/*! 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 5 The circumradius of a triangle i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 5

The circumradius of a triangle is at least twice the inradius.

Short Answer

Expert verified
The circumradius \(R\) of a triangle is always at least twice the inradius \(r\). This statement is proved by connecting the values of R and r to the triangle's area, and using the inequality that the semiperimeter \(s\) is always greater or equal to twice the circumradius. After canceling out common factors, we're left with the inequality \(R/r 鈮� 2\), confirming the initial statement.

Step by step solution

01

Rewrite the equation for area

Let's write down the area of a triangle using both the inradius and the circumradius: area = rs = abc/4R.
02

Isolate R to one side

Isolating R we have R = abc/4rs.
03

Insert inequality s 鈮� 2R

Now we know that s, the semi-perimeter of the triangle, is greater than or equal to 2R. So we replace s with 2R on the left side: abc/4r 鈮� 2abc/4R = 2Rs.
04

Cancel out common factors

Now we cancel out the common factors on both sides, so we're left with: R/r 鈮� 2.
05

Interpret the Result

This means that the circumradius R must be at least twice the inradius r.

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!

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

The internal angle bisectors of \(\triangle A B C\) are extended to meet the circumcircle at points \(L, M, N\), respectively. Find the angles of \(\triangle L M N\) in terms of the angles \(A, B\) and \(C\).

What is the (algebraically) smallest possible value that the power of a point can have with respect to a circle of given radius \(R\) ? Which point has this critical power?

The Simson lines of diametrically opposite points on the circumcircle are perpendicular to each other and meet on the nine-point circle.

Two circles are in contact internally at a point \(T\). Let the chord \(A B\) of the larger circle be tangent to the smaller circle at a point \(P\). Then the line \(T P\) bisects \(\angle A T B\).

Let \(A B C\) be an equilateral triangle inscribed in a circle with center \(O\), and let \(P\) be any point on the circle. Then the Simson line of \(P\) bisects the radius \(O P\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Statistics

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.