For a right angled triangle

75.jpg

 

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 - Definition - Part 2

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

Let’s do some questions

 

Find x

Pythagoras theorem - Definition - Part 3

Here,

  βˆ†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

Pythagoras theorem - Definition - Part 4

Here,

  βˆ†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

Pythagoras theorem - Definition - Part 5

Here,

  βˆ†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

Pythagoras theorem - Definition - Part 6

 

Find x

Pythagoras theorem - Definition - Part 7

Here,

  βˆ†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

Pythagoras theorem - Definition - Part 8

625 = 5 × 5 × 5 × 5

       = 25 × 25 

       = (25) 2

 

 

 

 

Find x

Pythagoras theorem - Definition - Part 9

In βˆ†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 βˆ†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

Pythagoras theorem - Definition - Part 10

 

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

Pythagoras theorem - Definition - Part 11

Let's look at one last example

Pythagoras theorem - Definition - Part 12

 

  1. Chapter 6 Class 7 Triangle and its Properties
  2. Concept wise

Transcript

∴ AD is median of BC i.e. D is mid-point of BC ∴ BD = DC = 𝐡𝐢/2 BD = DC = π‘₯/2 In βˆ†ADC, right angled at D. By Pythagoras Theorem 〖𝐴𝐡〗^2 = 〖𝐴𝐷〗^2 + 〖𝐡𝐷〗^2 γ€–(37)γ€—^2 + (12)^2 + (π‘₯/2)^2 1369 = 144 + π‘₯^2/4 1369 βˆ’ 144 = π‘₯^2/4 1225 = π‘₯^2/4 1225 Γ— 4 = π‘₯^2 4900 = π‘₯^2 π‘₯^2 = 4900 π‘₯^2 = 702 Cancelling squares π‘₯ = 70 cm ∴ π‘₯ = 70 cm Rough 4900 = 49 Γ— 100 = 7 Γ— 7 Γ— 10 Γ— 10 = (7 Γ— 10) Γ— (7 Γ— 10) = 70 Γ— 70 = γ€–70γ€—^2 Find x Here, CD = x So, we need to find CD In βˆ†ADB, right angled at D. By Pythagoras Theorem 〖𝐴𝐡〗^2 = 〖𝐴𝐷〗^2 + 〖𝐡𝐷〗^2 5^2 = 3^2 + 〖𝐡𝐷〗^2 25 = 9 + 〖𝐡𝐷〗^2 25 βˆ’ 9 = 〖𝐡𝐷〗^2 16 = 〖𝐡𝐷〗^2 〖𝐡𝐷〗^2 = 16 〖𝐡𝐷〗^2 = 4^2 BD = 4 Now, In βˆ†ABC, right angled at A. By Pythagoras Theorem 〖𝐡𝐢〗^2 = 〖𝐴𝐡〗^2 + 〖𝐴𝐢〗^2 〖𝐡𝐢〗^2 = 5^2 + γ€–12γ€—^2 〖𝐡𝐢〗^2 = 25 + 144 〖𝐡𝐢〗^2 = 169 〖𝐡𝐢〗^2 = γ€–13γ€—^2 BC = 13 Now, BC = CD + BD BC = x + BD 13 = x + 4 13 βˆ’ 4 = x 9 = x x = 9 Therefore, x = 9

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.