NCERT Class 9 Solutions: Constructions (Chapter 11) Exercise 11.2

Construction of triangle

Construction of Triangle

Construction of triangle

Q-1 Construct a triangle JAB in which Equation

Solution:

Image Title: Triangle JAB

Image Description: Triangle JAB, Equation

Give, Equation

Step of construction:

1. A line segment AB of 7 cm is drawn.

2. At point A, an angle Equation is constructed such that it is equal to Equation .

3. A line segment Equation cm is cut on AC (which is equal to Equation ).

4. DB is joined and Equation is made.

5. Let BE intersect AC at J.

Thus, required triangle is Equation

Q-2 Construct a triangle DAB in which Equation .

Solution:

Arc c Arc c: CircumcircularArc[I, J, K] Arc k Arc k: CircumcircularArc[W, Z, A_1] Arc p Arc p: CircumcircularArc[B_1, C_1, D_1] Arc q Arc q: CircumcircularArc[E_1, F_1, G_1] Arc r Arc r: CircumcircularArc[H_1, I_1, J_1] Arc s Arc s: CircumcircularArc[K_1, L_1, M_1] Arc t Arc t: CircumcircularArc[N_1, O_1, P_1] Arc c_1 Arc c_1: CircumcircularArc[Q_1, R_1, S_1] Arc d_1 Arc d_1: CircumcircularArc[T_1, R_1, U_1] Arc e_1 Arc e_1: CircumcircularArc[V_1, G, W_1] Arc d Arc d: CircumcircularArc[Z_1, A_2, B_2] Arc e Arc e: CircumcircularArc[C_2, A_2, D_2] Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [B, D] Segment h Segment h: Segment [E, F] Segment i Segment i: Segment [B, G] Segment j Segment j: Segment [A, H] Vector u Vector u: Vector[A, C] Vector u Vector u: Vector[A, C] Point A A = (-2.24, 0.16) Point A A = (-2.24, 0.16) Point A A = (-2.24, 0.16) Point B B = (2.84, 0.16) Point B B = (2.84, 0.16) Point B B = (2.84, 0.16) Point C C = (1.18, 5.04) Point C C = (1.18, 5.04) Point C C = (1.18, 5.04) Point D Point D: Point on u Point D Point D: Point on u Point D Point D: Point on u Point G Point G: Point on u Point G Point G: Point on u Point G Point G: Point on u Q text1 = "Q" P text2 = "P"

Triangle DAB

Triangle DAB,AB=8cm,∠A=45°andAD-DB=3.5cm

Give Equation

Steps of Construction:

1. A line segment Equation cm is drawn and at point A, make an angle of Equation i.e. Equation

2. Cut the line segment Equation (equal to Equation ) on ray Equation .

3. Join GB and draw the perpendicular bisector PQ of GB.

4. Let it intersect AC at point D. Join DB.

Thus, Equation is the required triangle.

Q-3 Construct a triangle EAR in which Equation

Solution:

Arc c Arc c: CircumcircularArc[G, F, H] Arc d Arc d: CircumcircularArc[M, N, O_1] Arc e Arc e: CircumcircularArc[P, Q, R_1] Arc k Arc k: CircumcircularArc[S, T, U] Arc p Arc p: CircumcircularArc[V, W, Z] Arc q Arc q: CircumcircularArc[A_1, W, B_1] Angle α Angle α: Angle between R, O, E Angle α Angle α: Angle between R, O, E Angle α Angle α: Angle between R, O, E Segment f Segment f: Segment [E, F] Segment g Segment g: Segment [I, J] Segment h Segment h: Segment [K, L] Vector u Vector u: Vector[A, B] Vector u Vector u: Vector[A, B] Vector v Vector v: Vector[A, C] Vector v Vector v: Vector[A, C] Vector w Vector w: Vector[A, D] Vector w Vector w: Vector[A, D] Point A A = (-1.7, 0.08) Point A A = (-1.7, 0.08) Point A A = (-1.7, 0.08) Point E Point E: Point on v Point E Point E: Point on v Point E Point E: Point on v Point R Point R: Intersection point of f, h Point R Point R: Intersection point of f, h Point R Point R: Intersection point of f, h Point O Point O: Intersection point of g, h Point O Point O: Intersection point of g, h Point O Point O: Intersection point of g, h Y text1 = "Y" X text2 = "X" B text3 = "B" Y' text4 = "Y'"

Triangle EAR

Triangle EAR, AR=6cm,∠A=60°andPR–EA=2cm.

Give, Triangle EAR, Equation

Steps of Construction:

1. A ray AX is drawn and cut off a line segment AR = 6 cm from it.

2. A ray AY is constructed making an angle of 60º with AR and YA is produced to form a line YAY'

3. Cut off a line segment AB = 2cm from AY'. RB is joined.

4. Draw perpendicular bisector of RB intersecting AY at a point E. ER is joined.

Thus, ΔEAR is the required triangle.

