## Note on Pythagoras Theorem

• Note
• Things to remember
• Exercise
• Quiz

#### Trigonometry

The word "Trigonometry" is derived from the Greek word "Tri-Gonia-Metron" where Tri means three, Gonia means angles and Metron mean measure. So, trigonometry is a branch of mathematics which concerned with the measurement of sides, angles and their relation to a triangle.

#### Pythagoras Theorem

The relationship between the three sides of a triangle is simply known as Pythagoras Theorem. The relation was given by the popular Mathematician Pythagoras which is called Pythagoras theorem.
In mathematics, the Pythagorean theorem, also known as Pythagoras's theorem, is a relation in Euclidean geometry among the three sides of a right triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
According to this theorem "In any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of squares of perpendicular and base".

By Pythagoras Theorem

Hypotenuse (h2) = Perpendicular (p2) + Base (b2)
or, h2= p2+ b2
From this theory we can derive,
h = $$\sqrt{p^{2} + b^{2}}$$
p = $$\sqrt{h^{2} - b^{2}}$$
b = $$\sqrt{h^{2} - p^{2}}$$

Pythagoras theorem states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Pythagoras Theorem

Hypotenuse (h2) = Perpendicular (p2) + Base (b2)

or, h2 = p2 + b2

Where, h = $$\sqrt{p^{2}+ b^{2}}$$
p = $$\sqrt{h^{2}- b^{2}}$$
b = $$\sqrt{h^{2}- p^{2}}$$

.

### Very Short Questions

Solution:

Here,

$$\angle$$ABC = $$\theta$$, angle of reference.

$$\angle$$ABC = 90°

Hence,

AC = h, AB = p and BC = b

Solution:

Here,

$$\angle$$QPR = α, angle of reference.

$$\angle$$PQR = 90°

Hence,

RP = h, RQ = p and PQ = b

Solution:

Here,

$$\angle$$EFG = β, angle of reference.

$$\angle$$GEF 90°

Hence,

FG = h, EG = p and EF = b.

Solution:

Here,

$$\angle$$ABC = 90°

Hence,

or, $$\theta$$ + 35° = 90°

or, $$\theta$$ = 90°− 35°

∴ $$\theta$$ = 55°

Solution:

$$\angle$$XYZ = 90°

Here,

or, β + 56° = 90°

or, β = 90°− 56°

∴ β = 34°

Solution:

$$\angle$$QPR = 90°

Here,

or, α + 28 = 90°

or, α = 90°− 28°

∴ α = 52°

Solution,

Here,

AB = P = 3cm

BC = b = 4cm

AC = h = ?

From the right angled ΔABC we have,

or, h = $$\sqrt{p^2+b^2}$$

or, h = $$\sqrt{3^2+4^2}$$

or, h = $$\sqrt{9+16}$$

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

∴ h = 5cm

Solution:

Here,

PQ = h = 13cm

RQ = p = 12cm

PR = b = ?

From right angled ΔPQR we have,

or, b = $$\sqrt{h^2−p^2}$$

or, b = $$\sqrt{13^2−12^2}$$

or, b = $$\sqrt{169−12}$$

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

∴ b = 5cm

Solution:

Here,

YZ = h = 10cm

XY = b = 6cm

XZ = p = ?

From right angled ΔXYZ we have,

or, p = $$\sqrt{h^2−b^2}$$

or, p = $$\sqrt{10^2−6^2}$$

or, p = $$\sqrt{100−36}$$

or, p = $$\sqrt{64}$$

∴ p = 8cm

Solution:

The three sides of a triangle are 5cm, 12cm and 13cm

Here,

or, 132 = 52 + 122

or, 169 = 25 + 144

∴ 169 = 169

i.e. h2 = p2+ b2

Hence, the triangle is right angle triangle.

Use the Pythagorean theorem, with a = 12 and b = 5.

or, c2= a2+ b2[Pythagorean theorem]

or, c2= 122+ 52 [Putting value of a & b]

or, c2= 144 + 25

or, c2= 169

or, $$\sqrt{c^2}$$ = $$\sqrt{169}$$ [Squaring on both sides]

$$\therefore$$ c =13

Hence, The length of the hypotenuse is 13 metres.

Solution:

Use the Pythagorean theorem, with a = 3 and b = 4.

or, c2= a2+ b2[Pythagorean theorem]

or, c2= 32+ 42[Putting value of a & b]

or, c2= 9 + 16

or, c2= 25

or, $$\sqrt{c^2}$$ = $$\sqrt{25}$$ [Squaring on both side]

$$\therefore$$ c = 5

Hence, The length of the hypotenuse is 5 metres.

0%

26.5
30.0
2.5
2.9

90.0
180.0
127.3
180.7

19
21
16
14

7.4
6.5
5.3
7.6

10
8
5
12

9.5
8.4
9.4
13.1

60
85
70
50

4.3
5
6
4

20
36
14
25

5
9
4
6

12
6
10
8

19
20
13
16

18
17
15
25

9
8
52
10

15
16
42
13
• ## You scored /15

Forum Time Replies Report
##### David

1 tan×tan30÷1-tan×tan30

1/√3 √3/2..........what is the answer plzz tell me

##### The A.D.

If cosA=2p/p^2 1 then prove tanA=p^2-1/2p.