MENU
❌
Home
Math Forum/Help
Problem Solver
Practice
Algebra
Geometry
Tests
College Math
History
Games
MAIN MENU
1 Grade
Adding and subtracting up to 10
Comparing numbers up to 10
Adding and subtracting up to 20
Addition and Subtraction within 10 or 20
2 Grade
Adding and Subtracting up to 100
Addition and Subtraction within 10 or 20
3 Grade
Addition and Subtraction up to 1000
Multiplication up to 5
Multiplication Table
Rounding
Dividing
Perimeter
4 Grade
Adding and Subtracting
Equivalent Fractions
Divisibility by 2, 3, 4, 5, 9
Area of Squares and Rectangles
Fractions
Equivalent Fractions
Least Common Multiple
Adding and Subtracting
Fraction Multiplication and Division
Operations
Mixed Numbers
Decimals
6 Grade
Percents
Signed Numbers
The Coordinate Plane
Equations
Polynomials
Symplifying Expressions
Polynomial Expressions
Factoring
7 Grade
Angles
Parametric Linear Equations
Word Problems
Exponentiation
Roots
Quadratic Equations
Vieta's Formulas
Progressions
Arithmetic Progressions
Geometric Progression
Progressions
Number Sequences
Reciprocal Equations
Logarithms
Logarithmic Expressions
Logarithmic Equations
Extremal value problems
Trigonometry
Geometry
Intercept Theorem
Law of Sines
Law of Cosines
Probability
Limits of Functions
Properties of Triangles
Pythagorean Theorem
Inverse Trigonometric Functions
Analytic Geometry
Easy
Normal
Pythagorean Theorem - Problems with Solutions
Problem 1
Given a triangle
ABC
with [tex]\angle C = 90 ^{\circ}[/tex],
AB=8
and
BC=5
, find the square of
AC
.
Solution:
By the Pythagorean theorem we have [tex]AB^2=BC^2+AC^2[/tex], so [tex]AC^2=AB^2-BC^2=8^2-5^2=64-25=39[/tex].
Problem 2
Could a right triangle have sides: 3, 4, 5?
Answers:
yes
or
no
.
Solution:
3
^{2}
+ 4
^{2}
= 5
^{2}
because 25=25 so a right triangle can have sides 3, 4, 5
Problem 3
Given a right triangle
ABC
, [tex]\angle C = 90 ^{\circ}[/tex], in which
AC=3
,
BC=4
. Determine the length of
AB
.
Solution:
By the Pythagorean theorem we have: [tex]AB^2=AC^2+BC^2=3^2+4^2=9+16=25[/tex]. Since [tex]AB^2=5^2[/tex], AB=5.
Problem 4
Given a right triangle
ABC
, [tex]\angle C = 90 ^{\circ}[/tex], in which
AC=7
,
AB=25
. Determine the length of
BC
.
Solution:
By the Pythagorean theorem we have:
[tex]AB^2=AC^2+BC^2[/tex]
[tex]25^2 = 7^2+BC^2[/tex]
[tex]625 - 49=BC^2[/tex]
Hence, [tex]BC=\sqrt{576}=24[/tex]
Problem 5
Given a right triangle
ABC
, [tex]\angle C = 90 ^{\circ}[/tex], in which
AC=8
,
BC=15
. Determine the length of
AB
.
Solution:
By the Pythagorean theorem we have: [tex]AB^2=AC^2+BC^2=8^2+15^2=64+225=289[/tex]. Since [tex]AB^2=17^2[/tex], AB=17.
Easy
Normal
Submit a problem on this page.
Problem text:
Solution:
Answer:
Your name(if you would like to be published):
E-mail(you will be notified when the problem is published)
Notes
: use [tex][/tex] (as in the forum if you would like to use latex).
Correct:
Wrong:
Solution seen:
Contact email:
Follow us on
Twitter
Google+
Facebook
Author
Copyright © 2005 - 2018.