Pythagoras Theorem

Chapter 6 Class 7 Triangle and its Properties
Concept wise

For a right angled triangle

Β

Side opposite to right angle is Hypotenuse

Side adjacent to right angle are Base & height

Β

Note :

Here, we can also take AB as base & BC as height.

It does not change our answer

LetΒ Β  AB = a

BC = b

AC = c

Pythagoras theorem says that

Β  Square of hypotenuse = Sum of square of other two sides

Β  Β  a 2 + b 2 = c 2

Β

There are a lot of interesting things that we can do with Pythagoras theorem

Like

Letβs do some questions

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Find x

Here,

Β  βABC is a right angled triangle, right angled at B

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And,

Β  AB = 3, BC = 4, AC = x

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By Pythagoras Theorem

AC 2 = AB 2 + BC 2

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x 2 = 3 2 + 4 2

x 2 = 9 + 16

x 2 = 25

x 2 = 5 2

Cancelling squares

Β  x = 5

Β

Therefore, x = 5

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Find x

Here,

Β  βABC is a right angled triangle, right angled at B

Β

And,

Β  AB = 6 cm, BC = 8 cm, AC = x

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By Pythagoras Theorem

AC 2 = AB 2 + BC 2

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x 2 = 6 2 + 8 2

x 2 = 36 + 64

x 2 = 100

x 2 = 10 2

Cancelling squares

Β  x = 10 cm

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Therefore, x = 10 cm

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Find x

Here,

Β  βPQR is a right angled triangle, right angled at P

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And,

Β  PR = 8 cm, PQ = 15 cm, QR = x

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By Pythagoras Theorem

QR 2 = PR 2 + PQ 2

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Find x

Here,

Β  βPQR is a right angled triangle, right angled at P

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And,

Β  PQ = 24 cm, QR = x, PR = 7 cm

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By Pythagoras Theorem

QR 2 = PQ 2 + PR 2

Β  Β  Β  Β  Β  x 2 = (24) 2 + 7 2

Β  Β  Β  Β  Β x 2 = 576 + 49

Β  Β  Β  Β  Β x 2 = 625

Β  Β  Β  Β  Β x 2 = (25) 2

Cancelling squares

Β  x = 25 cm

Β

Therefore, x = 25 cm

625 = 5 Γ 5 Γ 5 Γ 5

Β  Β  Β  Β = 25 Γ 25Β

Β  Β  Β  Β = (25) 2

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Β

Β

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Find x

In βABC,

Β  AB = AC

Hence, it is an isosceles triangle.

Β

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In isosceles triangle,

altitude and median are same

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β΄ AD is median of BC

i.e. D is mid-point of BC

β΄ BD = DC = BC/2

Β  Β BD = DC = x/2

Β

In β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

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1369 β 144 = x 2 /4

1225 = x 2 /4

Β  Β 1225 Γ 4 = x 2

Β  Β  Β  Β  Β 4900 = x 2

Β  Β  Β  Β  Β  Β  Β  x 2 = 4900

Β

Β  Β  Β  Β  Β  x 2 =Β  70 2

Cancelling squares

x = 70 cm

Β

β΄ x = 70 cm

Let's look at one last example

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### Transcript

β΄ AD is median of BC i.e. D is mid-point of BC β΄ BD = DC = π΅πΆ/2 BD = DC = π₯/2 In βADC, right angled at D. By Pythagoras Theorem γπ΄π΅γ^2 = γπ΄π·γ^2 + γπ΅π·γ^2 γ(37)γ^2 + (12)^2 + (π₯/2)^2 1369 = 144 + π₯^2/4 1369 β 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 β΄ 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 βADB, right angled at D. By Pythagoras Theorem γπ΄π΅γ^2 = γπ΄π·γ^2 + γπ΅π·γ^2 5^2 = 3^2 + γπ΅π·γ^2 25 = 9 + γπ΅π·γ^2 25 β 9 = γπ΅π·γ^2 16 = γπ΅π·γ^2 γπ΅π·γ^2 = 16 γπ΅π·γ^2 = 4^2 BD = 4 Now, In βABC, right angled at A. By Pythagoras Theorem γπ΅πΆγ^2 = γπ΄π΅γ^2 + γπ΄πΆγ^2 γπ΅πΆγ^2 = 5^2 + γ12γ^2 γπ΅πΆγ^2 = 25 + 144 γπ΅πΆγ^2 = 169 γπ΅πΆγ^2 = γ13γ^2 BC = 13 Now, BC = CD + BD BC = x + BD 13 = x + 4 13 β 4 = x 9 = x x = 9 Therefore, x = 9