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The larger container in (a) above has a volume of 875 cm³. Calculate the volume of the smaller container.

Mathematics

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Answer

c
(1, 2)

Step-by-step solution

3 steps

Part b

The problem states that the larger container has a volume of 875 cm3875 \text{ cm}^3. To calculate the volume of the smaller container, we need the linear scale factor between the two similar containers. This scale factor is typically provided in part (a) of the question, which is not visible here.

Assuming the linear scale factor from the smaller container to the larger container is kk, then the ratio of their volumes is k3k^3.
Let VLV_L be the volume of the larger container and VSV_S be the volume of the smaller container.

VLVS=k3\frac{V_L}{V_S} = k^3

Given VL=875 cm3V_L = 875 \text{ cm}^3, we would calculate VSV_S as:

VS=VLk3=875k3 cm3V_S = \frac{V_L}{k^3} = \frac{875}{k^3} \text{ cm}^3

Without the value of kk (the linear scale factor), a numerical answer cannot be provided.

Part c(1, 2)

Three conditions necessary for two solids to be similar are:

  1. All corresponding linear dimensions (e.g., lengths, widths, heights, radii) must be in the same constant ratio (scale factor).
  2. All corresponding angles (e.g., angles between faces, angles within faces) must be equal.
  3. The solids must have the same shape and the same number of corresponding faces, edges, and vertices.

Part 27(a):
We are given the vertices of triangle ABC and its image A'B'C' after an enlargement.
Original vertices: $A(1, 2)$, $B(3, 2)$, $C(3, 5)$
Image vertices: $A'(1, 2)$, $B'(7, 2)$, $C'(7, 11)$

  1. Observe the coordinates of the original point A and its image A'.

    We have $A(1, 2)$ and $A'(1, 2)$. Since the coordinates of A and A' are identical, point A is invariant under the enlargement.

  2. Identify the center of enlargement.

    If a point remains unchanged after an enlargement, that point is the center of enlargement.
    Therefore, the center of enlargement is $A(1, 2)$.

  3. (Optional) Verify the scale factor using other points.

    Let the center of enlargement be P(1,2)P(1, 2).
    For point B:
    Vector PB⃗=B−P=(3−1,2−2)=(2,0)\vec{PB} = B - P = (3-1, 2-2) = (2, 0)
    Vector PB′⃗=B′−P=(7−1,2−2)=(6,0)\vec{PB'} = B' - P = (7-1, 2-2) = (6, 0)
    The scale factor kk is found by PB′⃗=k⋅PB⃗\vec{PB'} = k \cdot \vec{PB}, so (6,0)=k(2,0)(6, 0) = k(2, 0), which implies 2k=6  ⟹  k=32k = 6 \implies k = 3.

    For point C:
    Vector PC⃗=C−P=(3−1,5−2)=(2,3)\vec{PC} = C - P = (3-1, 5-2) = (2, 3)
    Vector PC′⃗=C′−P=(7−1,11−2)=(6,9)\vec{PC'} = C' - P = (7-1, 11-2) = (6, 9)
    The scale factor kk is found by PC′⃗=k⋅PC⃗\vec{PC'} = k \cdot \vec{PC}, so (6,9)=k(2,3)(6, 9) = k(2, 3), which implies 2k=6  ⟹  k=32k = 6 \implies k = 3 and 3k=9  ⟹  k=33k = 9 \implies k = 3.
    The scale factor is consistently 3.

    The coordinates of the center of enlargement are (1,2)\boxed{(1, 2)}.

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