**NCERT Solution For Class 10, Maths, Chapter 11, Constructions, Exercise 11.1**, first we will learn how to divide the line segment and then find properties of triangles. **Class 10 maths, Chapter 11, Ex. 11.1** contains total seven questions to study which are given below.

Table of Contents

Toggle## Class 10, Maths, Chapter 11, Exercise 11.1, Solutions

**In each of the following, give the justification of the construction also:**

**Q.1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.**

**Ans:**

**Steps of construction: **

- Draw a line segment AB = 7.6 cm.
- Draw an acute angle ∠BAC.
- Locate (5+8 = 13) points equal line segments: A
_{1}, A_{2}, A_{3}, A_{4}, ……….. A_{13 }on line AC. Where, AA_{1 }= A_{1}A_{2}= A_{2}A_{3},……….= A_{12}A_{13}. - Join A
_{13}to B. - From point A
_{5}, draw a line A_{5}P||A_{13}B, by making an equal angle of ∠AA_{13}B. So, that P divides AB in the ratio 5: 8.

On measuring the two parts, we find AP = 2.9 cm and PB = 4.7 cm (approx.).

**Justification: **

**To prove: $\frac{AB}{DB}=\frac{5}{8}$**

In ∆ABA_{13}, PA_{5} || BA_{13}

∴ ∆ APA_{5} ~ ∆ ABA_{3} [by BPT]

⇒ **$\frac{AP}{PB}=\frac{A{{A}_{5}}}{{{A}_{5}}{{A}_{13}}}$** ………….. (i)

From our construction,

⇒ **$\frac{AP}{PB}=\frac{5}{8}$**………….. (ii)

From (i) and (ii), we get

⇒ **$\frac{AP}{PB}=\frac{5}{8}$**

Hence, AP: PB = 5 : 8

**Q.2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are ** $\frac{2}{3}$**of the corresponding sides of the first triangle.**

**Ans:**

**Steps of Construction:**

- Draw a line segment BC = 7 cm.
- Draw a ΔABC in which AB = 5 cm and AC = 4 cm and BC = 7 cm.
- below BC, make an acute angle ∠CBX.
- Along BX, mark off three points: B
_{1}, B_{2}and B_{3}that BB_{1}= B_{1}B_{2}= B_{2}B_{3}. - Join B
_{3}C. - From B
_{2}, draw B_{2}D || B_{3}C, meeting BC at D. - From D, draw ED || AC, meeting BA at E. then, EBD is the triangle whose sides are 2/3
^{rd}of the corresponding sides of ∆ ABC.

**Justification: **$\frac{BD}{BC}=\frac{BE}{AB}=\frac{ED}{AC}=\frac{2}{3}$

By construction,

Since, DE || CA

∠BDE = ∠BCA [Corresponding angles]

In Δ BDE and ΔBCA

∠BDE = ∠BCA

∠EBD = ∠ABC [common angle]

∴ ∆ BDE ~ ∆ BCA [By AA]

$\frac{BD}{BC}=\frac{BE}{AB}=\frac{ED}{AC}$……. (i)

In ΔBB_{2}D and ΔBB_{3}C

Since, B_{2}D || B_{3}C

∠BB_{2}D = ∠BB_{3}C [Corresponding angles]

∠DBB_{2} = ∠CBB_{3} [common angle]

So, ΔBB_{2}D ~ ΔBB_{3}C [by AA]

$\frac{BD}{BC}=\frac{B{{B}_{2}}}{B{{B}_{3}}}=\frac{2}{3}$……………. (ii) From (i) and (ii)

$\frac{BD}{BC}=\frac{BE}{AB}=\frac{ED}{AC}=\frac{2}{3}$ ** (Hence proved)**

Hence, we get the new triangle similar to the given triangle whose sides are equal to $\frac{2}{3}$ rd of the corresponding sides of ∆ ABC.

**Q.3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle. **

**Ans:** **Steps of Construction:**

- With the given data, construct ∆ ABC in which BC = 7 cm, CA = 4 cm and AB = 5 cm.
- Below BC, make an acute ∠CBX.
- Along BX, mark off seven points: B
_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}and B_{7}such that BB_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}= B_{4}B_{5}= B_{5}B_{6}= B_{6}B_{7}. - Join B
_{5}C - From B
_{7}, draw B_{7}D || B_{5}C, meeting BC produced at D. - From D, draw DE || CA, meeting BA produced at E. then, EBD is the required triangle whose sides are 7/5
^{th}of the corresponding sides of ∆ABC.

