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Chapter 28: Chapter 28

Problem 536

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Find, in radical form, the length of the radius of a circle circumscribed about an equilateral triangle, the length of whose side is 24 .

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The length of the radius of the circle circumscribed about the equilateral triangle with a side length of 24 is 24.
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Understanding the problem description

We are given that there is an equilateral triangle with side length 24. A circle is drawn such that it touches all the three corners of the equilateral triangle. The goal is to find the radius of this circumscribed circle.

Step 2: Drawing the diagram

First, draw an equilateral triangle and then draw a circumscribed circle touching all the three vertices of the triangle. Let the vertices of the triangle be A, B, and C, and the center of the circle be O.

Step 3: Deriving the relationship between radius and side length

Let's draw an altitude from point A of the triangle to the opposite side BC, and call the point where the altitude intersects BC as point D. Altitude AD will be perpendicular to BC. Since the triangle is equilateral, AD will also bisect side BC, with BD = DC = 12. Now, let's draw a radius from the center of the circle O to point A, creating triangle AOD. Since the triangle is equilateral, and the altitudes in equilateral triangles are also medians, it means that point D is the midpoint of side BC, and OD is the perpendicular bisector of BC. As the perpendicular bisector, OD is perpendicular to BC and angle ODA forms a right angle. So, triangle AOD is a right-angled triangle with angle AOD = 90 degrees. Now, we also know that angles in an equilateral triangle are all 60-degree angles. Thus, angle ADO is 60 degrees because it is an angle of triangle A. Therefore, in a right angle triangle with angles (60, 90, 30), the shortest side (here OD) is half the length of the hypotenuse (OA). Here, OA is the radius we want to find. So, we have: $$radius = OA = 2 * OD$$

Step 4: Finding the length of side OD

We already know that BD = DC = 12. And in right-angled triangle ADB, angle ABD = 90 degrees, angle ADB = 30 degrees, and angle BAD = 60 degrees. Using the Pythagorean theorem, we can find the length of AD: $$AD^2 = AB^2 - BD^2$$ $$AD^2 = 24^2 - 12^2$$ $$AD^2 = 576 - 144$$ $$AD^2 = 432$$ $$AD = 12\sqrt{3}$$ Now, in right-angled triangle AOD, angle ADO = 60 degrees, and hypotenuse OA is equal to the radius. We can use the sine rule: $$\sin{ADO} = \frac{AD}{radius}$$ $$\sin{60°} = \frac{12\sqrt{3}}{radius}$$ As we know, \(\sin{60°} = \frac{\sqrt{3}}{2}\), so: $$\frac{\sqrt{3}}{2} = \frac{12\sqrt{3}}{radius}$$

Step 5: Solve for the radius

Now we can solve for the radius: $$radius = \frac{2 \cdot 12\sqrt{3}}{\sqrt{3}}$$ $$radius = \frac{24\sqrt{3}}{\sqrt{3}}$$ $$radius = 24$$ So, the length of the radius of the circle circumscribed about the equilateral triangle with a side length of 24 is 24.

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Most popular questions from this chapter

Chapter 28

In the figure, let \(\mathrm{ABCDE}\) be a regular pentagon touching the circle at points \(\mathrm{M}_{1}, \mathrm{M}_{2}, \mathrm{M}_{3}, \mathrm{M}_{4}\) and \(\mathrm{M}_{5}\). Assume that \(\mathrm{OA}=\mathrm{OB}=\mathrm{OC}=\mathrm{OD}=\mathrm{OE}\). (a) Prove that \(\triangle \mathrm{AM}_{1} \mathrm{O} \cong \Delta \mathrm{BM}_{1} \mathrm{O}\) (b) Prove that $\triangle \mathrm{AOB} \cong \triangle \mathrm{BOC} \cong \ldots \triangle \mathrm{EOA}$. (c) Sow that \((\theta / 2)=36^{\circ}\). (d) Prove that \(\mathrm{AM}_{1}=\mathrm{r} \tan 36^{\circ}\). (e) Prove that the perimeter equals \(10 \mathrm{r} \tan 36^{\circ}\).
Chapter 28

Show that the perimeter of an n-sided polygon circumscribed about a circle whose radius has length \(\mathrm{r}\) is \(2 \mathrm{nr} \tan \pi / \mathrm{n}\).
Chapter 28

Find the perimeter of a regular polygon of \(\mathrm{n}\) sides which is circumscribed about a circle of radius length 10 when \(\mathrm{n}\) equals a) \(3 ;\) b) \(4 ;\) c) \(6 ;\) d \(9 ;\) e) \(12 ;\) f) \(18 ;\) g) 36 .
Chapter 28

Show that for a triangle of area \(\mathrm{A}\), and perimeter \(\mathrm{P}\), the radius of the inscribed circle, \(\mathrm{r}\), equals $2 \mathrm{~A} / \mathrm{P}$.
Chapter 28

Show that the perimeter of a square which circumscribes a circle of radius of length \(\mathrm{r}\) is \(8 \mathrm{r}\).
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