Question

1. the diagram shows two similar triangles.

a) write a proportion that relates the sides of the similar triangles.
b) determine the non-permissible values for the rational equation.
c) algebraically determine the value of x that makes the triangles similar.

Explanation:

Identify corresponding sides of similar triangles

The diagram shows a smaller triangle nested inside a larger triangle sharing the top vertex.
The sides of the smaller triangle from the shared vertex are \(x\) and \(x+3\).
The corresponding sides of the larger triangle from the shared vertex are:

• One side: \(x + x = 2x\)
• Other side: \((x+3) + 7 = x + 10\)

Since the triangles are similar, the ratio of corresponding sides must be equal:

$$\frac{x}{x+3} = \frac{x+3}{x+10}$$

Alternatively, comparing the smaller triangle's sides to the larger triangle's corresponding sides:

$$\frac{x}{x+3} = \frac{x+3}{x+10}$$

Determine non-permissible values

Using the Non-permissible Values knowledge point, we find the values of \(x\) that make any denominator in our rational equation equal to zero.
The denominators in the proportion are \(x+3\) and \(x+10\).
Set each denominator to zero:

$$x + 3 = 0 \implies x = -3$$

$$x + 10 = 0 \implies x = -10$$

Thus, the non-permissible values are \(x = -3\) and \(x = -10\).
Additionally, since \(x\) represents a physical length, we must have \(x > 0\).

Solve the rational equation for x

Cross-multiply the proportion to solve for \(x\):

$$x(x+10) = (x+3)^2$$

$$x^2 + 10x = x^2 + 6x + 9$$

Isolate the variable x

Subtract \(x^2\) from both sides:

$$10x = 6x + 9$$

Subtract \(6x\) from both sides:

$$4x = 9$$

Divide by 4:

$$x = \frac{9}{4} \quad (\text{or } 2.25)$$

Since \(2.25\) is positive and does not equal any non-permissible values, it is the valid solution.

Answer:

Question a

A proportion that relates the sides of the similar triangles is:

$$\frac{x}{x+3} = \frac{x+3}{x+10}$$

Question b

The non-permissible values for the rational equation are:

$$x = -3 \quad \text{and} \quad x = -10$$

Question c

The value of \(x\) that makes the triangles similar is:

$$x = \frac{9}{4} \quad (\text{or } 2.25)$$