What steps shall be required to rotate an object about the point P1 (as shown in fig.1) and its placement such that what was at P1 is now reduced and is at P2 (as shown in fig.2). Image I. Translate P1 to origin II. Scale as required III. Rotate IV. Translate to the final position P2. (A) I,II and III (B) II,III and IV (C) I,III and IV (D) All of the above

What steps shall be required to rotate an object about the point P1 (as shown in fig.1) and its placement such that what was at P1 is now reduced and is at P2 (as shown in fig.2).

image

I. Translate P1 to origin

II. Scale as required

III. Rotate

IV. Translate to the final position P2.

(A) I,II and III (B) II,III and IV

(C) I,III and IV (D) All of the above

by

1 Answer

Answer :

Answer: Marks given to all
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Description : What is the validity after supervisors course (TW-3)? (i) Normally, supervisors shall not be required to undergo the course again. However, they may be sent for course on performance basis.* (ii) 2 years (iii) 5 Years (iv) 10 Years

Last Answer : (i) Normally, supervisors shall not be required to undergo the course again. However, they may be sent for course on performance basis.*

Consider a unit square centred at origin. The coordinates of the square are translated by a factor (1/2, 1) and rotated by an angle of 90o . What shall be the coordinates of the new square ? (A) (-1/2, 0), (-1/2, 1), (-3/2, 1), (-3/2, 0) (B) (-1/2, 0), (1/2, 1), (3/2, 1), (3/2, 0) (C) (-1/2, 0), (1/2, 0), (-3/2, 1), (-3/2, 0) (D) (-1/2, 0), (1/2, 1), (-3/2, 1), (-3/2, 0)

Description : Consider a unit square centred at origin. The coordinates of the square are translated by a factor (1/2, 1) and rotated by an angle of 90o . What shall be the coordinates of the new square ? (A) (-1/2, 0), (-1/2, 1), (-3/2, 1), (-3/2, ... , 1), (-3/2, 0) (D) (-1/2, 0), (1/2, 1), (-3/2, 1), (-3/2, 0)

Last Answer : (A) (-1/2, 0), (-1/2, 1), (-3/2, 1), (-3/2, 0)

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]