**A-Level** **Mathematics**** Question Analysis Topic: Mathematics - **

**Exam Questions: **

1) In the figure, a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of $500 / sq. metre. Consider (π = 22 / 7)

2) Prove that the points (−3, 0), (1,−3) and (4,1) are the vertices of a right-angled isosceles triangle.

**Answers:**

For A-Level Mathematics, you should know:

1) The radius of the conical piece = radius of cylindrical part r = **3 / 2 m.**

The slant height of the cone l is **2.8 m**.

The height of the cylinder h is **2.1 m**.

Therefore, the **canvas needed to make the tent** = curved surface area of the cone + curved surface area of the cylinder.

Curved surface area of the cone = **πrl** = 22 / 7 * 3 / 2 * 2.8 = 13.2 m^{2}

On the other hand, the curved surface area of the cylinder is 2πrh = 2 x 22 / 7 x 3 / 2 x 2.1 = **19.8 m**^{2}

Hence, the total surface area = curved surface area of the cone + curved surface area of the cylinder = 13.2 + 19.8 = **33 m**^{2}

Because the cost of 1 m2 of canvas is $500, the cost of 33 m^{2} would be **$16,500**.

2) The solution to this is using the formula of the distance between two points: **√[(x**_{2}** - x**_{1}**)**^{2}** + (y**_{2}** - y**_{1}**)**^{2}**]**

AB = √[(1 + 3)^{2} + (-3 - 0)^{2}] = 5

BC = √[(4 - 1)^{2} + (1 + 3)^{2}] = 5

AC = √[(4 + 3)^{2} + (1 - 0)^{2}] = 5√2

AB = BC

From this, we're halfway there - we now know that ABC is an isosceles triangle.

**Now, let's use the Pythagoras' theorem**.

(AB)2 + (BC)2 = 5^{2} + 5^{2} = 50

and (AC)^{2 }= (5√2)^{2} = 50

Therefore, we have found that **AB**^{2}** + BC**^{2}** = AC**^{2}, which satisfies the theorem and is hence** right-angled**.

Work hard for your A-Level Mathematics examination!

End of analysis. Great!