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Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 13 |

Chapter Name | Constructions |

Exercise | 13 A |

Category | RS Aggarwal Solutions |

**RS Aggarwal Solutions Class 10 Chapter 13 Constructions Ex 3A**

**Question 1:****Steps of construction:**

Step 1 : Draw a line segment AB = 6.5 cm

Step 2: Draw a ray AX making ∠ BAX.

Step 3: Along AX mark (4+7) = 11 points

A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8}, A_{9}, A_{10}, A_{11}, such that

AA_{1} = A_{1}A_{2}

Step 4: Join A_{11} and B.

Step 5: Through A_{4} draw a line parallel to A_{11} B meeting AB at C.

Therefore, C is the point on AB, which divides AB in the ratio 4 : 7

On measuring,

AC = 2.4 cm

CB = 4.1 cm

**Question 2:****Steps of Construction:**

Step 1 : Draw a line segment PQ = 5.8 cm

Step 2: Draw a ray PX making an acute angle QPX.

Step 3: Along PX mark (5 + 3) = 8 points

A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7 }and A_{8} such that

PA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A6 = A_{6}A7 = A_{7}A_{8}

Step 4: Join A_{8}Q.

Step 5: From A_{5} draw A_{5}C || A_{8}Q meeting PQ at C.

C is the point on PQ, which divides PQ in the ratio 5 : 3

On measurement,

PC = 3.6 cm, CQ = 2.2 cm

**Question 3:****Steps of construction:**

Step 1: Draw a line segment BC = 6 cm

Step 2: With B as centre and radius equal to 5 cm draw an arc.

Step 3: With C as centre and radius equal to 7 cm draw another arc cutting the previous arc at A.

Step 4: Join AB and AC. Thus, ∆ABC is obtained.

Step 5: Below BC draw another line BX.

Step 6: Mark 7 points B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}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}

Step 7: Join B_{7}C.

Step 8: from B_{5}, draw B_{5}D || B_{7}C.

Step 9: Draw a line DE through D parallel to CA.

Hence ∆ BDE is the required triangle.

**Question 4:****Steps of construction:**

Step 1: Draw a line segment QR = 6 cm

Step 2: At Q, draw an angle RQA of 60◦.

Step 3: From QA cut off a segment QP = 5 cm.

Join PR. ∆PQR is the given triangle.

Step 4: Below QR draw another line QX.

Step 5: Along QX cut – off equal distances Q_{1}Q_{2}Q_{3}Q_{4}Q_{5}

QQ_{1} = Q_{1}Q2= Q_{2}Q_{3} = Q3Q_{4} = Q_{4}Q_{5}

Step 6: Join Q_{5}R.

Step 7: Through Q_{3} draw Q_{3}S || Q_{5}R.

Step 8: Through S, draw ST || PR.

∆ TQS is the required triangle.

**Question 5:****Steps of construction:**

Step 1: Draw a line segment BC = 6 cm

Step 2: Draw a right bisector PQ of BC meeting it at M.

Step 3: From QP cut – off a distance MA = 4 cm

Step 4: Join AB, AC.

∆ ABC is the given triangle.

Step 5: Below BC, draw a line BX.

Step 6: Along BX, cut – off 3 equal distances such that

BR_{1} = R_{1}R_{2}= R_{2}R_{3}

Step 7: Join R_{2}C.

Step 8: Through R_{3} draw a line R_{3}C_{1} || R_{2}C.

Step 9 : Through C_{1} draw line C_{1}A_{1 }|| CA_{ .}

∆ A_{1}BC_{1} is the required triangle.

**Question 6:****Steps of Construction:**

Step 1: Draw a line segment BC = 5.4 cm

Step 2. At B, draw ∠ CBM = 45°

Step 3: Now ∠ A = 105°, ∠ B = 45°, ∠ C = 180° – (105°+ 45°) = 30°

At C draw ∠ BCA = 30°.

∆ ABC is the given triangle.

Step 4: Draw a line BX below BC.

Step 5: Cut-off equal distances such that BR_{1} = R_{1}R_{2}= R_{2}R_{3 }= R_{3}R_{4}

Step 6: Join R_{3}C.

Step 7: Through R_{4}, draw a line R_{4}C_{1} || R_{3}C.

