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.
