# how to represent a square root of 5 on a number line

• -3

by the method of square root spiral

draw any line AB.

dRAW A LINE SEGMENT OD SUCH THAT OD IS EQUAL TO 2 UNITS.

from the  point D, draw 1 cm (upwards)

and join DO.

DO= ROOT 5

hope this helps u... thumbs up pls :P

• 2

we draw a number line and mark the integers −2, −1, 0, 1, 2, 3, 4, etc. on it, so that the distance between any two consecutive integers is one unit. On this number line, we mark O and A at the points 0 and 2 respectively. At A, we draw AB of unit length which is perpendicular to OA. Now, we join OB. Taking O as the centre and OB as the radius, we will draw an arc which cuts the number line at P.

Now, point P represents the irrational numberon the number line. Let us verify it.

Using Pythagoras theorem in ΔOAB,

OB2 = OA2 + AB2

= 22 + 12

= 4 + 1

= 5

But OP = OB

• 16

by using spiral system and pythagoras theorem

• 0

by the method of square root spiral

draw any line AB.

dRAW A LINE SEGMENT OD SUCH THAT OD IS EQUAL TO 2 UNITS.

from the point D, draw 1 cm (upwards)

and join DO.

DO= ROOT 5

hope this helps u... thumbs up pls :P and this may help you also we draw a number line and mark the integers 2, 1, 0, 1, 2, 3, 4, etc. on it, so that the distance between any two consecutive integers is one unit. On this number line, we mark O and A at the points 0 and 2 respectively. At A, we draw AB of unit length which is perpendicular to OA. Now, we join OB. Taking O as the centre and OB as the radius, we will draw an arc which cuts the number line at P.

Now, point P represents the irrational numberon the number line. Let us verify it.

Using Pythagoras theorem in ΔOAB,

OB2= OA2+ AB2

= 22+ 12

= 4 + 1

= 5

But OP = OB

• 1

1.

We know that,

And,

Mark a point A representing 2 on number line. Now, construct AB of unit length perpendicular to OA. Then, taking O as centre and OB as radius, draw

an arc intersecting number line at C.

2.Construction of √(3.5):1) Draw a right Triangle, in which the two perpendicular sides measure one unit each. The hypotenuse of this triangle measures √2 units. Bisect this into two equal parts, such that each measure (√2)/2 units.2) Draw another right triangle, in which the two perpendicular sides measure √2 and 1 unit. The hypotenuse of this triangle measure √3 units.3) Draw another right triangle, in which the two perpendicular sides measure √3 and (√2)/2 units. The hypotenuse of this triangle measures √(3.5) units.

do other parts with the help of these answers

• -2

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• -7

1425255522 no it is a not a sum

• -4

we draw a number line and mark the integers 2, 1, 0, 1, 2, 3, 4, etc. on it, so that the distance between any two consecutive integers is one unit. On this number line, we mark O and A at the points 0 and 2 respectively. At A, we draw AB of unit length which is perpendicular to OA. Now, we join OB. Taking O as the centre and OB as the radius, we will draw an arc which cuts the number line at P.

Now, point P represents the irrational numberon the number line. Let us verify it.

Using Pythagoras theorem in ΔOAB,

OB2= OA2+ AB2

= 22+ 12

= 4 + 1

= 5

But OP = OB

posted By ;- kaumudhi

• 7
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