Live practice

The exposed surface consists of the cylinder's curved surface and the curved surfaces of two hemispherical hollows. Thus,

With \(\pi=\frac{22}{7}\), \(A=374\text{ cm}^2\).

The circumference is (\pi d=\(\frac{22}{7}\)\(35\)=110) mm. The five diameters require (5\(35\)=175) mm. Hence total wire \(=110+175=285\) mm.

The radius is \(17.5\) mm, so one of ten equal sectors has area

Let the tangent touch the circle at \(P\), and let \(O\) be the centre. By the tangent-radius theorem, \(OP\) is perpendicular to the tangent at \(P\).

Only one perpendicular to a given line can pass through a specified point. Therefore the perpendicular erected at \(P\) is the line \(OP\), which passes through the centre \(O\).

Formula or theorem used: Draw and label a right triangle for each line of sight. Use \(\tan\theta=\frac{\text{vertical height}}{\text{horizontal distance}}\).

Working: Define the unknown height or distance, write a tangent equation for each observation, solve the equations, and include the observer's height when the question requires it.

After substitution and simplification:

216 m3

Formula or theorem used: In a right triangle use \(\sin\theta=\frac{P}{H}\), \(\cos\theta=\frac{B}{H}\), and \(\tan\theta=\frac{P}{B}\), together with the standard-angle values and identities.

Working: Write the relevant ratio or identity, substitute the known value, simplify one equality at a time, and state restrictions when a denominator could be zero.

After substitution and simplification:

(i) 1 (ii) 0

Formula or theorem used: Use \(PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) or the section formula \(\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)\).

Working: Substitute the coordinates before simplifying. For a shape, compare the required side lengths or diagonals and state the conclusion.

After substitution and simplification:

4, QR = 41, PR = 82 , 9 2±

Since \(AB\parallel PQ\), triangles \(OAB\) and \(OPQ\) are similar. Therefore,

Since \(AC\parallel PR\), triangles \(OAC\) and \(OPR\) are similar. Hence,

From \(1\) and \(2\),

Thus \(B\) and \(C\) divide \(OQ\) and \(OR\) in the same ratio. By the converse of the Basic Proportionality Theorem in \(\triangle OQR\), \(BC\parallel QR\).

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

(i) 34 (ii) 27

Formula or theorem used: Write the equation in the form \(ax^2+bx+c=0\). Factorise where possible; otherwise use \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\).

Working: Display the standard form, the factorisation or discriminant calculation, both roots, and reject any value that is impossible in the stated context.

After substitution and simplification:

(i) Real roots do not exist (ii) Equal roots; 2 2,
3 3
(iii) Distinct roots; 3 3
2
±

Formula or theorem used: Translate the information into two linear equations. Use the method named in the exercise, keeping equivalent equations aligned.

Working: Eliminate or substitute one variable, obtain the other variable, substitute back, and verify the ordered pair in both original equations.

After substitution and simplification:

x = –2, y = 5; m = –1