RS Agarwal Solution | Class 10th | Chapter-9 | Constructions | Edugrown

Exercise 9A

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, A₇ and 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 C₁A₁ || 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 = R₂R₃ = 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.
∆ A₁BC₁ is 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 = R₂R₃ = 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.

Exercise 9B

https://youtube.com/watch?v=HEdUc-q9_es%3Ffeature%3Doembed

Question 1:

Steps of construction:
Step 1: Draw a circle of radius 5 cm with centre O.
Step 2: A point P at a distance of 8cm from O is taken.
Step 3: A right bisector of OP meeting OP at M is drawn.
Step 4: With centre M radius OM a circle is drawn intersecting the previous circle at T₁ and T₂
Step 5: Join PT₁ and PT₂
PT₁ and PT₂ and the required tangents. Measuring PT₁ and PT₂
We find, PT₁= PT₂ =6.2 cm

Question 2:

Steps of construction:
Step 1: Two concentric circles with centre O and radii 4 cm and 6 cm are drawn.
Step 2: A point P is taken on outer circle and O, P are joined.
Step 3: A right bisector of OP is drawn bisecting OP at M.
Step 4: With centre M and radius OM a circle is drawn cutting the inner circle at T₁ and T₂
Step 5: Join PT₁ and PT₂
PT₁ and PT₂ are the required tangents. Further PT₁= PT₂ =4.8 cm

Question 3:

Steps of construction:
Step 1: Draw a circle with centre O and radius 3.5 cm
Step 2: the diameter P₁P₂ is extended to the points A and B such that AO = OB = 7 cm
Step 3: With centre P₁ and radius 3.5 cm draw a circle cutting the first circle at T₁ and T₂
Step 4: join AT₁ and AT₂
Step 5: With centre P₂ and radius 3.5 cm draw another circle cutting the first circle at T₃ and T₄
Step 6: Join BT₃ and BT₄ . Thus AT₁, AT₂ and BT₃, BT₄ are the required tangents to the given circle from A and B.

Question 4:

Steps of construction:
(i) A circle of radius 4.2 cm at centre O is drawn.
(ii) A diameter AB is drawn.
(iii) With OB as base, an angle BOC of 45° is drawn.
(iv) At A, a line perpendicular to OA is drawn.
(v) At C, a line perpendicular to OC is drawn.
(vi) These lines intersect each other at P.
PA and PC are the required tangents.

Question 5:

Steps of construction:
(i) A line segment AB = 8,5 cm is drawn.
(ii) Draw a right bisector of AB which meets AB at M.
(iii) With M as centre AM as radius a circle is drawn intersecting the given circles at T₁, T₂, T₃ and T₄
(iv) Join AT₃, AT₄ and BT₁, BT₂.
Thus AT₃, AT₄, BT₁, BT₂ are the required tangents.

Question 6:

Steps of construction:
(i) Draw a line segment AB = 7cm
(ii) Taking A as centre and radius 3 cm, a circle is drawn.
(iii) With centre B and radius 2.5 cm, another circle is drawn.
(iv) With centre A and radius more than 1/2 AB, arcs are drawn on both sides of AB.
(v) With centre B and the same radius, [as in step (iv)] arcs are drawn on both sides of AB intersecting previous arcs at P and Q.
(vi) Join PQ which meets AB at M.
(vii) With centre M and radius AM, a circle is drawn which intersects circle with centre A at T₁ and T₂ and the circle with centre B at T₃ and T₄
(viii) Join AT₃, AT₄, BT₁ and BT₂
Thus, AT₃, AT₄, BT₁, BT₂ are the required tangents.

Question 7:

Steps of construction:
(i) A circle of radius 3 cm with centre O is drawn.
(ii) A radius OC is drawn making an angle of 60° with the diameter AB.
(iii) At C, ∠OCP = 90° is drawn.
CP is required tangent.

Question 8:

Steps of construction :
(i) Draw a circle of radius 4 cm with centre O.
(ii) With diameter AB, a line OC is drawn making an angle of 30° i.e., ∠BOC = 30°
(iii) At C a perpendicular to OC is drawn meeting OB at P.
PC is the required tangent.