/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 An equilateral triangle has a ci... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 45

An equilateral triangle has a circle inscribed in it and is circumscribed by a circle. There is another equilateral triangle inscribed in the inner circle. Find the ratio of the areas of the outer circle and the inner equilateral triangle. (1) \(\frac{16 \pi}{3 \sqrt{3}}\) (2) \(\frac{8 \pi}{2 \sqrt{3}}\) (3) \(\frac{24 \pi}{3 \sqrt{3}}\) (4) None of these

Short Answer

Expert verified
Answer: (4) None of these (The correct ratio is $\frac{16\pi}{27}$)

Step by step solution

01

Assign variables

Let's denote the side length of the outer equilateral triangle as a, the radius of the outer circle as R, and the radius of the inner circle as r. Let the side length of the inner equilateral triangle be x.
02

Find the relationship between r and x

Let's draw an altitude from one vertex of the inner equilateral triangle to the midpoint of the opposite side. This altitude will divide the equilateral triangle into two 30-60-90 right triangles, with the longer leg (half the side length) being x/2, and the shorter leg being the radius (r). Using the properties of 30-60-90 triangles, we can write the relationship as: \[r = \frac{x\sqrt{3}}{4}\]
03

Find the relationship between R and a

By drawing an altitude from one vertex of the outer equilateral triangle to the midpoint of the opposite side, we can find a relationship between its radius, R, and its side length, a. As with the inner equilateral triangle, this will be a 30-60-90 triangle, with R being the longer leg. Therefore, we can write the relationship as: \[R = \frac{a\sqrt{3}}{6}\]
04

Find the relationship between x and a

To find the relationship between the side lengths x and a, notice that both the outer and inner equilateral triangles share the same altitude when drawn from the same vertex. This altitude divides both triangles into two 30-60-90 right triangles. Using the properties of 30-60-90 triangles, we can write: \[\frac{x}{2} = a - R - r = a - \frac{a\sqrt{3}}{6} - \frac{x\sqrt{3}}{4}\] Solving for x, we get: \[x = \frac{a(6 - \sqrt{3})}{2}\]
05

Find the areas of the outer circle and inner equilateral triangle

Using the expressions for R and x, the area of the outer circle (A_circle) and the inner equilateral triangle (A_triangle) can be expressed as: \[A_{circle} = \pi R^2 = \pi \left(\frac{a\sqrt{3}}{6}\right)^2 = \frac{\pi a^2 \sqrt{3}}{12}\] \[A_{triangle} = \frac{\sqrt{3}x^2}{4} = \frac{\sqrt{3}}{4} \left(\frac{a(6 - \sqrt{3})}{2}\right)^2 = \frac{9a^2 \sqrt{3}}{16}\]
06

