# RS Aggarwal Solutions Class 10 Chapter 13 Constructions

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## 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

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, Aand 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

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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 C1A|| 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= R2R= 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.
∆ A1BCis 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= R2R= 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 PTare 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, ATand 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, BTand BT2
Thus, AT3, AT4, BT1, BTare 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.

Complete RS Aggarwal Solutions Class 10

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