Check sibling questions

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,

  ∆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,

  ∆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,

  ∆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,

  ∆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 ∆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

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

 

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

∴ 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 = 702 Cancelling squares 𝑥 = 70 cm ∴ 𝑥 = 70 cm Rough 4900 = 49 × 100 = 7 × 7 × 10 × 10 = (7 × 10) × (7 × 10) = 70 × 70 = 〖70〗^2 Find x Here, CD = x So, we need to find CD In ∆ADB, right angled at D. By Pythagoras Theorem 〖𝐴𝐵〗^2 = 〖𝐴𝐷〗^2 + 〖𝐵𝐷〗^2 5^2 = 3^2 + 〖𝐵𝐷〗^2 25 = 9 + 〖𝐵𝐷〗^2 25 − 9 = 〖𝐵𝐷〗^2 16 = 〖𝐵𝐷〗^2 〖𝐵𝐷〗^2 = 16 〖𝐵𝐷〗^2 = 4^2 BD = 4 Now, In ∆ABC, right angled at A. By Pythagoras Theorem 〖𝐵𝐶〗^2 = 〖𝐴𝐵〗^2 + 〖𝐴𝐶〗^2 〖𝐵𝐶〗^2 = 5^2 + 〖12〗^2 〖𝐵𝐶〗^2 = 25 + 144 〖𝐵𝐶〗^2 = 169 〖𝐵𝐶〗^2 = 〖13〗^2 BC = 13 Now, BC = CD + BD BC = x + BD 13 = x + 4 13 − 4 = x 9 = x x = 9 Therefore, x = 9

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.