Notes on Pythagoras Theorem | Grade 8 > Compulsory Maths > Co-ordinate Geometry | KULLABS.COM

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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= b2+ c2

= 52+ 122

= 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), d2 = x2 + y2

Therefore, d = $$\sqrt{x^2 + y^2}$$

Example 2

How far from the origin is the point (4, -2)?

Solution:

d = $$\sqrt{4^2+(-2)^2}$$

= $$\sqrt{16+4}$$

=$$\sqrt{20}$$

• The side opposite to the right angle is called the hypotenuse.
• Pythagoras Theorem states the square of the hypotenuse(the side opposite the right angle) is equal to the sum of the squares of the other two sides.
• If three numbers are such that a + b= c2, then the numbers a, b, and c are called Pythagorean triples.

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### Very Short Questions

Solution:

Here, b=12 cm, h=13 cm, p=x=?

According to pythagoras theorem,

h2=p2+b2

or, 132=x2+122

or, x2=132-122

or, x2=169-144

or, x2=25

or, x=$$\sqrt{25}$$

$$\therefore$$ x=5 cm

Solution:

Here, p=14 cm, b=48 cm, h=x=?

According to pythagoras theorem,

h2=p2+b2

or,x2=142+482

or, x2=196+2304

or, x2=2500

or, x=$$\sqrt{2500}$$

$$\tharefore$$ x=50 cm

Solution;

Here, p=21 cm, b=72 cm, h=x=?

According to pythagoras theorem,

h2=p2+b2

or, x2=212+722

or, x2=441+5184

or, x2=5625

or, x=$$\sqrt{5625}$$

$$\therefore$$ x=75 cm

Solution;

Here, p= 6 inch, b=8 inch, h=x=?

According to pythagoras theorem

h2=p2+b2

or,x2=62+82

or, x2=36+64

or, x2=100

or, x=$$\sqrt{100}$$

$$\therefore$$ x=10 inch

Solution:

Here, p= 15ft, h=25ft and b=x=?

According to pythagoras theorem

h2=p2+b2

or, 252=152+x2

or, x2=252-152

or, x2=625-225

or, x2=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

h2=p2+b2

or, ($$\sqrt{8}$$)2=x2+ ($$\sqrt{5}$$)2

or, x2=8-5

or, x2 =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

h2=p2+b2

or, x2=82+152

or, x2=64+225

or, x2=289

or, x=$$\sqrt{289}$$

$$\therefore$$ x=17 cm

Solution;

Here,p=11 cm, b=x=?, h=16 cm

According to pythagoras theorem

h2=p2+b2

or, 162=112+x2

or, x2=162-112

or, x2=256-121

or, x2=135

or, x=$$\sqrt{135}$$cm

$$\therefore$$ x=$$\sqrt{135}$$cm

Solution;

Here,p=7cm, b=24 cm, h=x=?

According to pythagoras theorem

h2=p2+b2

or, x2=72+242

or, x2=49+576

or, x2=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

h2 = p2 + b2

or, 52 = 32 + 42

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

h2 = p2 + b2

or, 102 = 62 + 82

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

h2 = p2 + b2

or, 142 = 122 + 132

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

h2 = p2 + b2

or, 252 = 72 + 242

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

h2 = p2 + b2

or, 142 = 102 + 122

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

h2 = p2 + b2

or, 152 = 102 + 122

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,

h2 = p2 + b2

or,x2 = 152 +82

or, x2 = 225 +64

or, x2 = 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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