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Rotation Questions
1. Triangle PQR has vertices P(3,2), Q(-1,1) and R(-3,-1).
(a) Draw PQR on the grid provided. (1mk)
(b) Under a rotation the vertices of P1Q1R1 are P1(1,4), Q1(2,0) and R1(4,-1). Find the centre and angle of rotation using points P and Q. (4mks)
(c) Triangle PQR is enlarged with scale factor 3 centre O(0,0) to give triangle P2Q2R2. Draw triangle P2Q2R2 and state its co-ordinates. (2mks)
(d) Triangle P1Q1R1 undergoes reflection in line y = -x to give triangle P3Q3R3. draw P3Q3R3 and state its coordinates. (3mks)
2. The figure below shows part of a diagram of rotation symmetry order 3 about a point O. Complete the diagram. (3mks)
3. In the figure below, triangle AIBICI is the image of triangle ABC under a rotation, centre O.
By construction, find and label the centre O of the rotation.
Hence, determine the angle of the rotation. (3mks)
4. The ratio of the lengths of the corresponding sides of two similar rectangular water
tanks is 3: 5. The volume of the smaller tank is 8.1m3. Calculate the volume of the
larger tank
Similarities and enlargement Answers
1 | (a) L.S.F 4:5 (b) (c) |
M1
M1 M1 A1
B1 M1
A1
B1 M1 A1 |
A.S.F ex
V.S.F |
2. | Centre (x,y) A(1,4) A1(2,5)
3 – 3x = 2 – x x = ½ -12 – 3y = 5 – y y = -8 ½ centre ( ½ , =8 ½ ) |
M1
M1
A1 | |
03 | |||
3. | 10 = x + 6 5 6 60 = 5x + 30 30 = 5x 6 = x 10 = 5 + y 5 y 10y = 25 + 5y 5y = 25 Y = 5 | M1
A1
B1 | Application of L.S.F |
1. E.S.F = 4 – x = 3
0 – x
4 – x = -3x
2x = -4
x = -2
6 – y = 3 6 – y = 6 – 3y
2 – y
-2y = 0
y = 0
Centre of enlargement
= (-2, 0)
2. a) L.S.F = 1:500
Height in cm = (500 x 5)= 2500cm
Height in m = 2500/100 = 25m
b) A.S.F = 1:250000
= 1:25 (in m2)
if 25 = 36
= (36/25)m2 = 1.44m2
c) V.S.F = 1:500
1:125m3
Corresponding volume
= (125/120)m3
= 1.042 m3 = 10420cm3
3. Let DE = x cm
∴ AD = 3 + x
3 + x = 9
x 4
12 + 4x = 9x
x = 2.4 cm
DE = 2.4
4. L.S.F = 12 = 3
8 2
A.S.F = 9 = 336
4 x
x = 1491/3cm2
Area of QRTS = 336 – 1491/3
= 1862/3cm2
5. (a) 4 = 64
3 x
x = 48cm
(b) ¾ = 810
y
27 = 810
64 y
27y = 810 x 64
y = 1920grams
6. ABC is similar to ADE
DE = 7
4
DE = (7×8)cm
4
= 14cm = -7/23
7. Area scale factor = 12: 108
= 1: 9
Linear scale factor = 1 : 9
= 1 : 3
Volume scale factor = 13 : 33
= 1 : 27
Volume of the smaller cone = 810cm3 x 1
27
= 30cm2
8. ½ h (a + b) = Area of trap.
½ x 3 (DC + 4) = 15.6
DC + 4 = 15.6 x 2
3
DC = 6.4
DC = DA
BE EA
∴
3 + x = 6.4
x 4
12 + 4x = 6.4 x √
2.4x = 12 √
x = 5cm
