Check sibling questions

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

Let’s do some questions

 

Find x

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

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

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

 

Find x

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

625 = 5 × 5 × 5 × 5

       = 25 × 25 

       = (25) 2

 

 

 

 

Find x

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

 

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

 


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

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

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo