Constructions is an important chapter for class 10 students, which teaches students about the construction of different geometrical shapes with the help of a compass and a protractor. With the help of this chapter, students can learn how to construct different types of angles such as 30,45,60,90 and 120-degree angles. It is a critical concept for students studying mathematics and students can learn how to construct different geometrical figures such as angle bisector, line segment bisector, perpendicular lines, equilateral triangle, pentagon, and hexagon.

Check out the RS Aggarwal Class 10 Solutions Chapter 13 Constructions available below:

Q.1: Draw a line segment AB of length 7cm. Using ruler and compasses find a point P on AB such that \(\frac{AP}{AB} = \frac{3}{5}\)</amp-mathml/>

Steps of Construction:

Step 1: Draw a line segment AB = 6.5 cm

Step 2: Draw a ray AX making \(\angle\) 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

Q.2: Draw a line segment of length 7.6cm and divide it in the ratio 5:8. Measure the two parts.

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}A_{6} = A_{6}A_{7}= A_{7}A_{8}

Step 4: Join A_{8}Q.

Step 5: From A_{5} draw A_{5}C \(\left | \right |\) A_{8}Q meeting PQ at Â 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

Q.3: Construct a \(\triangle\)PQR, in which PQ = 6cm, QR = 7cm and PR = 8cm. Then, construct another triangle whose sides are \(\frac{4}{5}\) times the corresponding sides \(\triangle\)PQR.

Steps of construction:

Step 1: Draw a line segment BC = 6 cm

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

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

Step 4: Join AB and AC. Thus, \(\Delta\) ABC is obtained.

Step 5: Below BC draw another line BX.

Step 6: Mark 7 points B_{1} Â 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 \(\left | \right |\) B_{7}C

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

Hence \(\Delta\) BDE is the required triangle.

Q.4: Construct a triangle with sides 5cm, 6cm and 7cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.

Steps of Construction:

Step 1: Draw a line segment QR = 6 cm

Step 2: At Q, draw an angle RQA of 60^{o}

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

Join PR. \(\Delta\) PQR is the given triangle.

Step 4: Below QR draw another line QX.

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

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

Step 6: Join Q_{5}R

Step 7: Through Q_{3} draw Q_{3}S \(\left | \right |\) Q_{5}R

Step 8: Through S, draw ST \(\left | \right |\) PR

\(\Delta\) TQS is the required triangle.

Q.5: Construct a \(\Delta\)ABC with BC = 7cm, âˆ B = 60Â° and AB = 6cm. Construct another triangle whose sides are \(\frac{3}{4}\) times the corresponding sides of \(\Delta\)ABC.

Steps for Construction:

Step 1: Draw a line segment BC = 6cm

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

Step 3: From QP cut â€“ off a distance MA = 4 cm.

Step 4: Join AB, AC.

\(\Delta\) ABC is the given triangle.

Step 5: Below BC, draw a line BX

Step 6: Along BX, cut â€“ 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} \(\left | \right |\) R_{2}C.

Step 9: Through C_{1} draw a line C_{1}A_{1} \(\left | \right |\) CA.

\(\Delta\) A_{1}BC_{1} is the required triangle.

Q.6: Construct a \(\Delta\)ABC in which AB = 6cm, âˆ A = 30Â° and âˆ B = 60Â°. Construct another \(\Delta\)ABâ€™Câ€™ similar to \(\Delta\)ABC with base ABâ€™ = 8cm.

Steps of construction:

Step 1: Draw a line segment BC = 5.4 cm

Step 2: At B, draw \(\angle\) CBM = 45^{o}

Step 3: Now \(\angle\) A = 105^{o} , \(\angle\) B = 45^{ o}, \(\angle\) C = 180^{o} â€“ (105^{o} + 45^{o}) = 30^{o}

At C draw \(\angle\) BCA = 30^{o}

\(\Delta\) ABC is the given triangle.

Step 4: Draw a line BX below BC.

Step 5: Cut â€“ 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} \(\left | \right |\) R_{3}C

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

\(\Delta\) A_{1}BC_{1} is the required triangle.

Q.7: Construct a \(\Delta\)ABC in which BC = 8cm, âˆ B = 45Â° and âˆ C = 60Â°. Construct another triangle similar to \(\Delta\)ABC such that its sides are \(\frac{3}{5}\) of the corresponding sides of \(\Delta\)ABC.

Steps of Construction:

Step 1: Draw a line segment BC = 4 cm.

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

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

Step 4: Join AC.

\(\Delta\) ABC is the given triangle.

Step 5: Below BC draw a line BX.

Step 6: Along BX, cut- off 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}

\(\Delta\) A_{1}BC_{1} is the required triangle.

Q.8: To construct a triangle similar to \(\Delta\)ABC in which BC = 4.5cm, âˆ B = 45Â° and âˆ C= 60Â°, using a Â scale factor of \(\frac{3}{7}\), BC will be divided into ratio

(a) 3:4 Â Â (b) 4:7 Â (c) 3:10 Â (d) 3:7

Steps of Construction:

Step 1: Draw a line segment BC = 5cm

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

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

Step 4: Join AB and AC.

Step 5: \(\Delta\) ABC is the given triangle.

Step 6: Draw a line BX below BC.

Step 7: Cut â€“ 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 Câ€™.

Step 10: Through Câ€™ draw a line parallel to CA, cutting BA produced at Aâ€™.

Step 11: \(\Delta\) Aâ€™BCâ€™ is the required triangle.

Q.9: Construct an isosceles triangle whose base is 8cm and altitude 4cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.

Steps of Construction:

Step 1: Draw a line segment AB = 6.5 cm

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

Step 3: With center D and some radius draw another arc cutting the previous arc at E. \(\angle\) ABE = 60^{o}

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

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

Step 6: Join AC.

\(\Delta\) ABC is the required triangle.

Step 7: Draw a line AP below AB.

Step 8: Cut â€“ off 3 equal distances such that

A A_{1} = A_{1} A_{2} = A_{2} A_{3}

Step 9: Join BA_{2}

Step 10: Draw A_{3}Bâ€™ through A_{3} parallel to A_{3}B

Step 11: Draw a line parallel to BC through Bâ€™ intersecting AY at Câ€™.

\(\Delta\) ABâ€™Câ€™ is the required triangle.

Q.10: Draw a right triangle in which the sides are of lengths 4cm and 3cm. Then, construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.

Steps of construction:

Step 1: Draw a line segment BC = 6.5 cm

Step 2: Draw an angle of 60^{o} at B so that \(\angle\) XBC = 60^{0}

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

Step 4: Join AC.

\(\Delta\) ABC is the required triangle.

Step 5: Draw a line BY below BC.

Step 6: Cut â€“ 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}Câ€™ parallel to CB_{4}

Step 9: Draw Câ€™Aâ€™ parallel to CA through Câ€™ intersecting BA produced at Aâ€™

\(\Delta\) Aâ€™BCâ€™ is the required similar triangle.