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**Note - the problem I'm struggling with is how to calculate the area of APBQ (the last question)

Figure 1 on the right shows a right-angled triangle ABC where AB = 1 cm, AC = 2 cm, and angle BAC = 90°. Triangle PAB is an isosceles triangle where AP = AB and sides PA and BC are parallel. Assume point P is located opposite to point C with respect to line AB.

Answer the following questions.

〔Question 1 〕 Consider the case in Figure 1 where the magnitude of angle APB is a°. Find the magnitude of angle ACB in terms of a.

〔Question 2 〕 Figure 2 on the right shows the case in Figure 1 where a perpendicular line to side BC is drawn from vertex A. Let Q be the intersection of side BC and the perpendicular line.

Answer (1)and (2).

(1) Prove triangle ABQ is similar to triangle CAQ.

(2) Calculate the area of quadrilateral APBQ.

  • 1
    $\begingroup$ Do you have the working out for all the other parts? This would really help answer the question. $\endgroup$ – Toby Mak Aug 7 at 11:25
  • $\begingroup$ The first question is irrelevant, and I've already proved the second part (if you see the image you should be able to see the third question). This is what I'm really struggling with. $\endgroup$ – mathinjpm Aug 7 at 11:28
  • $\begingroup$ ABQ is without a doubt, similar to CAQ $\endgroup$ – mathinjpm Aug 7 at 11:29
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    $\begingroup$ In part (1) of question 2, you are asked to show that triangle ABQ is similar to triangle CAQ. Once you know that, and that side AB is half as long as side BC, you know that the area of CAQ is one fourth the area of ABQ. $\endgroup$ – user247327 Aug 7 at 11:31
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    $\begingroup$ You don't need $PB$ to find the area. One side and one altitude is enough. You have the side $AP = 1,$ and $AQ$ equals the altitude to that side. $\endgroup$ – David K Aug 7 at 18:23

Question (1):

  • If $\angle ~ APB = a^{\circ}$, then $\angle ~ PBA = a^{\circ} $, because $\Delta APB$ is isosceles and $AP = AB$.
  • If $\angle ~ APB = a^{\circ} $ and $\angle ~ PBA = a^{\circ} $, then $\angle ~ BAP = 180^{\circ}-2a^{\circ}$.
  • If $PA\mathbin{\|} BC$, then $ \angle ~CBA = 180^{\circ}-2a^{\circ}$ (alternate interior angles).
  • If $\angle ~CBA = 180^{\circ}-2a^{\circ}$ and $\angle~ BAC = 90^{\circ} $, then $\angle ~ ACB = 2a^{\circ}-90^{\circ}$.

Question (2):

The geometrical situation

Another way of doing things is $AQ/AB=AC/BC$, because $\Delta BQA \sim \Delta BAC $ are similar. With $BC=\sqrt{AB^2+AC^2}=\sqrt{5}$ we get $AQ = AB \cdot (AC/BC) = 2/\sqrt{5}$. From there we can calculate $BQ = \sqrt{AB^2-AQ^2}=\sqrt{1-(2/\sqrt{5})^2}=1/\sqrt{5}$ and the trapezoid area according to the above formula

$$ A_{APBQ} = 2\Phi/5 $$

where $\Phi=(1+\sqrt{5})/2$ is the golden ratio.

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    $\begingroup$ Thank you very much - I really appreciate it! $\endgroup$ – mathinjpm Aug 7 at 23:22

Following on from your comment:

$\angle CAB = \tan^{-1} (2)$, and since $PA$ and $BC$ are parallel, $\angle PAB = \tan^{-1} (2)$ as well.

Now if you split $\Delta PAB$ in half where $M$ is the midpoint of $PB$, you will have $\sin PAM = \frac{PM}{PA} = \frac{PM}{1}$. This gives me a value of $PM = \sqrt{\frac{2}{5+\sqrt5}}$ and $PB= \sqrt{2 - \frac{2}{\sqrt5}}$.


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