NCERT Class 9 Solutions: Constructions (Chapter 11) Exercise 11.1 – Part 1

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Construction of point angle at Equation

Construction of point angle at 120 degree

Construction of Point Angle at 120 Degree

Construction of point angle at 120 degree

Q-1 Construct an angle of Equation at the initial point of a given of ray and justify the construction.

Solution:

Arc c Arc c: CircumcircularArc[B, C, F] Arc d Arc d: CircumcircularArc[H, D, I] Arc e Arc e: CircumcircularArc[J, D, K] Arc k Arc k: CircumcircularArc[L, C, M] Arc p Arc p: CircumcircularArc[N, C, O] Arc q Arc q: CircumcircularArc[P, Q, R] Arc r Arc r: CircumcircularArc[S, T, U] Segment f Segment f: Segment [A, D] Segment g Segment g: Segment [D, C] Segment h Segment h: Segment [C, B] Segment i Segment i: Segment [C, A] Vector u Vector u: Vector[A, B_1] Vector u Vector u: Vector[A, B_1] Vector v Vector v: Vector[A, C_1] Vector v Vector v: Vector[A, C_1] Point A A = (-1.8, 1.2) Point A A = (-1.8, 1.2) Point A A = (-1.8, 1.2) Point B Point B: Point on v Point B Point B: Point on v Point B Point B: Point on v Point C C = (-1.23, 2.36) Point C C = (-1.23, 2.36) Point C C = (-1.23, 2.36) Point D Point D: Point on c Point D Point D: Point on c Point D Point D: Point on c P text1 = "P" 90 text2 = "90" 90 text2 = "90" S text3 = "S"

Ray as With the Initial Point a

Ray AS with the initial point A, an arc BCD, cutting AS at B

To construct an angle Equation at point A

Steps of construction:

1. Draw a ray AS.

2. With its initial point A as centre and any radius, draw an arc BCD, cutting AS at B.

3. With same radius and center at B, cut the arc intersecting the circle with center A at C

4. With centre C and same radius, draw an arc, cutting the circle with center A at D.

5. With D and C as centres, and any convenient radius ( Equation ) draw two arcs intersecting at P

6.Join AP. Then Equation

Justification:

  • AB and AC are the radius and AB = BC by construction, therefore, Equation

  • Therefore Equation is an equilateral triangle. So, Equation

  • Similarly, Equation , therefore Equation is also an equilateral triangle. So Equation

  • AP bisects Equation , so Equation

  • Now Equation

Q-2 Construct an angle of Equation at the initial point of a given ray and justify the construction.

Solution:

Arc c Arc c: CircumcircularArc[D_2, E, F_1] Arc f Arc f: CircumcircularArc[B_1, C_1, D_1] Arc g Arc g: CircumcircularArc[E_1, G, H_1] Arc h Arc h: CircumcircularArc[I_1, J_1, K_1] Arc k Arc k: CircumcircularArc[L_1, J_1, M_1] Arc p Arc p: CircumcircularArc[N_1, P_1, Q_1] Arc d Arc d: CircumcircularArc[C_3, C, S_1] Arc e Arc e: CircumcircularArc[D_3, D, U_1] Segment i Segment i: Segment [B, G] Segment j Segment j: Segment [G, F] Vector u Vector u: Vector[O, B_2] Vector u Vector u: Vector[O, B_2] Vector v Vector v: Vector[O, C_2] Vector v Vector v: Vector[O, C_2] Vector w Vector w: Vector[O, G_1] Vector w Vector w: Vector[O, G_1] Point O O = (-0.89, 0.76) Point O O = (-0.89, 0.76) Point O O = (-0.89, 0.76) Point A Point A: Point on v Point A Point A: Point on v Point A Point A: Point on v Point B Point B: Point on v Point B Point B: Point on v Point B Point B: Point on v Point F Point F: Point on c Point F Point F: Point on c Point F Point F: Point on c Point G Point G: Point on w Point G Point G: Point on w Point G Point G: Point on w Point N_1 Point N_1: Point on v Point N_1 Point N_1: Point on v Point Q_1 Point Q_1: Point on w Point Q_1 Point Q_1: Point on w Point C Point C: Point on c Point C Point C: Point on c Point C Point C: Point on c Point D Point D: Point on c Point D Point D: Point on c Point D Point D: Point on c E text1 = "E" 45° text2 = "45°"

Ray OA, O Is the Centre of the 45° Angle

Ray OA, O is the center of the 45° angle

Steps of Construction:

1. Draw a ray OA

2. With O as centre and any suitable radius draw an arc cutting OA at B.

3. With B as centre and same radius cut the previously drawn arc at C and then with C as centre and same radius cut the arc in step 1 at D.