**Justification: **

$\frac{EB}{AB}=\frac{BD}{BC}=\frac{DE}{CA}=\frac{7}{5}$

**By construction,**

Since, DE || CA

∠BDE = ∠BCA [Corresponding angles]

In Δ BCA and ΔBDE

∠BCA = ∠BDE

∠ABC = ∠EBD [common angle]

∴ ΔBCA ~ ΔBDE [By AA]

$\frac{EB}{AB}=\frac{BD}{BC}=\frac{DE}{CA}$…….(i)

In ΔBB_{5}C and ΔBB_{7}D

Since, B_{5}C || B_{7}D

∠BB_{5}C = ∠BB_{7}D [Corresponding angles]

∠CBB_{5} = ∠DBB_{7} [common angle]

So, ΔBB_{5}C ~ ΔBB_{7}D [By AA]

$\frac{BD}{BC}=\frac{B{{B}_{7}}}{B{{B}_{5}}}=\frac{7}{5}$…………. (ii)

From (i) and (ii)

$\frac{EB}{AB}=\frac{BD}{BC}=\frac{DE}{CA}=\frac{7}{5}$ (Hence proved)

Hence, we get the new triangle similar to the given triangle whose sides are equal to ${{\left( \frac{7}{5} \right)}^{rd}}$ of the corresponding sides of ∆ ABC.

**Q.4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are $1\frac{1}{2}$ times the corresponding sides of the isosceles triangle.**

**Ans:** **Step of Construction:**

Draw BC = 8 cm.

- Construct PQ, the perpendicular bisector of line segment AB at M.
- Along MP, cut off MA = 4 cm.
- Join AC and BC. Where AC = BC.
- draw an acute angle ∠BAA
_{3}. - locate 3 point A
_{1}, A_{2}, A_{3}. Where, AA_{1}= AA_{2}= AA_{3}. - Join A
_{2}B. - Draw A
_{2}B || A_{3}B’. - Draw B’C’||BC

**Justification:**

**$\frac{AC’}{AC}=\frac{AB’}{AB}=\frac{B’C’}{BC}=1\frac{1}{2}=\frac{3}{2}$**

**By construction,**

** **Since, B’C’ || BC

∠AB’C’ = ∠ABC [Corresponding angles]

In ΔABC and ΔAB’C’

∠ABC = ∠AB’C’

∠CAB = ∠C’AB’ [common angle]

∴ ΔAB’C’ ~ ΔABC [By AA]

** $\frac{AC’}{AC}=\frac{AB’}{AB}=\frac{B’C’}{BC}$**…….(i)

In ΔAA_{2}B and ΔAA_{3}B’

Since, A_{2}B || A_{3}B’

∠AA_{2}B = ∠AA_{3}B’ [Corresponding angles]

∠BAA_{2} = ∠B’AB_{3} [common angle]

So, ΔAA_{3}B ~ ΔAB_{3}B’ [By AA]

**$\frac{AB’}{AB}=\frac{A{{B}_{3}}}{A{{A}_{2}}}=\frac{3}{2}$** …………. (ii)

From (i) and (ii)

** $\frac{AC’}{AC}=\frac{AB’}{AB}=\frac{B’C’}{BC}=\frac{3}{2}$** (Hence proved)

Hence, we get the new triangle similar to the given triangle whose sides are $\frac{3}{2}$ , i.e., $1\frac{1}{2}$ times of the corresponding sides of the isosceles ∆ABC.

**Q.5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ****∠**** ABC = 60°. Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle ABC.**

**Ans:** **Steps of construction:**

- Construct ∆ ABC in which BC = 6 cm.
- Make an angle 60
^{0}at point B, ∠ABC = 60^{0}. - from point B take 5 cm and draw an arc to form point A.
- join A to C.
- Below BC, make an acute ∠CBX.
- Along BX, mark off 4 points: B
_{1}, B_{2}, B_{3}and B_{4 }such that BB_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}. - Join B
_{4}C. - Through B
_{3}draw a B_{3}D|| B_{4}C by making same angle as ∠BB_{4}C. - From D, draw ED || CA, meeting BA at E. then, EBD is the required triangle whose sides are
**$\frac{3}{4}$**th of the corresponding sides of ∆ABC.

** ****Justification:**

** $\frac{BD}{BC}=\frac{BE}{AB}=\frac{ED}{AC}=\frac{3}{4}$**

By construction,

Since, DE || CA

∠BDE = ∠BCA [Corresponding angles]

In Δ BDE and ΔBCA

∠BDE = ∠BCA

∠EBD = ∠ABC [common angle]

∴ ∆ BDE ~ ∆ BCA [By AA]

** $\frac{BD}{BC}=\frac{BE}{AB}=\frac{ED}{AC}$**……. (i)

In ΔBB_{3}D and ΔBB_{4}C

Since, B_{3}D || B_{4}C

∠BB_{3}D = ∠BB_{4}C [Corresponding angles]

∠DBB_{3} = ∠CBB_{4} [common angle]

So, ΔBB_{2}D ~ ΔBB_{3}C [by AA]

** $\frac{BD}{BC}=\frac{B{{B}_{3}}}{B{{B}_{4}}}=\frac{3}{4}$**……………. (ii)

From (i) and (ii)

** $\frac{BD}{BC}=\frac{BE}{AB}=\frac{ED}{AC}=\frac{3}{4}$** (Hence proved)

Hence, we get the new triangle similar to the given triangle whose sides are equal to ${{\left( \frac{3}{4} \right)}^{rd}}$of the corresponding sides of ∆ ABC.

