A circle of radius 4 is inscribed within an equilateral triangle. Find the area of the triangle.
I am unsure on how to find the height of the triangle and the base.
If an equilateral triangle has side length s, then any altitude divides it into two right triangles with hypotenuse of length s and one leg of length s/2. By the pythagorean theorem, the other leg, the altitude of the equilateral triangle has length b given by $\displaystyle b^2+ \frac{s^2}{4}= s^2$ so that $\displaystyle b^2= s^2- \frac{s^2}{4}= \frac{3s^2}{4}$ so that $\displaystyle b= \frac{\sqrt{3}}{2}s$.
Of course all three altitudes cross at a single point. Let x be the distance from the foot of one altitude to that point so the distance from that point to a vertex is $\displaystyle \frac{\sqrt{3}}{2}s- x$. Then we have a right triangle whose vertices are that point of intersection, the foot of an altitude, and one vertex on that base. That right triangle has one leg of length s/2, another leg of length x, and hypotenuse of length $\displaystyle \frac{\sqrt{3}}{2}s- x$. Put those into the Pythagorean theorem to find x, the radius of the circle, as a function of s.
Hello, johnsy123!
Here is another solution, using some clever formulas.
A circle of radius 4 is inscribed within an equilateral triangle.
Find the area of the triangle.The area of an equilateral triangle with side $\displaystyle x$ is: .$\displaystyle A \:=\:\tfrac{\sqrt{3}}{4}x^2$ .[1]Code:A * / \ / \ / \ / \ / \ / * * * \ /* *\ * * * * / \ /* *\ / * * * \ / * | * \ / | \ / * |4 * \ / * | * \ / * | * \ B *---------------*-*-*---------------* C : - - - - - - - - x - - - - - - - - :
The area of a triangle is given by: .$\displaystyle A \:=\:\tfrac{1}{2}pr$
. . where $\displaystyle p$ is the perimeter and $\displaystyle r$ is the radius of the inscribed circle.
So we have: .$\displaystyle A \:=\:\tfrac{1}{2}(3x)(4) \:=\:6x$ .[2]
Equate [1] and [2]: .$\displaystyle \tfrac{\sqrt{3}}{4}x^2 \:=\:6x \quad\Rightarrow\quad \sqrt{3}x^2 \:=\:24x$
Since $\displaystyle x \ne 0$, divide by $\displaystyle x\!:\;\sqrt{3}x \:=\:24 \quad\Rightarrow\quad x \:=\:\tfrac{24}{\sqrt{3}} \:=\:8\sqrt{3}$
Substitute into [1]: .$\displaystyle A \;=\;\tfrac{\sqrt{3}}{4}(8\sqrt{3})^2 \;=\;48\sqrt{3}$
How do you prove that the altitude is 12? you can't really just say that the distance from the tip of the circle to the sharp point of the triangle is the same as the radius. A proven statement would better clarify my understanding.
Hi johnsey123,
Draw an equilateral triangle and its three medians.These are perpendicular to each side.Note that you have created 6 triangles which are congruent to each other.Each one is a 30-60- 90 triangle with one side given (4).(incircle radius)The alttude has another segment which you derive from the property of 30-60-90 triangle. The ratio of its sides is 2-1-rad3 so the hyp is 2 times 4= 8 making the altitude 4+8 =12.The other leg of this triangle is 4rad3= 1/2 of the side of the equil tria so the area of the equilateral tri is 12 *4rad3 = 48rad3. You could also calculate the area by caculating the area of the small tri and multiplying by 6