Similarities and Enlargement Questions and Answers - Form 2 Topical Mathematics

Questions

1. 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 8.1m^³. Calculate the volume of the larger tank.
2. The image of P(0,2) under an enlargement with a scale factor 3 is P¹ (4,6). Find the co-ordinates of Q
3. A model of a building is made using a scale 1:500.
   1. Find the height of a room (in meters) in the building which is 5cm long on the model? A room has a floor area of 36m². What is the corresponding area on the floor of the model
   2. A room has a volume of 120m³. What is the corresponding volume of the model in cm³?
      
4. In the triangle ABD, BA is parallel, to CE, given that BA= 9cm, CE = 4cm and AE =3cm, find the length of DE
   
   
5. In the following figure, PR = 12cm, TR = 4cm and ST is parallel to QR. Given that the area of triangle PQR is 336cm², find the area of quadrilateral QRTS
   
   
6. Two dogs regarded similar with the length in ratio 4:3
   1. If the bigger dog has a tail 64cm long, find the length of the tail of the smaller dog
   2. If the smaller dog requires 810g of meat per day how much meat per day does the bigger dog require
7. In the figure below, ADE is a triangle and BC is parallel to DE, AB, BD and BC are 4cm, 3cm and 8cm respectively. Find the length of DE
   
   
8. The surface area of two similar bottles are 12cm² and 108cm² respectively. If the larger one has a volume of 810cm³. Find the volume of the smaller one
9. Given that the area of the trapezium CDEB is 15.6 cm², find the length EA marked X.
   
   

Answers

1. V.S.F = 3^³ : 5^³ = 27 : 125
   Volume of larger tank = ^{(8.1 x 125)}/₂₇
   = 37.5m^³
2.  E.S.F = ^{4 – x}/_{0 – x} = 3
   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)
3. 1. L.S.F = 1:500
      Height in cm = (500 x 5)= 2500cm
      ∴Height in m =2500 /100 = 25m
   2. A.S.F = 1:250000
      = 1:25 (in m2)
      ∴ if 25 = 36
      = (³⁶/₂₅)m² = 1.44m²
   3. V.S.F = 1:500
      1:125m³
      Corresponding volume
      = (¹²⁵/₁₂₀)m³
      = 1.042 m³ = 10420cm³
4. Let DE = x cm
   ∴ AD = 3 + x
   3 + x = 9
   x           4
   12 + 4x = 9x
   x = 2.4 cm
   DE = 2.4
   
5. L.S.F = ¹²/₈ = 3
   A.S.F = ⁹/₄ = ³³⁶/_x
   x = 149¹/₃cm²
   Area of QRTS = 336 – 149¹/₃
   = 186²/₃cm²
6. 1. 4 = 64
      3     x
      x = 48cm
   2. ¾ = ⁸¹⁰/_y
      27 = 810
      64      y
      27y = 810 x 64
      y = 1920grams
7. ΔABC is similar to ΔADE
   DE= ⁷/₄
   DE = ^(7x8)/₄cm
   = 14cm =
   
8. 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
   = 30cm³
9. 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}/_x= ^6.4/₄
   12 + 4x = 6.4 x
   2.4x = 12
   x = 5cm