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Home
Practice
Pythagorean Theorem
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
.
Yes
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.
Problem 6
A company must stretch a cable from the top of a tower that is 25 meters high to a point 50 meters away from the base of the tower.
Calculate the length of the cable.
Answer:
meters.
Solution:
The triangle that is formed by these three segments is clearly rectangular, therefore we can apply the Pythagorean theorem.
$c^{2} = a^{2} +b^{2}$
$a = 50 m$
$b = 25 m$
Let $c$ be length of the cable.
According to the Pythagorean theorem:
$c^{2} = 50^{2} +25^{2}$
$ \Longrightarrow c^{2} = 2500 +625 = 3125 \text{m}^{2}$
$ \Longrightarrow c = \sqrt{3125} = 55.9$
Problem 7
Given a square plot of land with distance between two opposite vertexes of $2\sqrt{2}$ kilometers.
Calculate the total area of the plot.
Answer:
square kilometers.
Solution:
It is given the length of the diagonal of the square. It divides it into two equal triangles. In addition, the two triangles are right and the legs of the same length.
Let x be the length of square side and by the Pythagorean theorem we get:
$x^{2} +x^{2}=\left (2\sqrt{2}\right )^{2}$
$ \Longrightarrow 2x^{2}=4\left (\sqrt{2}\right )^{2}$
$ \Longrightarrow 2x^{2}=8$
$x^{2}=\frac{8}{2} \Longrightarrow x^{2}=4$ $ \Longrightarrow x=\sqrt{4}$ $ \Longrightarrow x=2 \text{km}$
Therefore, the four sides of the plot measure 2 kilometers each and, consequently, its $A = x \times x$
(The area of a square is the product of its sides) So:
$A = 4 \text{km}^{2}$
The area of the land is 4 square kilometers.
Problem 8
At sunset, a tree projects a shade that is 2.5 meters of length. The distance from the tree top to the most distant end of the shade is 4 meters. What is height of the tree?
Answer:
meters.
Solution:
We imagine a right triangle with base $b$, that is the shade of the tree.
Its height $a$, is the height of the tree and its hypotenuse, $h$, is the distance from the top of the tree to the end of the shade.
Since the triangle is right, we apply the Pythagorean theorem to calculate its height
$h^{2} = a^{2} +b^{2}$
$ \Longrightarrow 4^{2} = a^{2} +\left (2.5\right )^{2}$
$ \Longrightarrow 16 -6.25 = a^{2}$
$ \Longrightarrow 9.75 = a^{2}$
So $a = \sqrt{9.75}$
$a = 3.12$ meters.
Easy
Normal
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