Constructions Exercise 13.1 |

**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 \(\angle\) 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 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} 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 – 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, ∠B = 60° 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 – 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 – 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, ∠A = 30° and ∠B = 60°. Construct another \(\Delta\)AB’C’ similar to \(\Delta\)ABC with base AB’ = 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} – (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 – 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, ∠B = 45° and ∠C = 60°. 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, ∠B = 45° and ∠C= 60°, using a scale factor of \(\frac{3}{7}\), BC will be divided into ratio **

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

**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: **\(\Delta\) A’BC’ 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 – 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’.

\(\Delta\) AB’C’ 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 centre 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 – 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’

\(\Delta\) A’BC’ is the required similar triangle.

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