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Pythagorean theorem is a theorem of a great mathematician namely Theorem. When a triangle has a right angle (90°) and squares are made on each of the three sides then the biggest square has the exact same area as the other two squares. Itis called It can be written in one short equation: a² = b² + c²
Example 1
Calculate the hypotenuse a when the sides are b = 5 cm and c = 12 cm
Solution:
a^{2 }= b^{2}+ c^{2}
= 5^{2}+ 12^{2}
= 25 + 144
= 169
Therefore, √169 = 13 cm
The distance d of a point (x,y) from the origin.
According to the Pythagorean theorem, and the meaning of the rectangular coordinates (x, y), d^{2} = x^{2} + y^{2}
Therefore, d = \(\sqrt{x^2 + y^2}\)
Example 2
How far from the origin is the point (4, -2)?
Solution:
d = \(\sqrt{4^2+(-2)^2}\)
= \(\sqrt{16+4}\)
=\(\sqrt{20}\)
.
Solution:
Here, b=12 cm, h=13 cm, p=x=?
According to pythagoras theorem,
h^{2}=p^{2}+b^{2}
or, 13^{2}=x^{2}+12^{2}
or, x^{2}=13^{2}-12^{2}
or, x^{2}=169-144
or, x^{2}=25
or, x=\(\sqrt{25}\)
\(\therefore\) x=5 cm
Solution:
Here, p=14 cm, b=48 cm, h=x=?
According to pythagoras theorem,
h^{2}=p^{2}+b^{2}
or,x^{2}=14^{2}+48^{2}
or, x^{2}=196+2304
or, x^{2}=2500
or, x=\(\sqrt{2500}\)
\(\tharefore\) x=50 cm
Solution;
Here, p=21 cm, b=72 cm, h=x=?
According to pythagoras theorem,
h^{2}=p^{2}+b^{2}
or, x^{2}=21^{2}+72^{2}
or, x^{2}=441+5184
or, x^{2}=5625
or, x=\(\sqrt{5625}\)
\(\therefore\) x=75 cm
Solution;
Here, p= 6 inch, b=8 inch, h=x=?
According to pythagoras theorem
h^{2}=p^{2}+b^{2}
or,x^{2}=6^{2}+8^{2}
or, x^{2}=36+64
or, x^{2}=100
or, x=\(\sqrt{100}\)
\(\therefore\) x=10 inch
Solution:
Here, p= 15ft, h=25ft and b=x=?
According to pythagoras theorem
h^{2}=p^{2}+b^{2}
or, 25^{2}=15^{2}+x^{2}
or, x^{2}=25^{2-}15^{2}
or, x^{2}=625-225
or, x^{2}=400
or, x=\(\sqrt{400}\)
\(\therefore\) x=20ft
Solution;
Here, b=\(\sqrt{5}\) cm, h=\(\sqrt{8}\)cm and p=x
According to pythagoras theorem
h^{2}=p^{2}+b^{2}
or, (\(\sqrt{8}\))^{2}=x^{2}+ (\(\sqrt{5}\))^{2}
or, x^{2}=8-5
or, x^{2} =3
or, x=\(\sqrt{3}\)cm
\therfore\)x=\(\sqrt{3}\)cm
Solution:
Here, p=8 cm, b=15 cm, h=x=?
According to pythagoras theorem
h^{2}=p^{2}+b^{2}
or, x^{2}=8^{2}+15^{2}
or, x^{2}=64+225
or, x^{2}=289
or, x=\(\sqrt{289}\)
\(\therefore\) x=17 cm
Solution;
Here,p=11 cm, b=x=?, h=16 cm
According to pythagoras theorem
h^{2}=p^{2}+b^{2}
or, 16^{2}=11^{2}+x^{2}
or, x^{2}=16^{2}-11^{2}
or, x^{2}=256-121
or, x^{2}=135
or, x=\(\sqrt{135}\)cm
\(\therefore\) x=\(\sqrt{135}\)cm
Solution;
Here,p=7cm, b=24 cm, h=x=?
According to pythagoras theorem
h^{2}=p^{2}+b^{2}
or, x^{2}=7^{2}+24^{2}
or, x^{2}=49+576
or, x^{2}=625 cm
or, x=\(\sqrt{625}\)
\(\therefore\) x=25
Solution:
Here given data 3, 4, 5
Let, h = 5, p = 3 and b = 4
By Pythagoras Theorem
h^{2} = p^{2} + b^{2}
or, 5^{2} = 3^{2} + 4^{2}
or, 25 = 9+16
or, 25 = 25
Hence, The given number is Pythagorean triple.
Solution:
Here given data 6, 8,10
Let, h = 10, p = 6and b =8
By Pythagoras Theorem
h^{2} = p^{2} + b^{2}
or, 10^{2} = 6^{2} + 8^{2}
or, 100 = 36 + 64
or, 100 =100
Hence, The given number is Pythagorean triple.
Solution:
Here given data 12, 13,14
Let, h = 14, p = 12 and b =13
By Pythagoras Theorem
h^{2} = p^{2} + b^{2}
or, 14^{2} = 12^{2} + 13^{2}
or, 196 = 144 + 169
or, 196≠ 313
Hence, The given number is not Pythagorean triple.
Solution:
Here given data 7, 24, 25
Let, h = 25, p = 7 and b = 24
By Pythagoras Theorem
h^{2} = p^{2} + b^{2}
or, 25^{2} = 7^{2} + 24^{2}
or, 625= 49 +576
or, 625 = 625
Hence, The given number is Pythagorean triple.
Solution:
Here given data 10, 12,14
Let, h = 14, p = 10 and b =12
By Pythagoras Theorem
h^{2} = p^{2} + b^{2}
or, 14^{2} = 10^{2} + 12^{2}
or, 196 =100 + 144
or, 196 ≠ 244
Hence, The given number is not Pythagorean triple.
Solution:
Here given data 10, 12,15
Let, h = 15, p = 10 and b =12
By Pythagoras Theorem
h^{2} = p^{2} + b^{2}
or, 15^{2} = 10^{2} + 12^{2}
or, 255=100 +144
or, 255 ≠ 244
Hence, The given number is not Pythagorean triple.
Solution:
From the given triangle,
hypotenuse (h) =x [It is opposite to right angle]
perpendicular (p) = 15cm [It is opposite to reference angle]
base (b) =8cm [It is opposite to remaining angle]
Now, By Pythagoras Theorem,
h^{2} = p^{2} + b^{2}
or,^{x}^{2} = 15^{2} +8^{2}
or, x^{2} = 225 +64
or, x^{2} = 289
or, x = √289
or, x = 17
\(\therefore\) x = 17cm
Solution:
We know that hypotenuse is the longest side. If 9 cm, 11 cm and 6 cm are the lengths of angled triangle, then 11 cm will be the hypotenuse.
Using the converse of Pythagoras theorem, we get
(11)^{2} = (9)^{2} + (6)^{2}
⇒ 121 = 81 + 36
⇒ 121 ≠ 117
Since, both the sides are not equal therefore 9 cm, 11 cm and 6 cm are not the side of the right angled triangle.
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