A triangle has one side of length 10 cm and another side of length x cm - Junior Cycle Mathematics - Question 7 - 2019

For $x = 9$ cm:

1. Using the same formula:

   $10 + 9 + \text{third side} = 26$

2. Calculate: $10 + 9 + \text{third side} = 26 \\ \text{third side} = 26 - 19 = 7\ cm$

Thus, for Diagram B, the length of the third side is 7 cm.

For the triangle to be isosceles:

1. Two sides must be equal. We can have:
   • Case 1: $10 = x$, hence $x = 10$ cm.
   • Case 2: $10 = 10$, hence $x = 8$ cm.
   • Case 3: $x + x = 10$, hence $2x = 10$ and $x = 5$ cm.

The three values are: $x = 10 cm, 8 cm,$ or $5 cm$.

From the graph, the area corresponding to $x = 4$ is approximately 18 cm². Hence:

The estimated area of the triangle in Diagram A is: 18 cm².

You will plot the point corresponding to $x = 9$, using the same method as point A. Ensure the point is properly labeled as Point B on the graph.

The graph is symmetrical about $x = 8$. Thus the equation of the axis of symmetry is:

Equation: x = 8.

Using Pythagoras' theorem, if it was a right triangle:

• The two shorter sides squared should equal the longest side squared: $(10)^2 + (5)^2 \neq (11)^2$
• Calculate: $100 + 25 = 125 \neq 121$
• So, the triangle with sides 10 cm, 5 cm, and 11 cm is not a right-angled triangle.

For the triangle with $x = 8$, the dimensions are: 10 cm and 8 cm.

1. Calculate the height using:

   • Let h be the height: $(8)^2 = (h)^2 + (5)^2$
   • Solve: $64 = h^2 + 25 \\ h^2 = 39 \\ h = \sqrt{39}$

2. Use the area formula: $Area = \frac{1}{2} \times base \times height = \frac{1}{2} \times 10 \times \sqrt{39} = 5\sqrt{39} ext{ cm}^2$

Thus, the area can be expressed as: $5 \sqrt{39} \text{ cm}^2$.