Represent root 5.7 on a number line
Answers
Answered by
13
Heya!!! ✌️✌️
Here is your answer dear...
Let K = 5.7 cm
Draw a right angle triangle with base of length (K-1)cm and hypotenuse (K+1)cm . The third side of the triangle represents twice the square root of K . i.e. 2√k . Thus we can bisect the line obtained to get √k
Explanation :-
This can be proved by Pythagoras theorem .
Length of third side = √{(K+1)^2-(K-1)^2} =2√K
The triangle can be constructed as
explained :-
1) Draw the base line of length (K-1) cm .
2) Draw perpendicular line from one end of the base .
3) Cut an arc of length (K+1) cm from the other end of the base line to the perpendicular drawn.
4) Joint the point obtained to complete the triangle.
I HOPE THIS WILL HELP YOU OUT....
HAVE A GREAT DAY DEAR.....
#Bhavana ☺️
Here is your answer dear...
Let K = 5.7 cm
Draw a right angle triangle with base of length (K-1)cm and hypotenuse (K+1)cm . The third side of the triangle represents twice the square root of K . i.e. 2√k . Thus we can bisect the line obtained to get √k
Explanation :-
This can be proved by Pythagoras theorem .
Length of third side = √{(K+1)^2-(K-1)^2} =2√K
The triangle can be constructed as
explained :-
1) Draw the base line of length (K-1) cm .
2) Draw perpendicular line from one end of the base .
3) Cut an arc of length (K+1) cm from the other end of the base line to the perpendicular drawn.
4) Joint the point obtained to complete the triangle.
I HOPE THIS WILL HELP YOU OUT....
HAVE A GREAT DAY DEAR.....
#Bhavana ☺️
rohansbanik2003rohan:
Can u give me a picture
Answered by
27
Draw a line of 5.7 cm + 1 cm i.e.. 6.7 cm
Mark half of 6.7 i.e..3.35 on the line
From the point draw an arc of radius 3.35 that meets both the ends of the line
Extent the line
Take 5.7 cm as centre and draw a perpendicular that meets the previous arc at a point D
Consider BC as a radius and draw an arc that meets the extended line
Mark the point at which the arc meets as √5.7
Hope this helps and also refer to my attachment for further details
Mark half of 6.7 i.e..3.35 on the line
From the point draw an arc of radius 3.35 that meets both the ends of the line
Extent the line
Take 5.7 cm as centre and draw a perpendicular that meets the previous arc at a point D
Consider BC as a radius and draw an arc that meets the extended line
Mark the point at which the arc meets as √5.7
Hope this helps and also refer to my attachment for further details
Attachments:
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