Question

3 copy and complete this enlargement with scale factor 4 and centre of enlargement shown.

Explanation:

Step1: Identify the original triangle's vertices

Let's assume the grid has coordinates. The original triangle has vertices: let's say the top vertex (with the cross) is \( A(x_1,y_1) \), the bottom - left vertex is \( B(x_2,y_2) \), and the bottom - right vertex is \( C(x_3,y_3) \). From the grid, if we consider the cross as \( A(1,3) \), \( B(1,1) \), and \( C(2,1) \) (assuming the bottom - left corner of the grid is the origin or a suitable reference point).

Step2: Recall the formula for enlargement

The formula for enlargement with center \( (x_0,y_0) \) and scale factor \( k \) is \( (x',y')=(x_0 + k(x - x_0),y_0 + k(y - y_0)) \). Here, the center of enlargement is the cross (let's assume its coordinates are \( (1,3) \)) and the scale factor \( k = 4 \).

Step3: Enlarge each vertex

• For vertex \( A(1,3) \):

Using the enlargement formula, \( x'=1+4(1 - 1)=1 \), \( y'=3+4(3 - 3)=3 \). So \( A'=(1,3) \) (since it is the center of enlargement).

• For vertex \( B(1,1) \):

\( x'=1+4(1 - 1)=1 \), \( y'=3+4(1 - 3)=3+4\times(- 2)=3 - 8=-5 \)? Wait, maybe my coordinate system is wrong. Let's re - establish the coordinate system. Let's take the bottom - left cell's bottom - left corner as \( (0,0) \). Then the cross (center) is at \( (1,3) \), \( B \) is at \( (1,1) \), and \( C \) is at \( (2,1) \). The vector from the center \( A(1,3) \) to \( B(1,1) \) is \( (0,-2) \). When we enlarge with scale factor 4, the new vector from \( A \) to \( B' \) is \( 4\times(0,-2)=(0,-8) \). So \( B'=(1,3)+(0,-8)=(1,-5) \). The vector from \( A(1,3) \) to \( C(2,1) \) is \( (1,-2) \). When we enlarge with scale factor 4, the new vector is \( 4\times(1,-2)=(4,-8) \). So \( C'=(1,3)+(4,-8)=(5,-5) \).

Step4: Draw the enlarged triangle

Now, plot the points \( A'(1,3) \), \( B'(1,-5) \), and \( C'(5,-5) \) and connect them to form the enlarged triangle. The base of the original triangle (distance between \( B \) and \( C \)) is \( 2 - 1 = 1 \) unit. After enlargement with scale factor 4, the base length becomes \( 1\times4 = 4 \) units. The height of the original triangle (distance between \( A \) and the base \( BC \)) is \( 3 - 1 = 2 \) units. After enlargement, the height becomes \( 2\times4 = 8 \) units. Which matches our coordinate calculations (the vertical distance between \( A'(1,3) \) and \( B'(1,-5) \) is \( 3-(-5)=8 \) units, and the horizontal distance between \( B'(1,-5) \) and \( C'(5,-5) \) is \( 5 - 1 = 4 \) units).

Step5: Draw the triangle

Using the new coordinates of \( A' \), \( B' \), and \( C' \), we can draw the enlarged triangle. The original triangle has a base of length 1 (horizontal) and height of length 2 (vertical). The enlarged triangle should have a base of length \( 1\times4 = 4 \) and height of length \( 2\times4 = 8 \), with the same shape (since it's a similar triangle, as enlargement preserves the shape) and the center of enlargement as the common vertex (the cross).

Answer:

To draw the enlarged triangle:

1. Identify the center of enlargement (the cross) and the vertices of the original triangle.
2. For each vertex of the original triangle, find the vector from the center of enlargement to that vertex.
3. Multiply each vector by the scale factor (4) to get the vectors from the center to the corresponding vertices of the enlarged triangle.
4. Add these new vectors to the coordinates of the center of enlargement to get the coordinates of the vertices of the enlarged triangle.
5. Plot these new vertices and draw the triangle. The enlarged triangle will have a base length 4 times the original base length and a height 4 times the original height, and it will be similar to the original triangle with the center of enlargement as a common vertex.

(Note: Since this is a drawing - based problem, the final answer is the correctly drawn enlarged triangle as per the above steps. If we were to describe the dimensions, the original triangle (assuming base = 1 unit, height = 2 units) has an enlarged triangle with base = 4 units and height = 8 units, with the center of enlargement at the top vertex (the cross).)