**Q.6. Draw a triangle ABC with side BC = 7 cm, ****∠**** B = 45°, ****∠**** A = 105°. Then, construct a triangle whose sides are $\frac{3}{4}$ times the corresponding sides of Δ ABC.**

**Ans:** Step of Construction:

- Construct ∆ ABC in which BC = 7 cm, ∠B = 45
^{0}, ∠C = 180^{0}– (∠A+∠B)

⇒ ∠C = 180° – (105°+45°) = 180° – 150° = 30°

- Below BC, make an acute ∠CBX.
- Along BX, mark off 4 points: B
_{1}, B_{2}, B_{3}and B_{4 }such that BB_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}. - Join B
_{3}C. - From B
_{4}, draw B_{4}D || B_{3}C, meeting BC produced at D. by making same angle as ∠BB_{3}C. - From D, draw ED || CA, meeting BA at E. then, EBD is the required triangle whose sides are th times of the corresponding sides of ∆ABC.

**Justification: **

** $\frac{EB}{AB}=\frac{BD}{BC}=\frac{DE}{CA}=\frac{4}{3}$**

**By construction,**

Since, DE || CA

∠BDE = ∠BCA [Corresponding angles]

In Δ BCA and ΔBDE

∠BCA = ∠BDE

∠ABC = ∠EBD [common angle]

∴ ΔBCA ~ ΔBDE [By AA]

** $\frac{EB}{AB}=\frac{BD}{BC}=\frac{DE}{CA}$**…….(i)

In ΔBB_{3}C and ΔBB_{4}D

Since, B_{3}C || B_{4}D

∠BB_{3}C = ∠BB_{4}D [Corresponding angles]

∠CBB_{3} = ∠DBB_{4} [common angle]

So, ΔBB_{3}C ~ ΔBB_{4}D [By AA]

** $\frac{BD}{BC}=\frac{B{{B}_{4}}}{B{{B}_{3}}}=\frac{4}{3}$**…………. (ii)

From (i) and (ii)

** $\frac{EB}{AB}=\frac{BD}{BC}=\frac{DE}{CA}=\frac{4}{3}$** (Hence proved)

Hence, we get the new triangle similar to the given triangle whose sides are equal to ${{\left( \frac{4}{3} \right)}^{rd}}$of the corresponding sides of ∆ ABC.

**Q.7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are ** ** times the corresponding sides of the given triangle.**

**Ans:**

** ****Steps of Construction:**

- Construct ∆ABC in which BC = 4 cm, ∠B = 90
^{0}, BA = 3 cm - construct a line BC = 4 cm.
- Make an angle 90
^{0}at B and take 3cm to draw an arc where AB = 3 cm. - join AC.
- Below BC, make an acute ∠CBX.
- Along BX, mark off 5 points: B
_{1}, B_{2}, B_{3, }B_{4}and B_{5 }such that BB_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}= B_{4}B_{5}. - Join B
_{3}C. - From B
_{5}, draw B_{5}D || B_{3}C, meeting BC produced at D. - From D, draw ED || CA, meeting BA at E. then, EBD is the required triangle whose sides are th times of the corresponding sides of ∆ABC.

**Justification:**

** $\frac{EB}{AB}=\frac{BD}{BC}=\frac{DE}{CA}=\frac{5}{3}$**

**By construction,**

Since, DE || CA

∠BDE = ∠BCA [Corresponding angles]

In Δ BCA and ΔBDE

∠BCA = ∠BDE

∠ABC = ∠EBD [common angle]

∴ ΔBCA ~ ΔBDE [By AA]

** $\frac{EB}{AB}=\frac{BD}{BC}=\frac{DE}{CA}$**…….(i)

In ΔBB_{3}C and ΔBB_{5}D

Since, B_{3}C || B_{5}D

∠BB_{3}C = ∠BB_{5}D [Corresponding angles]

∠CBB_{3} = ∠DBB_{5} [common angle]

So, ΔBB_{3}C ~ ΔBB_{5}D [By AA]

** $\frac{BD}{BC}=\frac{B{{B}_{5}}}{B{{B}_{3}}}=\frac{5}{3}$**…………. (ii)

From (i) and (ii)

** $\frac{EB}{AB}=\frac{BD}{BC}=\frac{DE}{CA}=\frac{5}{3}$** (Hence proved)

Hence, we get the new triangle similar to the given triangle whose sides are equal to ${{\left( \frac{5}{3} \right)}^{rd}}$of the corresponding sides of ∆ ABC.