Q-4 Construct a triangle HFI in which Equation

Solution:

Arc c Arc c: CircumcircularArc[O, P, Q] Arc d Arc d: CircumcircularArc[R, S, T] Arc e Arc e: CircumcircularArc[U, V, W] Arc p Arc p: CircumcircularArc[Z, A_1, B_1] Arc q Arc q: CircumcircularArc[C_1, D_1, E_1] Arc r Arc r: CircumcircularArc[F_1, G_1, H_1] Arc s Arc s: CircumcircularArc[I_1, J_1, K_1] Arc t Arc t: CircumcircularArc[L_1, M_1, N_1] Arc c_1 Arc c_1: CircumcircularArc[O_1, P_1, Q_1] Arc d_1 Arc d_1: CircumcircularArc[R_1, S_1, T_1] Arc e_1 Arc e_1: CircumcircularArc[U_1, V_1, W_1] Arc f_1 Arc f_1: CircumcircularArc[Z_1, A_2, B_2] Arc g_1 Arc g_1: CircumcircularArc[C_2, D_2, E_2] Arc h_1 Arc h_1: CircumcircularArc[F_2, G_2, H_2] Arc k_1 Arc k_1: CircumcircularArc[I_2, J_2, K_2] Arc p_1 Arc p_1: CircumcircularArc[L_2, M_2, N_2] Arc q_1 Arc q_1: CircumcircularArc[O_2, P_2, Q_2] Arc r_1 Arc r_1: CircumcircularArc[R_2, S_2, T_2] Arc s_1 Arc s_1: CircumcircularArc[U_2, V_2, W_2] Arc t_1 Arc t_1: CircumcircularArc[Z_2, A_3, B_3] Arc c_2 Arc c_2: CircumcircularArc[C_3, D_3, E_3] Segment f Segment f: Segment [A, C] Segment g Segment g: Segment [C, E] Segment h Segment h: Segment [F, G] Segment i Segment i: Segment [H, I] Segment j Segment j: Segment [A, J] Segment k Segment k: Segment [K, L] Segment l Segment l: Segment [M, N] Vector u Vector u: Vector[A, B] Vector u Vector u: Vector[A, B] Vector v Vector v: Vector[C, D] Vector v Vector v: Vector[C, D] Point A A = (3.12, 0.22) Point A A = (3.12, 0.22) Point A A = (3.12, 0.22) Point B B = (3.14, 4.8) Point B B = (3.14, 4.8) Point B B = (3.14, 4.8) Point C C = (-3.48, 0.22) Point C C = (-3.48, 0.22) Point C C = (-3.48, 0.22) Point D D = (-1.28, 2.08) Point D D = (-1.28, 2.08) Point D D = (-1.28, 2.08) Point F Point F: Point on f Point F Point F: Point on f Point F Point F: Point on f Point H Point H: Intersection point of g, h Point H Point H: Intersection point of g, h Point H Point H: Intersection point of g, h Point I Point I: Point on f Point I Point I: Point on f Point I Point I: Point on f V text1 = "V" P text2 = "P" Q text3 = "Q" R text4 = "R" 30° text5 = "30°"

Triangle HFI

Triangle HFI,∠F=30°,∠I=90°andHF+FI+IH=11cm.

Give, Triangle HFI Equation

Steps of Construction:

1. A line segment Equation cm is drawn. ( Equation )

An angle, Equation and an angle Equation

2. Equation And Equation are bisected. The bisectors of these angles intersect each other at point H.

3. Perpendicular bisectors PQ of CH and RV of AH are constructed.

4. Let PQ intersect CA at F and RV intersect CA at I. FH and HI are joined.

Thus, Equation is the required triangle.

Q-5 Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Solution:

Arc c Arc c: CircumcircularArc[F, G, H] Arc d Arc d: CircumcircularArc[M, N, O] Arc e Arc e: CircumcircularArc[P_1, Q_1, R] Arc k Arc k: CircumcircularArc[S, T, U] Arc p Arc p: CircumcircularArc[V, W, Z] Arc q Arc q: CircumcircularArc[A_1, J, B_1] Arc r Arc r: CircumcircularArc[C, I, C_1] Arc s Arc s: CircumcircularArc[D_1, E_1, F_1] Arc t Arc t: CircumcircularArc[G_1, H_1, I_1] Segment f Segment f: Segment [A, P] Segment g Segment g: Segment [P, D] Segment h Segment h: Segment [Q, J] Segment i Segment i: Segment [J, P] Segment j Segment j: Segment [K, L] Vector u Vector u: Vector[A, B] Vector u Vector u: Vector[A, B] Vector v Vector v: Vector[D, E] Vector v Vector v: Vector[D, E] Point A A = (-1.8, -0.18) Point A A = (-1.8, -0.18) Point A A = (-1.8, -0.18) Point P P = (-1.8, 2.08) Point P P = (-1.8, 2.08) Point P P = (-1.8, 2.08) Point Q Point Q: Intersection point of c, g Point Q Point Q: Intersection point of c, g Point Q Point Q: Intersection point of c, g Point J Point J: Point on u Point J Point J: Point on u Point J Point J: Point on u Y text1 = "Y" X text2 = "X"

Triangle PAJ

Triangle PAJ,AJ=12cm,AQ=18cm

Give, right triangle base Equation and othrside is Equation

Steps of Construction:

1. A ray AX is drawn and a cut off a line segment Equation is made on it.

2. Equation is constructed.

3. Cut off a line segment Equation cm is made on AY. JQ is joined.

4. Perpendicular bisector of JQ is constructed intersecting AQ at P. PJ is joined.

Thus, Equation is the required triangle.

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