \sqrt{(21)(21 - 13)(21 - 14)(21 - 15)}$ $\Rightarrow Area = \sqrt{(21)(8)(7)(6)} = 84$ .

This question asks for the area of a triangle with side lengths of 13 , 14 , and 15 . Given all three side lengths, one can calculate the area using Heron's Formula , which is $\sqrt{s(s-a)(s-b)(s-c)}$ where s is the semiperimeter ( half the perimeter ), and a, b, and c are the side lengths. To find the semiperimeter, add all side lengths and divide by two: $\frac {a+b+c}{2}$ , which when the lengths are plugged in: $\frac {13+14+15}{2}$ , gives $\frac {42}{2}$ = 21 . One can then plug this answer in along with the three side lengths into Heron's formula: $\sqrt{21(21-13)(21-14)(21-15)}$ . Simplifying, each step is: $\sqrt{21(8)(7)(6)}$ , then $\sqrt{7056}$ , which leaves $\pm 84$ . As a triangle, the area cannot be negative, so the answer is 84 units squared

By the law of cosines, in a triangle with angles A, B, and C and sides a, b, and c opposite their respective angle, c^2=a^2 + b^2 - 2 * a * b * cos(C). Substituting 14 for a, 13 for b, and 15 for c, 15^2=14^2 + 13^2 - 2 * 14 * 13 * cos(C). Arithmetic will then show that cos(C)=0.384615 and C = 67.380135 degrees.

It is also known that in the same triangle ABC as mentioned above, (a * b * sin(C))/2=Area of triangle. Plugging in the values we found earlier, we find that the area of the triangle is 84.

Since we are given the three sides of a triangle here, we can use the Heron's formula to determine the area of the triangle.

The Heron's Formula: \sqrt{s(s-a)(s-b)(s-c)} with a, b, and c being the respective sides of the triangle and s being the semi perimeter of the triangle: \frac {a+b+c}{2}.

In this case, the semi perimeter is {13+14+15}{2} which comes out to 21.

Now, plugging values into the equation, we get \sqrt{21(21-13)(21-14)(21-15)}, which comes out to 84, the are of the triangle.

Recall Heron's formula, in a triangle of side lengths a, b and c, Area= \sqrt{P(P-a)(P-b)(P-c)} where P=\frac {a+b+c}{2}

In the question P=\frac {13+14+15}{2}=21 Area=\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{7056}=84

$Area = \sqrt{s(s-a)(s-b)(s-c)} = 84$ where $s = \frac{a+b+c}{2} = 14$

The area of any triangle can be found by taking the square root of s(s-a)(s-b)(s-c), where semiperimeter s = (a+b+c)/2. Substitution of a,b,c = 13,14,15 in no particular order gives the answer of 84.

Heron's formula: Let a, b, c be side lengths of a triangle.

The area of the triangle is (s(s-a)(s-b)(s-c))^(1/2), where s is the semiperimeter of the triangle, s = (a+b+c)/2.

Substitute a = 13, b = 14, c = 15 into Heron's Formula and the answer is 84.

• Reference (Wikipedia) : http://en.wikipedia.org/wiki/Heron's_formula

According to Heron's Formula, the area $A$ , of a triangle with sides $a, b$ and $c$ is given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter of the triangle which is equal to $\dfrac{a+b+c}{2}$ .

So using Heron's Formula with $a=13, b=14, c=15$ :

$s=\dfrac{13+14+15}{2}=21$

Hence,

$A=\sqrt{21(21-13)(21-14)(21-15)}=84$

Heron's Formula is used to solve for an area of a triangle when we know the value of the sides of the triangle.

It states that, given a triangle with sides $x , y$ and $z$ , we can get the area by substituting these values into this formula: $\sqrt{s \times (s-x) \times (s-y) \times (s-z)}$ , where $s = \frac{x + y + z}{2}$ .

$A = \sqrt{p(p-a)(p-b)(p-c)}$ where $a, b, c$ are the sides and $p = \frac{a+b+c}{2}$ . That's all there is to it. A proof can be found in wikipedia and possibly in most geometry textbooks.

Solution 1: Let triangle $ABC$ have lengths $AB=13, BC=14$ and $CA=15$ . From $A$ , drop a perpendicular intersecting $BC$ at $D$ . Now consider triangles $A'B'D'$ and $A'C'D'$ , with right angle at $D'$ and $B', C'$ on opposite sides of $A'D'$ . Let $B'D'=5, D'A'=12$ and $D'C'=9$ . By the Pythagorean theorem, $A'B' = \sqrt{5^2 + 12^2} = 13$ and $A'C' = \sqrt{9^2 + 12^2} = 15$ . Since $\angle B'D'A' + \angle C'D'A' = 90^\circ + 90^\circ = 180^\circ$ , thus $B'D'C'$ is a straight line, and so $B'C' = 5 + 9 = 14$ . Thus, triangles $ABC$ and $A'B'C'$ are congruent by side-side-side, hence they have area $\frac {1}{2}\times12\times14 = 84$ .

Solution 2: By Heron's Formula, the semi-perimeter is $\frac {13+14+15}{2} = 21$ and the area of the triangle is $\sqrt{21 \times (21-13) \times (21-14) \times (21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = 84$ .