## RS Aggarwal Solutions Class 10 Chapter 13 Constructions

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 13 Constructions

**Exercise 13A**

**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

**Read More:**

- Construction of an Equilateral Triangle
- Construction Of Similar Triangle As Per Given Scale Factor
- Construction Of A Line Segment
- Construction Of The Bisector Of A Given Angle
- Construction Of Perpendicular Bisector Of A Line Segment
- Construction Of An Angle Using Compass And Ruler

**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

**More Resources**

**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.

**Exercise 13B**

**Question 1:**

**Steps of construction:**

Step 1: Draw a circle of radius 5 cm with centre O.

Step 2: A point P at a distance of 8cm from O is taken.

Step 3: A right bisector of OP meeting OP at M is drawn.

Step 4: With centre M radius OM a circle is drawn intersecting the previous circle at T_{1} and T_{2}

Step 5: Join PT_{1} and PT_{2}

PT_{1} and PT_{2} and the required tangents. Measuring PT_{1} and PT_{2}

We find, PT_{1}= PT_{2} =6.2 cm

**Question 2:**

**Steps of construction:**

Step 1: Two concentric circles with centre O and radii 4 cm and 6 cm are drawn.

Step 2: A point P is taken on outer circle and O, P are joined.

Step 3: A right bisector of OP is drawn bisecting OP at M.

Step 4: With centre M and radius OM a circle is drawn cutting the inner circle at T_{1} and T_{2}

Step 5: Join PT_{1} and PT_{2}

PT_{1} and PT_{2 }are the required tangents. Further PT_{1}= PT_{2} =4.8 cm

**Question 3:**

**Steps of construction:**

Step 1: Draw a circle with centre O and radius 3.5 cm

Step 2: the diameter P_{1}P_{2} is extended to the points A and B such that AO = OB = 7 cm

Step 3: With centre P_{1} and radius 3.5 cm draw a circle cutting the first circle at T_{1} and T_{2}

Step 4: join AT_{1} and AT_{2}

Step 5: With centre P_{2} and radius 3.5 cm draw another circle cutting the first circle at T_{3} and T_{4}

Step 6: Join BT_{3} and BT_{4} . Thus AT_{1}, AT_{2} and BT_{3}, BT_{4} are the required tangents to the given circle from A and B.

**Question 4:**

**Steps of construction:**

(i) A circle of radius 4.2 cm at centre O is drawn.

(ii) A diameter AB is drawn.

(iii) With OB as base, an angle BOC of 45° is drawn.

(iv) At A, a line perpendicular to OA is drawn.

(v) At C, a line perpendicular to OC is drawn.

(vi) These lines intersect each other at P.

PA and PC are the required tangents.

**Question 5:**

**Steps of construction:**

(i) A line segment AB = 8,5 cm is drawn.

(ii) Draw a right bisector of AB which meets AB at M.

(iii) With M as centre AM as radius a circle is drawn intersecting the given circles at T_{1}, T_{2}, T_{3} and T_{4}

(iv) Join AT_{3}, AT_{4 }and BT_{1}, BT_{2}.

Thus AT_{3}, AT_{4}, BT_{1}, BT_{2} are the required tangents.

**Question 6:**

**Steps of construction:**

(i) Draw a line segment AB = 7cm

(ii) Taking A as centre and radius 3 cm, a circle is drawn.

(iii) With centre B and radius 2.5 cm, another circle is drawn.

(iv) With centre A and radius more than 1/2 AB, arcs are drawn on both sides of AB.

(v) With centre B and the same radius, [as in step (iv)] arcs are drawn on both sides of AB intersecting previous arcs at P and Q.

(vi) Join PQ which meets AB at M.

(vii) With centre M and radius AM, a circle is drawn which intersects circle with centre A at T_{1} and T_{2} and the circle with centre B at T_{3} and T_{4}

(viii) Join AT_{3}, AT_{4}, BT_{1 }and BT_{2}

Thus, AT_{3}, AT_{4}, BT_{1}, BT_{2 }are the required tangents.

**Question 7:**

**Steps of construction:**

(i) A circle of radius 3 cm with centre O is drawn.

(ii) A radius OC is drawn making an angle of 60° with the diameter AB.

(iii) At C, ∠OCP = 90° is drawn.

CP is required tangent.

**Question 8:**

**Steps of construction :**

(i) Draw a circle of radius 4 cm with centre O.

(ii) With diameter AB, a line OC is drawn making an angle of 30° i.e., ∠BOC = 30°

(iii) At C a perpendicular to OC is drawn meeting OB at P.

PC is the required tangent.

Hope given RS Aggarwal Solutions Class 10 Chapter 13 Constructions are helpful to complete your math homework.

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