Pythagorean Theorem: Problems with Solutions

Solution:
According to the Pythagorean Theorem: [tex]AB^2=BC^2+AC^2[/tex], so [tex]AC^2=AB^2-BC^2=8^2-5^2=64-25=39[/tex].

Solution:
[tex]3^{2} + 4^{2} = 5^{2}[/tex]
because [tex]25=25[/tex] so a right triangle can have sides 3, 4, 5

Solution:
According to the Pythagorean Theorem: [tex]AB^2=AC^2+BC^2=3^2+4^2=9+16=25[/tex]. Since [tex]AB^2=5^2[/tex], AB=5.

Solution:
The Pythagorean Theorem states that:
[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]

Solution:
According to the Pythagorean Theorem: [tex]AB^2=AC^2+BC^2=8^2+15^2=64+225=289[/tex]. Since [tex]AB^2=17^2[/tex], AB=17.

Solution:
The triangle that is formed by these three segments is clearly rectangular, therefore we can apply the Pythagorean theorem.
$c^{2} = a^{2} +b^{2}$
$a = 50\text{ ft}$
$b = 25\text{ ft}$
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{ ft}^{2}$
$ \Longrightarrow c = \sqrt{3125} = 55.9$

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{mi}$
Therefore, the four sides of the plot measure 2 miles each and, consequently, its $A = x \times x$
(The area of a square is the product of its sides) So:
$A = 4 \text{mi}^{2}$
The area of the land is 4 square miles.

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 $a = \sqrt{9.75}$
$a = 3.12$ feet.

Solution:
In a right triangle the leg opposite the [tex]30^\circ[/tex] angle is half the hypotenuse, so [tex]AC = \frac{AB}{2}[/tex].
[tex]AC = \frac{6}{2} = 3[/tex]
By the Pythagorean Theorem:
[tex]AB^2 = BC^2 + AC^2[/tex]
[tex]6^2 = BC^2 + 3^2[/tex]
[tex]BC^2 = 36 - 9 = 27[/tex]
[tex]BC = \sqrt{27} = 3\sqrt{3}[/tex]

Solution:
We can divide the triangle into two right triangles. (Look at the image below.)
By the Pythagorean Theorem we get the equations

$\left( \sqrt{2}\right)^{2}=x^{2}+a^{2}$
$\left( \sqrt{5}\right)^{2}=y^{2}+a^{2}$
And we know that $x + y = 3$

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Now
$\left( \sqrt{2}\right)^{2} -x^{2} =a^{2}$
$\left( \sqrt{5}\right)^{2}- y^{2}=a^{2}$

So $\left( \sqrt{2}\right)^{2} -x^{2} = \left( \sqrt{5}\right)^{2}- y^{2}$

We get 2 simultaneous equations:
$\left\{ \begin{array}{c} 2-x^{2}=5-y^{2} \\ x+y=3 \end{array} \right. $

After substituting $x=3-y$ into the first equation and get

$2-\left( 3-y\right)^{2}=5-y^{2}\Longrightarrow 2-9+6y-y^{2}=5-y^{2}$

So $6y=5+7\Longrightarrow y=2$
Since $5=y^{2}+a^{2}$

Then $a=\sqrt{5-y^{2}}=\sqrt{5-4}=1$

Solution:
The center of the circumscribed circle is the intersection point of the diagonals, so the radius is half of a diagonal.
By the Pythagorean theorem: [tex]d^2=6^2+6^2=72 \Rightarrow d=\sqrt{72}=6\sqrt{2}[/tex]
[tex]R=\dfrac{d}{2}=3\sqrt{2}[/tex]

Solution:
The height to the base is also a median, so it divides the base into two segments of length [tex]12[/tex]:
[tex]h^2=13^2-12^2=169-144=25 \Rightarrow h=5[/tex]
[tex]S=\dfrac{24\times 5}{2}=60[/tex]

Solution:
The height divides the triangle into two right triangles with hypotenuse [tex]6[/tex] and leg [tex]3[/tex]:
[tex]h^2=6^2-3^2=27 \Rightarrow h=\sqrt{27}=3\sqrt{3}[/tex]
[tex]S=\dfrac{6\times 3\sqrt{3}}{2}=9\sqrt{3}[/tex]

Solution:
The two tops and the point of the taller pole at height [tex]20[/tex] form a right triangle with legs [tex]40[/tex] and [tex]50-20=30[/tex]:
[tex]d^2=40^2+30^2=1600+900=2500 \Rightarrow d=50[/tex]

Solution:
The heights from the ends of the shorter base cut off from the longer base two equal segments of length
[tex]\dfrac{21-11}{2}=5[/tex]
By the Pythagorean theorem: [tex]x^2=12^2+5^2=144+25=169 \Rightarrow x=13[/tex]

Solution:
The perpendicular from the center to the chord bisects the chord. We get a right triangle with hypotenuse [tex]10[/tex] (a radius) and leg [tex]8[/tex] (half of the chord):
[tex]d^2=10^2-8^2=100-64=36 \Rightarrow d=6[/tex]

Solution:
By the Pythagorean theorem: [tex]c^2=12^2+16^2=144+256=400 \Rightarrow c=20[/tex]
The median to the hypotenuse equals half of the hypotenuse (the midpoint of the hypotenuse is the center of the circumscribed circle):
[tex]m=\dfrac{20}{2}=10[/tex]

Solution:
The side of the rhombus is [tex]100\div 4=25[/tex]. The diagonals of a rhombus are perpendicular and bisect each other, so a side and the halves of the diagonals form a right triangle:
[tex]\left(\dfrac{d}{2}\right)^2=25^2-15^2=625-225=400 \Rightarrow \dfrac{d}{2}=20 \Rightarrow d=40[/tex]

Solution:
North and east are perpendicular directions, so the distance is the hypotenuse of a right triangle with legs [tex]9[/tex] and [tex]12[/tex]:
[tex]d^2=9^2+12^2=81+144=225 \Rightarrow d=15[/tex]