**1. Represent √𝟖, √𝟏𝟎 and √𝟏𝟑 on the number line.**

**Solution:**

**Representation of √8:**

Let* O* be the origin on the line *l*. Let *A *be on the line such that *OA = 2. *Draw *AP *perpendicular to *OA *at *A* such that * AP = 2. *

In Δ AOP, by Pythagoras’ theorem, we have, OP² = AP² + OA²

OP² = 2² + 2² = 4 + 4

OP² = 8

OP = √8

Hence OP = √8. With O as centre and OP as radius, cut *l *at C. Then OC = OP = √8. Thus C represnts √8 on the line *l.*

Therefore, OC represents √8.

**Representation of √10:**

Let* O* be the origin on the line *l*. Let *A *be on the line such that *OA = 1. *Draw *AP *perpendicular to *OA *at *A* such that * AP = 3. *

In Δ AOP, by Pythagoras’ theorem, we have, OP² = AP² + OA²

OP² = 3² + 1² = 9 + 1 = 10

OP² = 10

OP = √10

Hence OP = √10. With O as centre and OP as radius, cut *l *at C. Then OC = OP = √10. Thus C represnts √10 on the line *l.*

Therefore, OC represents √10.

**Representation of √13:**

Let* O* be the origin on the line *l*. Let *A *be on the line such that *OA = 3. *Draw *AP *perpendicular to *OA *at *A* such that * AP = 2. *

In Δ AOP, by Pythagoras’ theorem, we have, OP² = AP² + OA²

OP² = 2² + 3² = 4 + 9 = 13

OP² = 13

OP = √13

Hence OP = √13. With O as centre and OP as radius, cut *l *at C. Then OC = OP = √13. Thus C represnts √13 on the line *l.*

Therefore, OC represents √13.

**2. Represent 1 +√5, 2 + 2√3 on the number line.**

**Solution:**

**Representation of 1 + √5**

*Let **O* be apoint on the number line *l *and A be the origin on the line *l*. Let *OA = 1 and* B be another point on the number line such that AB = 1 on the number line *l.** ** *Draw BC* *perpendicular to AB* *at *B* such that * BC = 2. *

In Δ ABC, by Pythagoras’ theorem, we have, AC² = AB² + BC²

AC² = 1² + 2² = 1 + 4 = 5

AC² = 5

AC = √5

Hence AC = √5. With A as centre and AC as radius, cut *l *at D. Then AC = AD = √5. Thus D represnts √5 on the line *l.* Since OA = 1. Thus OD = OA + AD = 1 + √5

Therefore, OD represents 1 + √5 on the number line *l.*

**Representation of 2 + 2√3**

Let *O* be apoint on the number line *l *and A be the origin on the line *l*. Let *OA = 2 and* B be another point on the number line such that AB = √6 on the number line *l. ** *Draw BC* *perpendicular to AB* *at *B* such that * BC = √6. *

In Δ ABC, by Pythagoras’ theorem, we have, AC² = AB² + BC²

AC² = (√6)² + (√6)² = 6 + 6 = 12

AC² = 12

AC = √(12) = √(4×3) = 2√3

Hence AC = 2√3. With A as centre and AC as radius, cut *l *at D. Then AC = AD = 2√3. Thus D represnts 2√3 on the line *l.* Since OA = 2. Thus OD = OA + AD = 2 + 2√3

Therefore, OD represents 2 + 2√3 on the number line *l.*