Math, asked by arjunpradeep3949, 7 months ago

please solve this in a papar and give me that​

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Answers

Answered by anshikasharma09
0

Answer:

a) in triangle abc

by Pythagoras therom

h^2=p^2+b^2

h^2=1+1

h^2=2

h=√2

b) in ∆ acd,

by Pythagoras therom

h^2=p^2+b^2

h^2=√2+1

h^2=2.41

h=1.55

like this u can solve another part

Answered by Anonymous
18

Question given :

  • A spiral of square roots

Given :

  • AB = 1 cm
  • BC = 1 cm
  • CD = 1 cm
  • DE = 1 cm
  • EF = 1 cm

To find :

  • What is the length of AC ?
  • Find the length of AD and AF
  • What is the difference between the perimeters of triangle ADE and AEF ?

Solution :

In traingle ABC ,

We have

  • (AB)² + (BC)² = (AC)² by Pythagoras theorem

So therefore ,

  • 1² + 1² = (AC)²
  • 1 + 1 = (AC)²
  • 2 = (AC)²
  • \sf\sqrt{2}\:=\:AC

_____________________________________

In traingle ACD ,

We have

  • (AC)² + (CD)² = (AD)² by Pythagoras theorem

So therefore ,

  • (√2)² + 1² = (AD)²
  • 2 + 1 = (AD)²
  • 3 = (AD)²
  • \sf\sqrt{3}\:=\:AD

_____________________________________

In traingle ACD ,

We have

  • (AD)² + (DD)² = (AE)² by Pythagoras theorem

So therefore ,

  • (√3)² + 1² = (AE)²
  • 3 + 1 = (AE)²
  • 4 = (AE)²
  • \sf\sqrt{4}\:=\:AE

_____________________________________

In traingle ACD ,

We have

  • (AE)² + (EF)² = (AF)² by Pythagoras theorem

So therefore ,

  • (√4)² + 1² = (AF)²
  • 4 + 1 = (AF)²
  • 5 = (AF)²
  • \sf\sqrt{5}\:=\:AF

_____________________________________

b ) Perimeter of ∆ ADE = \sf\sqrt{3}\:+\:1\:+\:\sqrt{4}

= \sf\sqrt{3}\:+\:1\:+\:2

= \sf\sqrt{3}\:+\:3\:cm

Perimeter of ∆ AEF = \sf\sqrt{4}\:+\:1\:+\:\sqrt{5}

= \sf\sqrt{5}\:+\:1\:+\:2

= \sf\sqrt{5}\:+\:3\:cm

Difference between Perimeters

= \sf\sqrt{5}\:+\:3\:cm - \sf\sqrt{3}\:+\:3\:cm

= \sf\sqrt{5}\:+\:6\:-\:\sqrt{3}\:cm

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