Question

please solve this in a papar and give me that​

Answer 1

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

Answer 2

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

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

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