Question

present under root 7 on number line with construction
$\sqrt{7}$

Answer 1

$\mathfrak{\huge{\green{\boxed{Solution}}}}$

$\bold\red{Representation\: of \:root \:7\: on\: number\: line}$

Let OA = 2units.

Draw AB perpendicular at A and take AB = 1 unit.

Join OB.

Then,

According to pythogoras theorm.

OB = √(OA^2 + AB^2)

= √(2^2 + 1^2)

= √5.

Now,

Draw BC perpendicular at OB and BC = 1 unit.

Join OC.

Again according to pythagoras theorm,

OC = √OB^2 + √BC^2

= √(√5^2 + 1^2)

= √6

Now,

Draw CD perpendicular to OC and CD = 1 unit.

Join OD.

Now again using pythagoras theorm.

OD = √(OC^2 + CD^2)

= √(√6^2 + 1^2)

= √7.

With O as a centre and OD as a radius, draw an arc meeting OX ar P.

Then,

OP = OD = √7.

(See the attachment).