Foundation January 2023 Paper 1R Q24
24 The diagram shows an isosceles triangle \(ABC\)
Diagram NOT accurately drawn
\(AB = 7\) cm
\(AC = BC = y\) cm
The area of the triangle is 42 cm2
Work out the value of \(y\)
(4)
| Scheme | Marks |
|---|---|
\(\dfrac{1}{2} \times 7 \times h = 42\) oe or \((h =)\;\dfrac{42 \times 2}{7}\;(= 12)\) oe or \(3.5^2 + h^2 = y^2\) or \(h = \sqrt{y^2 - 3.5^2}\) oe | M1 |
| \(y^2 = \left(\dfrac{7}{2}\right)^2 + (\text{“}12\text{”})^2\) oe or \(\dfrac{1}{2} \times 7 \times \text{“}\sqrt{y^2 - 3.5^2}\text{”} = 42\) oe | M1 |
| \(y = \sqrt{\left(\dfrac{7}{2}\right)^2 + (\text{“}12\text{”})^2}\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 12.5 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: A correct equation involving the height or a correct expression for height – could be in terms of \(y\)
M1: (indep) use of their height (any found value that they have called ‘height’)
M1: all values must come from a correct method
A1: oe eg \(\dfrac{25}{2}\)