Area of triangle ABC = bh = 6 3
= 9 cm2

The properties of this triangle tell us that the perpendicular line of an isosceles triangle meets the base at angles of 900 and thus we have two right angles triangles. Therefore, we can find the lengths of the two sides using Pythagoras’ theorem. AC and BC are both considered as hypotenuses of their right angled triangles. We know that the perpendicular height of the triangle is 3 cm and AD and BD are 3 cm as well. Thus, AC2 = 33 + 32 = 18, BC2 = 32 + 32 = 18
AC = = , BC = =
The perimeter of triangle ABC is the sum of AB + AC + BC
P = + + 6
= + 6
14.5 cm
* Note: We reject the negative values for the figures above as there cannot be a negative value for a geometric shape thus we only apply the positive value.
Task 3 Find the area enclosed by the sector DCE shown below.
In order to find the area of sector DCE, the formula: where r is radius. In this case, both CD and CE can be counted as a radius, as like AC and BC, they are the same length. We can see this by subtracting the bigger line to the smaller one. Thus, AE – AC = BD – BC
This can be seen in the following way: CD = 6  ≈ 1.76 cm, CE = 6  ≈ 1.76cm
Therefore, 6 will be used as the value for r.
To solve the area of this sector, all that is now required is angle. In order to have this, we must find angles ACO and BCO. Since it has been justified that the triangle is an isosceles triangle, the angles of ACO and BCO are the same. ∠ = tan= ∴ = = = 450
Thus angle ACE is 900 and therefore is right angled. As ∠ ACB and DCE are vertically opposing angles, ∠ DCE is also 900. We can now substitute the information we have to determine the area of the sector: where is 900 and r is 6 .
2 ≈ 2.43cm2
Task 4 Find the area of the sector BAE in the diagram below.
To find sector BAE in the circle, we also must apply the formula: . In the diagram displayed below, two internal angles of the two right angled triangles have been found. Therefore, the quickest method to determine ∠BAE is to subtract 1800 from the two known angles, as it is known that all internal angles of any triangle is equal to 1800.
∴ ∠BAE = 180 – 90 – 45
= 45 0
The second value needed in calculating this sector is the radius. ∠BAE falls into both larger circles of the diagram. The length AB is the radius of these two circles, which we know to be 6cm in length. Therefore, the radius for determining the area of this sector is 6 cm. Thus, we simply substitute into the formula.
2
≈ 14.14 cm2
Task 5 Look for the shape of an egg in the enclosed areas of the circles shown below. This is called Moss’s Egg. Find the area and perimeter of Moss’s Egg.

In order to determine the area of Moss’s Egg, we must find the areas of the segments it is made up of. The area of the egg is therefore the sum of the semicircle AB, sector DCE, sector ABD, sector BAE as well as the subtraction of the triangle ABC as this area has been included when calculating sectors ABD and EAB. Therefore, the following operations would need to take place when calculating the area of Moss’s Egg:
A Moss’s Egg = A small ⊙ AB + A sector BAE+ A sector ABD + A sector DCE – A ΔABC
∴ A Moss’s Egg = + 2 + 2 + 2 – 9
≈ 35.8 cm2

Determining the perimeter of Moss’s Egg involves similar processes as that in finding its area, as this too, involves breaking up Moss’s Egg into its smaller segments. The perimeter of Moss’s Egg would therefore be the sum of half the circumference of circle AB as well as the arcs of ABD, BAE and DCE. To find the arc lengths of the sectors above, we employ the formula: l . Thus, Arc AD =
Arc BE =
Arc DE = =
To find the circumference of the semi circle, we use the formula and then divide by two in order to obtain the circumference for only half the circle.
∴ C semicircle = = 3
Thus, the perimeter of the egg can be calculated by the following processes:
C semicircle + Arc AD + Arc BE + Arc DE
∴ P Moss’s Egg = 3 + +
+
≈ 21.6 cm
Task 6 If the radii of the two large circles shown below are r, find a formula for the area and perimeter of the egg in terms of r.

The information that we are provided states that AB equals r. Therefore, it can be said that the radius of the smaller circle equals. Through this, we deduce that AC will equal by using Pythagoras’s Theorem. The radius of sector DCE would then therefore equal to:
Sector DCE = = r
Thus, in order to find a formula for determining the area of Moss’s Egg, the following equation must be used, with substituted figures in terms of r.
A Moss’s Egg = A small ⊙ AB + A sector BAE+ A sector ABD + A sector DCE – ΔABC
∴ A Moss’s Egg = 2 + r2 + r2 + 2
= 2 +2 + 2  2
= r2 ( + + )
= r2 ( )
≈ 0.995r2
To test this formula, the radius 6cm can be substituted into r, which allows the answer calculated by this formula to be compared with that of question 5a.
0.995(6)2 = 35.82 cm
Therefore, the suggested formula has been supported.
 Much like section A of this question, finding the perimeter of the egg consists of substituting the calculated figures in the previous questions into their relationships in terms of r. Therefore, a formula for the perimeter of Moss’s Egg can be calculated using the method displayed below.
C semicircle + Arc AD + Arc BE + Arc DE
= 2 + + ()
= x
= =
≈ 3.6r
To test this formula, substitute r to 6 and compare with answer calculated from question 5b.
3.6 x 6 = 21.6 cm
Therefore the formula above has been supported.
Bibliography
Haese, S et al. 2006, Mathematics For The International Student. Haese & Harris Publications, Australia