Solutions

Solutions are only provided for even-numbered questions.

1. There is no solution provided for this question.
2. The following triangles are both isosceles but they are not similar. The also both contain one side length equal to \(1\). Therefore, statements i) and ii) are false.

   

   The following two isosceles triangles each contain at least one \(20^{\circ}\) angle but are not similar. Therefore, statement iii) is false.

   

   For statement iv), we first note that a triangle can contain at most one \(120^{\circ}\) angle since the sum of the angles in a triangle is \(180^{\circ}\). Given that the triangle is isosceles, this further implies that the two remaining angles must be equal to one another and must sum to \(60^{\circ}\). That is, they must both be equal to \(30^{\circ}\). Since all three angles are fixed by the statement, then similarity will be guaranteed by the angle-angle similarity rule. Therefore, statement iv) is true.

3. There is no solution provided for this question.
4. 1. Solutions will vary but one approach is to recognize that \(\angle B = \angle D\) since both are right angles and \(\angle ACB = \angle DCE\) are equal since they are opposite angles. Therefore, \(\triangle ABC \sim \triangle EDC\) by the angle-angle similarity rule.
   2. We know from the ordering of vertices in the similarity relation (i.e., \(\triangle ABC \sim \triangle EDC\)) that side \(AB\) corresponds to side \(DE\). Therefore, the ratio \(\dfrac{AB}{DE} = \dfrac{3}{2}\) tells us that a side length in \(\triangle ABC\) is a factor of \(\dfrac{3}{2}\) times that of the corresponding side in \(\triangle EDC\). Conversely, a side in \(\triangle EDC\) is a factor of \(\dfrac{2}{3}\) times that of the corresponding side in \(\triangle ABC\). Since sides \(AC\) and \(CE\) are corresponding, we can deduce that \(\dfrac{CE}{AC} = \dfrac{2}{3}\).
   3. We first note that \(\triangle EDC\) is a right-angled triangle. This means that we can use the Pythagorean Theorem to solve for \(DE\).

      \( DE^2 = CE^2 - CD^2 = 20^2 - 16^2 = 144 \quad \implies \quad DE = 12\)

      Next, we recall from part b) that side lengths in \(\triangle ABC\) are \(\dfrac{3}{2}\) times that of the corresponding side lengths in \(\triangle EDC\). It follows that:

      • \(AB = \dfrac{3}{2} DE = \dfrac{3}{2}(12) = 18\)
      • \(BC = \dfrac{3}{2} CD = \dfrac{3}{2}(16) = 24\)
      • \(AC = \dfrac{3}{2} CE = \dfrac{3}{2}(20) = 30 \)

      Therefore, the side lengths of \(\triangle ABC\) are \(AB = 18\), \(BC = 24\), and \(AC = 30\).

5. There is no solution provided for this question.
6. In each case, the flagpole, the shadow, and a line connecting the top of the flagpole to the end of its shadow form a right-angled triangle. Moreover, since the Sun's rays come from the same angle, these right-angled triangles will be similar by the angle-angle similarity rule.

   

   Since the two triangles are similar, corresponding side lengths will be in equal proportions. If we let \(x\) denote the length of the shadow cast by the Jeddah flagpole, then this means

   \( \dfrac{x}{74.1} = \dfrac{171}{45.7}\)

   Multiplying both sides by \(74.1\) and simplifying, we find

   \( x = \dfrac{171}{45.7}(74.1) \approx 277.2 \)

   Therefore, the shadow case by the Jeddah flagpole would be approximately \(277\) m long.

7. There is no solution provided for this question.
8. Let us begin by sketching the situation.

   

   Let us label the student's head with the point \(A\) and the student's feet with point \(B\). Also, let's label the point where the student looks through the floor with \(C\) and the floor at the opposite end of the walkway with \(D\). Finally, let us label the point at the base of the building that the student is staring at with \(E\). This allows us to represent the situation with the following diagram.

   

   Note that we are given \(BD = 17\) and \(BC\ = 2.0\) from which we can deduce that \(CD = 15\).

   Observe that \(\angle B = \angle D\) since both are right angles. Also, \(\angle ACB = \angle DCE\) since these are opposite angles. By the angle-angle similarity rule, we can conclude that \(\triangle ABC \sim \triangle EDC\).

   By taking the ratio of the lengths of corresponding sides \(BC\) and \(CD\), we can find the scale factor from \(\triangle ABC\) to \(\triangle EDC\).

   \[ s = \dfrac{CD}{BC} = \dfrac{15}{2.0} = 7.5\]

   We now use this scale factor and the fact that sides \(AB\) and \(DE\) are corresponding to compute the length of \(DE\).

   \[ s = \dfrac{DE}{AB} \quad \implies \quad DE = s \cdot AB =7.5(1.6) = 12\]

   Therefore, the floor of the walkway is roughly \(12\) metres above the ground.

9. There is no solution provided for this question.
10. Recall the camera has dimensions \(21\) cm by \(34\) cm by \(34\) cm. Let's imagine the camera taking a picture of Daniel from a distance \(x\) away from the front of the camera with the image of Daniel filling up the photosensitive paper from top to bottom.

    

    In this optimal scenario, the light rays from the top of Daniel's head travel through the pinhole to the bottom of the photosensitive paper. Likewise, the light rays from the bottom of Daniel's feet travel through the pinhole to the top of the photosensitive paper. These rays along with Daniel and the photosensitive paper form a pair of isosceles triangles. Drawing a horizontal line through these triangles and passing through the pinhole, we can further form two pairs of congruent right-angled triangles.

    

    Observe that, by the angle-angle similarity rule, each right-angled triangle inside the camera is similar to each right-angled triangle outside of the camera. Thus, the side lengths of a right-angled triangle inside the camera must all be in the same proportion to the side lengths of a right-angled triangle outside of the camera . This means that the ratio of half of Daniel's height (\(90.0\) cm) to half of the camera height (\(17.0\) cm) must be the same as the ratio of the depth of the camera (\(21.0\) cm) to the distance from the camera to Daniel.

    \[ \frac{90.0}{17.0} = \frac{x}{21.0} \]

    Solving for \(x\), we find \(x \approx 111.2\). Therefore, the closest that the camera can be to Daniel is approximately \(111\) cm.