# RS Aggarwal Solutions Class 10 Chapter 13 Constructions Ex 13A

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 13 Constructions Ex 13A Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 13 Constructions Ex 13A Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 13 Constructions Ex 13A Maths book pdf download. Now you will get step by step solution to each question.

## 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
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

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.

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