Higher June 2022 Paper 1 Q17
17. The diagram shows an equilateral triangle with side \(2x\) cm.
Diagram not drawn to scale
The height of the triangle is \(h\) cm.
Find and simplify an expression for \(h\) in terms of \(x\). [4]
| Scheme | Marks | Comments |
|---|---|---|
| \((h =)\ \sqrt{3}\,x\) | B4 | B3 for \(h^2 = 3x^2\) or \((h =)\ \sqrt{3x^2}\) OR \((h =)\ \dfrac{2x\sqrt{3}}{2}\) B2 for \((2x)^2 - x^2 = h^2\); accept e.g. \((h =)\sqrt{(2x)^2 - x^2}\) OR \(\dfrac{h}{2x} = \dfrac{\sqrt{3}}{2}\) OR \(h^2 = (2x)^2 + x^2 - \dfrac{2(2x)(x)}{2}\) or \((h =)\sqrt{(2x)^2 + x^2 - \dfrac{2(2x)(x)}{2}}\) allow e.g. \((h =)\sqrt{2x^2 - x^2}\) or \(h^2 = 2x^2 - x^2\) or \((2x)^2 - y^2 = h^2\) OR \(h^2 = 2x^2 + x^2 - \dfrac{2(2x)(x)}{2}\) or \((h =)\sqrt{2x^2 + x^2 - \dfrac{2(2x)(x)}{2}}\) B1 for \(x^2 + h^2 = (2x)^2\); allow e.g. \(x^2 + h^2 = 2x^2\) or \(y^2 + h^2 = (2x)^2\) or \(2x^2 - y^2 = h^2\) or \((h =)\sqrt{2x^2 - y^2}\) OR \(\sin60^\circ = \dfrac{\sqrt{3}}{2}\) or \(\sin60^\circ = \dfrac{h}{2x}\) OR \(\cos30^\circ = \dfrac{\sqrt{3}}{2}\) or \(\cos30^\circ = \dfrac{h}{2x}\) OR \(h^2 = (2x)^2 + x^2 - 2(2x)(x)\cos60\) or \((h =)\sqrt{(2x)^2 + x^2 - 2(2x)(x)\cos60}\) or \(h^2 = 2x^2 + x^2 - 2(2x)(x)\cos60\) or \((h =)\sqrt{2x^2 + x^2 - 2(2x)(x)\cos60}\) |
| (4) |