RS Aggarwal Solutions Class 10 Chapter 13 Constructions

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
A₁, A₂, A₃, A₄, A₅, A₆, A₇, A₈, A₉, A₁₀, A₁₁, such that
AA₁ = A₁A₂
Step 4: Join A₁₁ and B.
Step 5: Through A₄ draw a line parallel to A₁₁ 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
A₁, A₂, A₃, A₄, A₅, A₆, A₇ and A₈ such that
PA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A6 = A₆A7 = A₇A₈
Step 4: Join A₈Q.
Step 5: From A₅ draw A₅C || A₈Q 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

• RS Aggarwal Solutions Class 10
• RD Sharma Class 10 Solutions
• NCERT Solutions

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 B₁B₂B₃B₄B₅B₆B₇ such that
BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇
Step 7: Join  B₇C.
Step 8: from B₅, draw B₅D || B₇C.
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 Q₁Q₂Q₃Q₄Q₅
QQ₁ = Q₁Q2 = Q₂Q₃ = Q3Q₄ = Q₄Q₅
Step 6: Join Q₅R.
Step 7: Through Q₃ draw Q₃S || Q₅R.
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
BR₁ = R₁R₂ = R₂R₃ 
Step 7: Join R₂C.
Step 8: Through R₃ draw a line R₃C₁ || R₂C.
Step 9 : Through C₁ draw line C₁A₁ || CA _.
∆ A₁BC₁ 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  BR₁ = R₁R₂ = R₂R₃ = R₃R₄
Step 6: Join R₃C.
Step 7: Through R₄, draw a line R₄C₁ || R₃C.
Step 8: Through C₁ draw a line C₁A₁ parallel to CA.
∆ A₁BC₁ 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
BR₁ = R₁R₂ = R₂R₃ = R₃R₄ = R₄R₅ = R₅R₆ = R₆R₇
Step 7: Join R₅C.
Step 8: Through R₇ draw a line parallel to R₅C cutting BC produced at C₁
Step 9: Through C₁ draw a line parallel to CA cutting BA at A₁
∆ A₁BC₁ 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
BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇
Step 8: join B₅C.
Step 9: Draw a line through B₇ parallel to B₅C 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
AA₁ = A₁A₂ = A₂A₃
Step 9: Join BA₂
Step 10: Draw A₃B’ through A₃ parallel to A₃B.
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 BB₁ = B₁B₂ = B₂B₃ = B₃B₄
Step 7: Join CB₄
Step 8: draw B₃C’ parallel to CB₄
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
BB₁ = B₁B₂ = B₂B₃ = B₃B₄
Step 9: Join CB₄
Step 10: Draw C’B₃ parallel to CB₄
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 T₁ and T₂
Step 5: Join PT₁ and PT₂
PT₁ and PT₂ and the required tangents. Measuring PT₁ and PT₂
We find, PT₁= PT₂ =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 T₁ and T₂
Step 5: Join PT₁ and PT₂
PT₁ and PT₂ are the required tangents. Further PT₁= PT₂ =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 P₁P₂ is extended to the points A and B such that AO = OB = 7 cm
Step 3: With centre P₁ and radius 3.5 cm draw a circle cutting the first circle at T₁ and T₂
Step 4: join AT₁ and AT₂
Step 5: With centre P₂ and radius 3.5 cm draw another circle cutting the first circle at T₃ and T₄
Step 6: Join BT₃ and BT₄ . Thus AT₁, AT₂ and BT₃, BT₄ 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 T₁, T₂, T₃ and T₄
(iv) Join AT₃, AT₄ and BT₁, BT₂.
Thus AT₃, AT₄, BT₁, BT₂ 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 T₁ and T₂ and the circle with centre B at T₃ and T₄
(viii) Join AT₃, AT₄, BT₁ and BT₂
Thus, AT₃, AT₄, BT₁, BT₂ 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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