Find the ratio of the areas

Now, we just need to find the ratio of A_circle to A_triangle: \[\frac{A_{circle}}{A_{triangle}} = \frac{\frac{\pi a^2 \sqrt{3}}{12}}{\frac{9a^2 \sqrt{3}}{16}} = \frac{16\pi}{9(3)}\] Comparing with the answer choices, we see that the ratio of the areas of the outer circle and the inner equilateral triangle is given by: \[\boxed{\frac{16\pi}{27}}\] Unfortunately, this result is none of the given options. The correct answer is (4) None of these.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Equilateral Triangle
An equilateral triangle is a special type of triangle where all three sides are of the same length, and all three angles are equal to 60 degrees. Since all sides and angles are equal, equilateral triangles are a classic example of geometric symmetry and simplicity. This type of triangle forms the basis of many geometric constructions and properties. For instance:
  • The height of an equilateral triangle can be calculated as \(\frac{\sqrt{3}}{2}a\), where \(a\) is the side length.
  • Each altitude (or height) equally bisects the opposite side, forming two 30-60-90 triangles within the equilateral triangle.
  • Its area can be determined using the formula: \(\frac{\sqrt{3}}{4}a^2\).
Its uniformity makes it the perfect candidate for inscribing and circumscribing circles, crucial in many geometric problem-solving scenarios.
Inscribed Circle
An inscribed circle, or incircle, of a triangle is the largest circle that fits inside the triangle and touches all three sides. The center of this circle is the incenter, which is also the point where the angle bisectors of the triangle's angles meet.In an equilateral triangle:
  • The radius \(r\) of the inscribed circle can be found using the formula: \(r = \frac{a\sqrt{3}}{6}\).
  • The diameter of the incircle is half the height of the triangle.
  • An incircle maximizes its area within the confines of the triangle, effectively showing the efficiency of the equilateral triangle's design.
An inscribed circle serves as a foundational concept in understanding nested geometric shapes and has numerous applications in real-world engineering and design.
Circumscribed Circle
A circumscribed circle, or circumcircle, is defined as the circle that passes through all the vertices of a polygon, in this case, a triangle. The center of this circle is known as the circumcenter, a significant point that can be located by intersecting the perpendicular bisectors of the triangle's sides.For an equilateral triangle:
  • The formula for the radius \(R\) of the circumcircle is \(R = \frac{a\sqrt{3}}{3}\).
  • The circumcenter is equidistant from all three vertices and coincides with the centroid and incenter in equilateral triangles due to its symmetry.
Understanding circumscribed circles helps in solving complex problems involving multiple shapes, like nested circles and triangles, and their configurations.
30-60-90 Triangle
The 30-60-90 triangle is a right triangle with angles measuring 30, 60, and 90 degrees. It鈥檚 a specific type of right triangle that arises naturally when you draw an altitude in an equilateral triangle. Key properties include:
  • The sides of a 30-60-90 triangle are in the ratio 1:鈭�3:2.
  • The shortest side is opposite the smallest angle (30 degrees), the longest side (the hypotenuse) is twice as long as the shortest side, and the medium side is 鈭�3 times the shortest side.
  • This consistent ratio makes calculations involving these triangles straightforward and is used in many geometry problems.
The 30-60-90 triangle is often used to simplify problems involving equilateral triangles and circles, as seen in the exercise of inscribing and circumscribing circles within and around equilateral triangles.
Area Ratio
Finding the ratio of areas is a common task in geometry, especially when dealing with figures like triangles and circles where comparisons provide insight into their relative sizes. In the problem at hand:
  • The outer circle is the circumcircle of the larger equilateral triangle.
  • The inner equilateral triangle fits snugly within the incircle of the larger equilateral triangle, which provides a base point for calculation.
  • Using the known formulas for the areas of circles and triangles allows for the derivation of an area ratio.
This exercise of comparing areas gives rise to deeper understanding of space, shape hierarchy, and the relationships between different geometric shapes, highlighting the elegance and complexity of geometry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Tivo hemispherical vessels can hold \(10.8\) litres and 50 litres of liquid respectively. The ratio of their inner curved surface areas is (1) \(16: 25\) (2) \(25: 9\) (3) \(9: 25\) (4) \(4: 3\)

What is the difference in the areas of the regular hexagon circumscribing a circle of radius \(10 \mathrm{~cm}\) and the regular hexagon inscribed in the circle? (1) \(50 \mathrm{~cm}^{2}\) (2) \(50 \sqrt{3} \mathrm{~cm}^{2}\) (3) \(100 \sqrt{3} \mathrm{~cm}^{2}\) (4) \(100 \sqrt{3} \mathrm{~cm}^{2}\)

A metal cube of edge \(\frac{3 \sqrt{2}}{\sqrt{5}} \mathrm{~m}\) is melted and formed into three smaller cubes. If the edges of the two smaller cubes are \(\frac{3}{\sqrt{10}} \mathrm{~m}\) and \(\frac{\sqrt{5}}{\sqrt{2}} \mathrm{~m}\), find the edge of the third smaller cube. (1) \(\frac{3}{\sqrt{7}} \mathrm{~m}\) (2) \(\frac{6}{\sqrt{15}} \mathrm{~m}\) (3) \(\frac{5}{\sqrt{11}} \mathrm{~m}\) (4) \(\frac{4}{\sqrt{10}} \mathrm{~m}\)

The volume of a hemisphere is \(2.25 \pi \mathrm{cm}^{3}\). What is the total surface area of the hemisphere. (1) \(2.25 \pi \mathrm{cm}^{2}\) (2) \(5 \pi \mathrm{cm}^{2}\) (3) \(6.75 \pi \mathrm{cm}^{2}\) (4) \(4.5 \pi \mathrm{cm}^{2}\)

The base of a right prism is a square of perimeter \(20 \mathrm{~cm}\) and its height is \(30 \mathrm{~cm}\). What is the volume of the prism? (1) \(700 \mathrm{~cm}^{3}\) (2) \(750 \mathrm{~cm}^{3}\) (3) \(800 \mathrm{~cm}^{3}\) (4) \(850 \mathrm{~cm}^{3}\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.