RS Aggarwal Solutions Chapter 13 Construction Exercise 13A Class 10 Maths
Chapter Name  RS Aggarwal Chapter 13 Construction 
Book Name  RS Aggarwal Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 13A Solutions
1. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that AP/AB = 3/5.
Solution
Steps of Construction:
Step 1: Draw a line segment AB = 7 cm
Step 2: Draw a ray AX, making an acute angle ∠BAX.
Step 3: Along AX, mark 5 points (greater of 3 and 5) A_{1}, A_{2}, A_{3}, A_{4} and A_{5} such that
AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A3A4 = A_{4}A_{5}
Step 4: Join A_{5}B.
Step 5: From A_{3}, draw A_{3}P parallel to A5B (draw an angle equal to ∠AA5B), meeting AB in P.
2. Draw a line segment of length 7.6 cm and divide it in the ration 5: 8. Measure the two parts.
Solution
Steps of Construction:
Step 1: Draw a line segment AB = 7.6 cm
Step 2: Draw a ray AX, making an acute angle ∠BAX.
Step 3: Along AX, mark (5 + 8 = 13) points A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8}, A_{9}, A_{10}, A_{11}, A_{12} and A_{13} such that
AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{6}A_{7} = A_{8}A_{9} = A_{9}A_{10} = A_{10}A_{11} = A_{11}A_{12} – A_{12}A_{13}.
Step 4: Join A_{13}B.
Step 5: From A_{5}, draw A_{5}P parallel to A_{13}B (draw an angle equal to ∠AA_{13}B), meeting AB in P.
∴ Length of AP = 2.9 cm (Approx)
Length of BP = 4.7 cm (Approx)
3. Construct a Î”PQR, in which PQ = 6 cm, QR = 7 cm and PR = 8 cm. Then, construct another triangle whose sides are 4/5 times the corresponding sides of Î”PQR.
Solution
Steps of Construction
Step 1: Draw a line segment QR = 7 cm.
Step 2: With Q as center and radius 6 cm, draw an arc.
Step 3: With R as center and radius 8 cm, draw an arc cutting the previous arc at P
Step 4: Join PQ and PR. Thus, Î”PQR is the required triangle.
Step 5: Below QR, draw an acute angle ∠RQX.
Step 6: Along OX, mark five points R_{1}, R_{2}, R_{3}, R_{4} and R_{5} such that
QR_{1} = R_{1}R_{2} = R_{2}R_{3} = R_{3}R_{4} = R_{4}R_{5}.
Step 7: Join RR_{5}
Step 8: From R_{4}, draw R_{4}R’ ∥ RR_{5} meeting QR at R’.
Step 9: From R’, draw P’R’ ∥ PR meeting QR at R’.
Here, Î”P’QR’ is the required triangle, each of whose sides are 4/5 times the corresponding sides of Î”PQR.
4. Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
Solution
Steps of construction:
Step 1: Draw a line segment BC = 4 cm.
Step 2: With B as center, draw an angle of 90°.
Step 3: With B as center and radius equal to 3 cm, cut an arc at the right angle and name it A.
Step 4: Join AB and AC.
Thus, Î”ABC is obtained.
Step 5: Extend BC to D, such that BD = 7/5.BC = 75(4) cm = 5.6 cm
Step 6: Draw DE ∥ CA, cutting AB produced to E.
Thus, Î”EBD is the required triangle, each of whose sides is 7/5 the corresponding sides of Î”ABC.
5. Construct a Î”ABC with BC = 7 cm, ∠B = 60° and AB = 6 cm. Construct another triangle whose sides are 3/4 times the corresponding sides of Î”ABC.
Solution
Steps of construction
Step 1: Draw a line segment BC = 7 cm
Step 2: At B, draw ∠XBC = 60°.
Step 3: With B as center and radius 6 cm, draw an arc cutting the ray BX at A.
Step 4: Join AC, Thus, Î”ABC is the required triangle.
