   Chapter 10.3, Problem 5E ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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# In Exercises 5 to 10, the real numbers a, b, c, and d are positive.Consider the triangle with vertices at A 0 ,   0 , B a ,   0 , and C a ,   b . Explain why ∆ A B C is a right triangle. To determine

To explain:

ABC is a right triangle.

Explanation

The given figure is shown below.

To prove ABC is a right triangle, we have to prove the following statement.

The midpoint of the hypotenuse of a right triangle is equidistant from the three vertices of the triangle.

The given vertices are A0, 0, Ba, 0, and Ca, b.

The endpoints of the hypotenuse line are A0, 0 and Ca, b.

Find the midpoint of the hypotenuse line.

x1,y1=0, 0

x2,y2=a, b

Formula for midpoint of the line segment joining two points.

M=x1+x22, y1+y22

Substituting the x and y co-ordinates.

M=0+a2, 0+b2

M=a2, b2

Let this midpoint be named as Da2, b2.

Now, find the distance between this midpoint D and A.

x1,y1=Da2, b2

x2,y2=A0, 0

Formula for distance between two points

dDA=x2-x12+y2-y12

Substituting the x and y co-ordinates

dDA=0-a22+0-b22

On solving this,

dDA=-a22+-b22

Squaring,

dDA=a24+b24

dDA=a2+b24

Taking square root,

dDA=a2+b24

dDA=a2+b22

Now, find the distance between this midpoint D and B

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