Higher June 2022 Paper 1 Q13
13 A right-angled triangle is formed by the diameters of three semicircular regions, A, B and C as shown in the diagram.
Show that
area of region A = area of region B + area of region C (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| Relationship shown | M1 | for use of Pythagoras’ theorem, eg \(d_A^{\,2} = d_B^{\,2} + d_C^{\,2}\) or \(a^2 = b^2 + c^2\) or \((2x)^2 = (2y)^2 + (2z)^2\) or \(a = \sqrt{b^2 + c^2}\) or uses a 3, 4, 5 triangle |
| M1 | for forming correct expressions for the areas of at least 2 of the 3 semicircles, eg at least two of \(\dfrac{1}{2}\pi\left(\dfrac{a}{2}\right)^2,\ \dfrac{1}{2}\pi\left(\dfrac{b}{2}\right)^2,\ \dfrac{1}{2}\pi\left(\dfrac{c}{2}\right)^2\) or at least two of \(\dfrac{1}{2}\pi x^2,\ \dfrac{1}{2}\pi y^2,\ \dfrac{1}{2}\pi z^2\) or at least two of \(\dfrac{1}{2}\pi\left(\dfrac{5}{2}\right)^2,\ \dfrac{1}{2}\pi\left(\dfrac{3}{2}\right)^2,\ \dfrac{1}{2}\pi\left(\dfrac{4}{2}\right)^2\) | |
| C1 | for a fully correct and convincing chain of reasoning, eg showing that eg \(\dfrac{1}{2}\pi\left(\dfrac{a}{2}\right)^2 = \dfrac{1}{2}\pi\left(\dfrac{b}{2}\right)^2 + \dfrac{1}{2}\pi\left(\dfrac{c}{2}\right)^2\) can be reduced to \(a^2 = b^2 + c^2\) or that \((2x)^2 = (2y)^2 + (2z)^2\) is the same as \(\dfrac{1}{2}\pi x^2 = \dfrac{1}{2}\pi y^2 + \dfrac{1}{2}\pi z^2\) |
Additional guidance
May be seen at any stage
Where \(d_A\), \(a\), \(2x\), etc are their diameters
Could be any Pythagorean triple
Where \(a\), \(b\), \(c\) are their diameters
Where \(2x\), \(2y\), \(2z\) are their diameters
Where 3, 4, 5 are their diameters