21M.1.SL.TZ2.6

pestleMathematicsAASLPaper 121M· sl-3-2-2d-and-3d-trig-sine-rule-cosine-rule-areasource ↗

The following diagram shows triangle $ABC$ , with $AB = 10$ , $BC = x$ and $AC = 2 x$ .

Given that $\cos \hat{C} = \frac{3}{4}$ , find the area of the triangle.

Give your answer in the form $\frac{p \sqrt{q}}{2}$ where $p, q \in ℤ^{+}$ .

Markscheme / solution

METHOD 1

attempt to use the cosine rule to find the value of $x$ (M1)

$100 = x^{2} + 4 x^{2} - 2 (x) (2 x) (\frac{3}{4})$ A1

$2 x^{2} = 100$

$x^{2} = 50$ OR $x = \sqrt{50} (= 5 \sqrt{2})$ A1

attempt to find $\sin \hat{C}$ (seen anywhere) (M1)

$\sin^{2} \hat{C} + {(\frac{3}{4})}^{2} = 1$ OR $x^{2} + 3^{2} = 4^{2}$ or right triangle with side $3$ and hypotenuse $4$

$\sin \hat{C} = \frac{\sqrt{7}}{4}$ (A1)

Note: The marks for finding $\sin \hat{C}$ may be awarded independently of the first three marks for finding $x$ .

correct substitution into the area formula using their value of $x$ (or $x^{2}$ ) and their value of $\sin \hat{C}$ (M1)

$A = \frac{1}{2} \times 5 \sqrt{2} \times 10 \sqrt{2} \times \frac{\sqrt{7}}{4}$ or $A = \frac{1}{2} \times \sqrt{50} \times 2 \sqrt{50} \times \frac{\sqrt{7}}{4}$

$A = \frac{25 \sqrt{7}}{2}$ A1

METHOD 2

attempt to find the height, $h$ , of the triangle in terms of $x$ (M1)

$h^{2} + {(\frac{3}{4} x)}^{2} = x^{2}$ OR $h^{2} + {(\frac{5}{4} x)}^{2} = 10^{2}$ OR $h = \frac{\sqrt{7}}{4} x$ A1

equating their expressions for either $h^{2}$ or $h$ (M1)

$x^{2} - {(\frac{3}{4} x)}^{2} = 10^{2} - {(\frac{5}{4} x)}^{2}$ OR $\sqrt{100 - \frac{25}{16} x^{2}} = \frac{\sqrt{7}}{4} x$ (or equivalent) A1

$x^{2} = 50$ OR $x = \sqrt{50} (= 5 \sqrt{2})$ A1

correct substitution into the area formula using their value of $x$ (or $x^{2}$ ) (M1)

$A = \frac{1}{2} \times 2 \sqrt{50} \times \frac{\sqrt{7}}{4} \sqrt{50}$ OR $A = \frac{1}{2} (2 \times 5 \sqrt{2}) (\frac{\sqrt{7}}{4} 5 \sqrt{2})$

$A = \frac{25 \sqrt{7}}{2}$ A1

[7 marks]

Examiners’ report

[N/A]