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

76.jpg

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 (link)
  • Proof of Pythagoras theorem – The normal way (link) & the interesting way (link)

Let’s do some questions

 

Find x

77.jpg

Here,

  โˆ†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

78.jpg

Here,

  โˆ†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

79.jpg

Here,

  โˆ†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

80.jpg

 

Find x

81.jpg

Here,

  โˆ†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

82.jpg

 

Find x

83.jpg

In โˆ†ABC,

  AB = AC

Hence, it is an isosceles triangle.

 

Now, AD ⊥ BC

 

In isosceles triangle,

altitude and median are same

84.jpg 85.jpg

 

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

โˆด AD is median of BC i.e. D is mid-point of BC โˆด BD = DC = ๐ต๐ถ/2 BD = DC = ๐‘ฅ/2 In โˆ†ADC, right angled at D. By Pythagoras Theorem ใ€–๐ด๐ตใ€—^2 = ใ€–๐ด๐ทใ€—^2 + ใ€–๐ต๐ทใ€—^2 ใ€–(37)ใ€—^2 + (12)^2 + (๐‘ฅ/2)^2 1369 = 144 + ๐‘ฅ^2/4 1369 โˆ’ 144 = ๐‘ฅ^2/4 1225 = ๐‘ฅ^2/4 1225 ร— 4 = ๐‘ฅ^2 4900 = ๐‘ฅ^2 ๐‘ฅ^2 = 4900 ๐‘ฅ^2 = 702 Cancelling squares ๐‘ฅ = 70 cm โˆด ๐‘ฅ = 70 cm Rough 4900 = 49 ร— 100 = 7 ร— 7 ร— 10 ร— 10 = (7 ร— 10) ร— (7 ร— 10) = 70 ร— 70 = ใ€–70ใ€—^2

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.