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

Steps of Construction:

Step 1: Draw a line segment AB = 6.5 cm

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

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 \(\left | \right |\)_{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

Q.3: Construct a \(\triangle\)

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, \(\Delta\)

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 \(\left | \right |\)_{7}C

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

Hence \(\Delta\)

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

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. \(\Delta\)

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}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 \(\left | \right |\)_{5}R

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

\(\Delta\)

Q.5: Construct a \(\Delta\)

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 – off a distance MA = 4 cm.

Step 4: Join AB, AC.

\(\Delta\)

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} \(\left | \right |\)_{2}C.

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

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

Q.6: Construct a \(\Delta\)

Steps of construction:

Step 1: Draw a line segment BC = 5.4 cm

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

Step 3: Now \(\angle\)^{o} , \(\angle\)^{ o}, \(\angle\)^{o} – (105^{o} + 45^{o}) = 30^{o}

At C draw \(\angle\)^{o}

\(\Delta\)

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} \(\left | \right |\)_{3}C

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

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

Q.7: Construct a \(\Delta\)

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.

\(\Delta\)

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}

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

Q.8: To construct a triangle similar to \(\Delta\)

(a) 3:4 (b) 4:7 (c) 3:10 (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: \(\Delta\)

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: \(\Delta\)

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. \(\angle\)^{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.

\(\Delta\)

Step 7: Draw a line AP below AB.

Step 8: Cut – 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}B’ through A_{3} parallel to A_{3}B

Step 11: Draw a line parallel to BC through B’ intersecting AY at C’.

\(\Delta\)

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

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 \(\angle\)^{0}

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

Step 4: Join AC.

\(\Delta\)

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’

\(\Delta\)