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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
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