please solve this in a papar and give me that
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Answered by
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
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
_____________________________________
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
_____________________________________
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
_____________________________________
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
_____________________________________
b ) Perimeter of ∆ ADE =
=
=
Perimeter of ∆ AEF =
=
=
Difference between Perimeters
= -
=
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