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IN the trapezium ABCD, the diagonals intercept at M. Let AM= a, BM= b, Cm = c and DM = d, and let Angle AMB be $\theta$.







$\cos(\theta) = -\frac{1}{4}$ and $\sin (\theta) = \frac{(15^{0.5})}{4}$ Find the area of the trapezium in exact form.

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I am just wondering how $\sin \theta$ can be greater than one. – Mhenni Benghorbal Sep 21 '13 at 13:44
What did you try? A hint: $\triangle AMB\cong \triangle DMC$ so you know proportion between basis and parts of height so... – user65985 Sep 21 '13 at 13:45
@CoarguAliquis I found the area of CMD and ABM using trigonometry, however, i could not find DMA or CMB. – Berry Sep 21 '13 at 13:50
First, Mhenni is right, so either you or your teacher had a little mistake. try cosine law in AMD and BMC if you want it in this way... – user65985 Sep 21 '13 at 14:09
There isn't a mistake $$\sin {\theta} = \frac{15^{0.5}}{4} = \frac{15^{\frac 12}}{4} = \frac {\sqrt{15}}{4} < 1$$ – Stefan4024 Sep 21 '13 at 14:14
up vote 1 down vote accepted

It's obvious that $\angle AMB + \angle AMD = 180^{\circ}$, so we have: $\angle AMD = 180^{\circ} - \theta$. Now we want to find the cosine of $\angle AMD$.

$$\cos {\angle AMD} = \cos ({180^{\circ} - \theta}) = \cos {180^{\circ}} \cdot \cos {\theta} + \sin {180^{\circ}} \cdot \sin {\theta}$$

We know that $\sin {180^{\circ}} = 0$ and $\cos {180^{\circ}} = -1$. So we substitute:

$$\cos {\angle AMD} = (-1)\cdot(-\frac 14) + 0 \cdot \frac{\sqrt{15}}{4} = \frac 14$$

Note that also $\cos {\angle BMC} = \frac 14$

Now apply the law of cosine on $\triangle AMD$ and $\triangle BMC$.

$$AD^2 = AM^2 + DM^2 - 2\cdot AM \cdot DM \cdot \cos {\angle AMD}$$ $$AD^2 = 6^2 + 2^2 - 2\cdot 6\cdot 2 \cdot \frac 14$$ $$AD^2 = 36 + 4 - 6 = 34 \implies AD = \sqrt{34}$$

$$BC^2 = BM^2 + CM^2 - 2\cdot BM \cdot CM \cdot \cos {\angle BMC}$$ $$BC^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \frac 14$$ $$BC^2 = 16 + 9 - 6 = 19 \implies BC =\sqrt{19}$$

Now just apply Heron's formula for area of the triangle on $\triangle ACD$ and $\triangle BAC$. The area of the trapezium will be the sum of the areas of these two triangles. You can do this by yourself.

The final answer should be $\approx 26.14$ or in exact form $\frac{27 \cdot \sqrt{15}}{4}$


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