For a right angled triangle
Side opposite to right angle is Hypotenuse
Side adjacent to right angle are Base & height
Note :
Here, we can also take AB as base & BC as height.
It does not change our answer
Let AB = a
BC = b
AC = c
Pythagoras theorem says that
Square of hypotenuse = Sum of square of other two sides
a 2 + b 2 = c 2
There are a lot of interesting things that we can do with Pythagoras theorem
Like
- Pythagoras triplets
-
Proof of Pythagoras theorem – The
normal way
Let’s do some questions
Find x
Here,
∆ABC is a right angled triangle, right angled at B
And,
AB = 3, BC = 4, AC = x
By Pythagoras Theorem
AC 2 = AB 2 + BC 2
x 2 = 3 2 + 4 2
x 2 = 9 + 16
x 2 = 25
x 2 = 5 2
Cancelling squares
x = 5
Therefore, x = 5
Find x
Here,
∆ABC is a right angled triangle, right angled at B
And,
AB = 6 cm, BC = 8 cm, AC = x
By Pythagoras Theorem
AC 2 = AB 2 + BC 2
x 2 = 6 2 + 8 2
x 2 = 36 + 64
x 2 = 100
x 2 = 10 2
Cancelling squares
x = 10 cm
Therefore, x = 10 cm
Find x
Here,
∆PQR is a right angled triangle, right angled at P
And,
PR = 8 cm, PQ = 15 cm, QR = x
By Pythagoras Theorem
QR 2 = PR 2 + PQ 2
Find x
Here,
∆PQR is a right angled triangle, right angled at P
And,
PQ = 24 cm, QR = x, PR = 7 cm
By Pythagoras Theorem
QR 2 = PQ 2 + PR 2
x 2 = (24) 2 + 7 2
x 2 = 576 + 49
x 2 = 625
x 2 = (25) 2
Cancelling squares
x = 25 cm
Therefore, x = 25 cm
625 = 5 × 5 × 5 × 5
= 25 × 25
= (25) 2
Find x
In ∆ABC,
AB = AC
Hence, it is an isosceles triangle.
Now, AD ⊥ BC
In isosceles triangle,
altitude and median are same
∴ AD is median of BC
i.e. D is mid-point of BC
∴ BD = DC = BC/2
BD = DC = x/2
In ∆ADC, right angled at D.
By Pythagoras Theorem
(AB) 2 = (AD) 2 + (BD) 2
(37) 2 + (12) 2 + (x/2) 2
1369 = 144 + x 2 /4
1369 − 144 = x 2 /4
1225 = x 2 /4
1225 × 4 = x 2
4900 = x 2
x 2 = 4900
x 2 = 70 2
Cancelling squares
x = 70 cm
∴ x = 70 cm
Let's look at one last example