Step 8: Through C_{1} draw a line C_{1}A_{1} parallel to CA.

∆ A_{1}BC_{1 }is the required triangle.

**Question 7:**

**Steps of Construction:**

Step 1: Draw a line segment BC = 4 cm

Step 2: Draw a right- angle CBM at B.

Step 3: Cut-off BA = 3cm from BM.

Step 4: Join AC.

ΔABC is the given triangle.

Step 5: Below BC draw a line BX.

Step 6: Along BX, cut-off 7 equal distances such that

BR_{1} = R_{1}R_{2}= R_{2}R_{3 }= R_{3}R_{4} = R_{4}R_{5} = R_{5}R_{6} = R_{6}R_{7}

Step 7: Join R_{5}C.

Step 8: Through R_{7} draw a line parallel to R_{5}C cutting BC produced at C_{1}

Step 9: Through C_{1} draw a line parallel to CA cutting BA at A_{1}

∆ A_{1}BC_{1} is the required triangle.

**Question 8:****Steps of Construction:**

Step 1: draw a line segment BC = 5 cm

Step 2: With B as centre and radius 7cm an arc is drawn.

Step 3: With C as centre and radius 6 cm another arc is drawn intersecting the previous arc at A.

Step 4: Join AB and AC.

Step 5: ∆ ABC is the given triangle.

Step 6: Draw a line BX below BC.

Step 7: Cut- off equal distances from DX 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}

Step 8: join B_{5}C.

Step 9: Draw a line through B_{7} parallel to B_{5}C cutting BC produced at C’.

Step 10: Through C’ draw a line parallel to CA, cutting BA produced at A’.

Step 11: ∆ A’BC’ is the required triangle.

**Question 9:****Steps of construction:**

Step 1: Draw a line segment AB = 6.5 cm

Step 2: With B as centre and some radius draw an arc cutting AB at D.

Step 3: With centre D and same radius draw another arc cutting previous arc at E. ∠ ABE = 60°

Step 4: Join BE and produce it to a point X.

Step 5: With centre B and radius 5.5 cm draw an arc intersecting BX at C.

Step 6: Join AC.

∆ ABC is the required triangle.

Step 7: Draw a line AP below AB.

Step 8: Cut- off 3 equal distances such that

AA_{1} = A_{1}A_{2} = A_{2}A_{3}

Step 9: Join BA_{2}

Step 10: Draw A_{3}B’ through A_{3} parallel to A_{3}B.

Step 11: Draw a line parallel to BC through B’ intersecting AY at C’.

∆ AB’C’ is the required triangle.

**Question 10:****Steps of construction:**

Step 1: Draw a line segment BC = 6.5 cm

Step 2: Draw an angle of 60° at B so that ∠ XBC = 60°.

Step 3: With centre B and radius 4.5cm, draw an arc intersecting XB at A.

Step 4: Join AC.

∆ ABC is the required triangle.

Step 5: Draw a line BY below BC.

Step 6: Cut- off 4 equal distances from BY.

Such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}

Step 7: Join CB_{4}

Step 8: draw B_{3}C’ parallel to CB_{4}

Step 9: Draw C’A’ parallel to CA through C’ intersecting BA produced at A’.

∆ A’BC’ is the required similar triangle.

**Question 11:****Steps of Construction:**

Step 1: Draw a line segment BC = 9 cm

Step 2: with centre B and radius more than 1/2 BC, draw arcs on both sides of BC.

Step 3: With centre C and same radius draw other arcs on both sides of BC intersecting previous arcs at P and Q.

Step 4: join PQ and produce it to a point X. PQ meets BC at M.

Step 5: With centre M and radius 5 cm, draw an arc intersecting MX at A.

Step 6: Join AB and AC.

∆ ABC is the required triangle.

Step 7: Draw a line BY below BC.

Step 8: Cut off 4 equal distances from BY so that

BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}

Step 9: Join CB_{4}

Step 10: Draw C’B_{3} parallel to CB_{4}

Step 11: Draw C’A’ parallel to CA, through C’ intersecting BA at A’.

∆ A’BC’ is the required similar triangle.

**All Chapter RS Aggarwal Solutions For Class 10 Maths**

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**All Subject NCERT Solutions For Class 10**

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