Step 5: Below BC, draw an acute angle ∠YBC.
Step 6: Along BY, mark four points B_{1}, B_{2}, B_{3} and B_{4} such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.
Step 7: Join CB_{4}.
Step 8: From B, draw B_{3}C’ ∥ CB_{4} meeting BC at C’’.
Step 9: From C’, Draw A’C’ ∥ AC meeting AB in A’.
Here, Î”A’BC’ is the required triangle whose sides are 3/4 times the corresponding sides of Î”ABC.
6. Construct a Î”ABC in which AB = 6 cm, ∠A = 30° and ∠AB = 60°, Construct another Î”AB’C’ similar to Î”ABC with base AB’ = 8 cm.
Solution
Steps of Construction
Step 1: Draw a line segment AB = 6 cm
Step 2: At A, draw ∠XAB = 30°.
Step 3: At B, draw ∠YBA = 60°. Suppose AX and BY intersect at C.
Thus, Î”ABC is the required triangle.
Step 4: Produce AB to B’ such that AB’ = 8 cm
Step 5: From B’, draw B’C’ ∥ BC meeting AX at C’.
Solution
Steps of Construction:
Step 1: Draw a line segment BC = 8 cm.
Step 2: At B, draw ∠XBC = 45°.
Step 3: At C, draw ∠YCB = 60°. Suppose BX and CY intersect at A.
Thus, Î”ABC is the required triangle.
Step 4: Below BC, draw an acute angle ∠ZBC.
Step 5: Along BZ, mark five points Z_{1}, Z_{2}, Z_{3}, Z_{4} and Z_{5} such that
BZ_{1} = Z_{1}Z_{2} = Z_{2}Z_{3} = Z_{3}Z_{4} = Z_{4}Z_{5}.
Step 6: Join CZ_{5}.
Step 7: From Z_{3}, draw Z_{3}C’ ∥ CZ_{5} meeting BC at C’.
Step 8: From C’, draw A’C ∥ AC meeting AB in A’.
8. To construct a triangle similar to Î”ABC in which BC = 4.5 cm, ∠B = 45° and ∠C = 60°, using a scale factor of 3/7, BC will be divided in the ratio
(a) 3 : 4
(b) 4 : 7
(c) 3 : 10
(d) 3 : 7
Solution
(a) 3 : 4
To construct a triangle similar to Î”ABC in which BC = 4.5 cm, ∠B = 45° and ∠C = 60°, using a scale factor of 3/7, BC will be divided in the ratio 3 : 4.
Here, Î”ABC ∼ Î”A’BC’
BC’: C’C = 3 : 4
Or BC’: BC = 3 : 7
Hence, the correct answer is option A.
9. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.1/2 times the corresponding sides of the isosceles triangle.
Solution
Steps of Construction:
Step 1: Draw a line segment BC = 8 cm.
Step 2: Draw the perpendicular bisector XY to BC, cutting BC at D.
Step 3: With D as center and radius 4 cm, draw an arc cutting XT at A.
Step 4: Join AB and AC. Thus, an isosceles Î”ABC whose base is 8 cm and altitude 4 cm is obtained.
Step 5: Extend BC to E such that BE = 3/2 BC = 3/2 × 8 cm = 12 cm
Step 6: Draw EF ∥ CA, cutting BA produced in F.
Here, Î”BEF is the required triangle similar to Î”ABC such that each side of Î”BEF is 1.1/2 (or 3/2) times the corresponding side of Î”ABC.
10. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm, Then, Construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
Solution
Steps of Construction
Step 1: Draw a line segment BC = 3 cm.
Step 2: At B, draw ∠XBC = 90°.
Step 3: With B as center and radius 4 cm, draw an arc cutting BX at A.
Step 4: Join AC. Thus, a right Î”ABC is obtained.
Step 5: Extend BC to D such that BD = 5/3 BC = 5/3 × 3 cm = 5 cm.
Step 6: Draw a DE ∥ CA, cutting BX in E.