4. With C as centre and radius more than half CD draw an arc.

5. With D as centre and same radius draw another arc to cut the previous arc at E

6. Join OE. Then Equation (we proved this above). Let OE cut the circle with center O at F

Now we will draw the bisector OG of Equation Then Equation

1. With B as center and radius more than half of BF draws an arc.

2. With F as center and radius more than half of BF draws another arc intersecting previous arc at G.

3. Now OG is the bisector of angle Equation . Since Equation is the bisector of Equation . So, Equation

Justification:

1. We already proved that Equation constructed this way would be Equation .

2. Now, in Equation and Equation ,

OG is common

Equation (by construction)

Equation (radius of the same circle at O)

3. Therefore, by SSS Equation

4. Therefore, Equation

5. Also, Equation

6. Therefore, Equation

Q-3 Construct the angle of the following measurements:

1. Equation

2. Equation

3. Equation

Solution:

i) Steps of Construction:

Arc c Arc c: CircumcircularArc[P, E, F] Arc d Arc d: CircumcircularArc[G, H, I] Arc e Arc e: CircumcircularArc[D, J, K] Arc f Arc f: CircumcircularArc[L, M, N] 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] Point A A = (-0.78, 1.38) Point A A = (-0.78, 1.38) Point A A = (-0.78, 1.38) Point P Point P: Point on u Point P Point P: Point on u Point P Point P: Point on u 30° text1 = "30°" D text2 = "D" Q text3 = "Q" B text4 = "B"

∠QAB=30°

∠QAB=30° ray AQ, initial point A cut AQ at P

1. Draw a ray AQ.

2. With its initial point A as centre and any radius, draw an arc, cutting AQ at P.

3. With centre P and same radius. Draw an arc, cutting the arc of step 2 in D

4. With P and D as centres, and any convenient radius ( Equation ), draw two arcs intersecting at B.

5. Join OB. Then Equation

ii) Step of construction:

Arc c Arc c: CircumcircularArc[F, G, H] Arc d Arc d: CircumcircularArc[I, J, K] Arc e Arc e: Semicircle through L and M Arc f Arc f: CircumcircularArc[N, O_1, P_1] Arc g Arc g: CircumcircularArc[Q, R, S] Arc h Arc h: CircumcircularArc[T, U, V] Arc k Arc k: CircumcircularArc[A, W, Z] Arc p Arc p: Semicircle through A_1 and B_1 Arc q Arc q: Semicircle through C_1 and D_1 Vector u Vector u: Vector[O, B] Vector u Vector u: Vector[O, B] Vector v Vector v: Vector[O, C] Vector v Vector v: Vector[O, C] Vector w Vector w: Vector[O, D] Vector w Vector w: Vector[O, D] Vector a Vector a: Vector[O, E] Vector a Vector a: Vector[O, E] Point O O = (-0.58, 1.14) Point O O = (-0.58, 1.14) Point O O = (-0.58, 1.14) Point P Point P: Point on u Point P Point P: Point on u Point P Point P: Point on u D text1 = "D" C text2 = "C" B text3 = "B" 22.5° text4 = "22.5°"

POB=90°

POB=90and∠POD=22.5°

1. Draw an angle Equation

2. Draw the bisector OC of Equation , then Equation

3. Bisect Equation , such that Equation

4. Thus, Equation

iii) Steps of Construction:

Arc c Arc c: CircumcircularArc[F, G, H] Arc d Arc d: CircumcircularArc[I, J, K] Arc e Arc e: CircumcircularArc[L, M, N] Arc f Arc f: CircumcircularArc[O, P, Q] Arc g Arc g: CircumcircularArc[R, S, T] Arc h Arc h: CircumcircularArc[U, V, W] Vector u Vector u: Vector[A, B_1] Vector u Vector u: Vector[A, B_1] 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] Vector a Vector a: Vector[A, E] Vector a Vector a: Vector[A, E] Point A A = (-0.18, 0.9) Point A A = (-0.18, 0.9) Point A A = (-0.18, 0.9) Point B Point B: Point on u Point B Point B: Point on u Point B Point B: Point on u C text3 = "C" 15° text4 = "15°" D text1 = "D" E text2 = "E"

Ray AB at ∠CAB=15°

Ray AB at ∠CAB=15° ,∠BAC=15°

1. Construct an Equation

2. Bisect Equation so that Equation

3. Bisect Equation ,so that Equation

4. Equation

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