Maths 4 Marks Questions

Draw a number line and mark a point O, representing zero, on it. Suppose point A represents 1 as shown. Then, OA = 1.

Draw a right triangle OAB such that AB = OA = 1.

By pythagoras theorem, we have

(OB)² = (OA)² +(AB)²

⇒ OB² = 1² + 1² = OB2 = 1 +1 = 2

$\Rightarrow\text{OB}=\sqrt{2}$

Now, draw a circle with centre O and radius OB. We find that the arcle cuts the number line at A.

Clearly, A₁ = OB = Radius of the cirde $=\sqrt{2}$

Thus, A₁ represents $\sqrt{2}$ on the number line.

Now, draw a right triangle OBB₁, such that BB₁ = 2.

Again by pythagoras theorem, we have,

$\text{OB}^2_1=\text{OB}^2+\text{BB}^2_1$

$\Rightarrow\text{OB}^2_1=\big(\sqrt{2}\big)^2+(2)^2$

$\Rightarrow\text{OB}^2_1=6$

$\Rightarrow\text{OB}_1=\sqrt{6}$

Now, draw a circle with centre O and radius OB₁. We find that the circle cuts the number line at A₂.

Clearly, OA₂ = OB₂ = Radius of circle $=\sqrt{6}$

Thus, A₂ represents $\sqrt{6}$ on the number line.

Now, draw a right angle triangle OB₁B₂ such that B₁B₂ = 1.

By pythagoras theorem, we have,

$\text{OB}^2_2=\text{OB}^2_1+\text{B}_1\text{B}^2_2$

$\Rightarrow\text{OB}^2_2=\big(\sqrt{6}\big)^2+(1)^2$

$\Rightarrow\text{OB}^2_2=6+1=7$

$\Rightarrow\text{OB}_2=\sqrt{7}$

Now, draw a circle with centre O and radius OB₂. We find that the circle cuts the number line at A₃.

Clearly, OA₃ = OB₂ = Radius of circle $=\sqrt{7}$

Thus, A₃ represents $\sqrt{7}$ on the number line.

Now, again draw a right triangle OB₂B₃ such that B₂B₃ = 1.

By pythagoras theorem, we have,

$\text{OB}^2_3=\text{OB}^2_2+\text{B}_2\text{B}^2_3$

$\Rightarrow\text{OB}^2_3=\big(\sqrt{7}\big)^2+(1)^2$

$\Rightarrow\text{OB}^2_3=7+1=8$

$\Rightarrow\text{OB}_3=\sqrt{8}$

Now, draw a circle with centre O and radius OB₃. We find that the circle cuts the number line at A₄.

Clearly, OA₄ = OB₃ = Radius of circle $=\sqrt{8}$

Thus, A₄ represents $\sqrt{8}$ on the number line.

In order to represent $\sqrt{3.5}$ on number line, we follow the following steps:

1. Draw a line and mark a point A on it.
2. Mark a point B on the line drawn in step 1 such that AB = 3.5cm.
3. Mark a point c on AB produced such that BC = 1unit.
4. Find mid-point of AC. Let the mid-point be O.
5. Taking O as the centre and OC = OA as radius draw a semi-circle; also draw a line passing through B perpendicular to OB. Suppose it cuts the semi-circle at D.
6. Taking B as the centre and BD as radius draw an arc cutting OC produced at E. Point E so obtained represent $\sqrt{3.5}.$

In order to represent $\sqrt{9.4}$ on number line, we follow the following steps:

1. Mark a point F on the line drawn such that AF = 9.4cm
2. Mark a point G on AF produced such that FG = 1unit.
3. Find mid-point of AG. Let the mid-point be O₁.
4. Taking O₁ as the centre and O₁A = O₁G as radius draw a semi-circle. Also, draw a line passing through F perpendicular to O₁F. Suppose it cuts the semi-circle at H
5. Taking F as the centre and FH as radius draw an arc cutting O₁G produced at I. Point I so obtained represents $\sqrt{9.4}.$

In order to represent $\sqrt{10.5}$ on number line, we follow the following steps:

1. Mark a point J on the line such that AJ = 10.5cm.
2. Mark a point K on AJ produced such that JK = 1unit.
3. Find mid-point of AK. Let the mid-point be O₂.
4. Taking O₂ as the centre and O₂A = O₂K as radius draw a semi-circle. Also, draw a line passing through J perpendicular to O₂J. Suppose it cuts the semi-circle at L.
5. Taking J as the centre and JL as radius draw an arc cutting O₂K produced at M. Point M so obtained represents $\sqrt{10.5}$