PT3 Mathematics 2016, Question 6

Question 6 (a):
(i) Fill in the blanks with the correct answer.



(ii) Diagram 6.1 shows a set of triangles.


On the answer space, mark (\/) for the right-angled triangles and mark (x) for the not right-angled triangles.

Answer:




Solution:
(i)
$\begin{array}{l}{k}^2={\boxed{m}}^2+{\boxed{n}}^2\\ \\ x=\sqrt{{\boxed{5}}^2-{\boxed{4}}^2}\end{array}$

(ii)
In ∆ABE,
AE² = AB² + BE²
= 8 ² + 17²
= 353
≠ 19²
Therefore, ∆ABE is not a right-angled triangle.

In ∆BCE,
BE² = BC² + CE²
= 9 ² + 12²
= 15²
≠ 17²
Therefore, ∆BCE is not a right-angled triangle.

In ∆CDE,
CD² = CE² + DE²
= 12 ² + 5²
= 169
= 13²
Therefore, ∆CDE is a right-angled triangle.

Question 6 (b):
Diagram 6.1 shows a polygon.


(a) If the polygon is redrawn using the scale 1 : 500, calculate the length of side drawn for the side 15 m.
(b) On the square grids in the answer space, redraw the polygon using the scale 1 : 500. The grid has equal squares with sides of 1 cm.

Solution:
(a)
$\begin{array}{l}\text{Length of side of the drawing}\\ =\frac{15}{5}\\ =3\text{ cm}\end{array}$

(b)

Question 6 (c):
Diagram 6.3 shows a solid cuboid and a solid cube. The base of the cuboid is a square with sides 9 cm.


The total surface of area of the cuboid and the cube are equal.
Calculate the length of the side of the cube.

Solution:
$\begin{array}{l}\text{Total surface area of cuboid}\\ =4\left(\text{area of rectangle}\right)+2\left(\text{area of cube}\right)\\ =4\left(19.5\times 9\right)+2\left(9\times 9\right)\\ =702+162\\ =864{\text{ cm}}^2\\ \\ \text{Total surface area of cube}=864{\text{ cm}}^2\\ \text{Area 1 surface of cube}\\ =\frac{864}{6}\\ =144{\text{ cm}}^2\\ \\ \text{Length of side of the cube}=x^2\\ x^2=144\\ x=\sqrt{144}\\ x=12\text{ cm}\end{array}$