WAEC 2019 · Paper 2 · Q13

(a)
The curved surface areas of two cones are equal. The base radius of one is $5\text{ cm}$ and its slant height is $12\text{ cm}$ . Calculate the height of the second cone if its base radius is $6\text{ cm}$ .
(b)
Given the matrices $\mathbf A = \begin{pmatrix} 2 & 5 \\ -1 & -3 \end{pmatrix}$ and $\mathbf B = \begin{pmatrix} 3 & -2 \\ 4 & 1 \end{pmatrix}$ , find: (i) $\mathbf{BA}$ ; (ii) the determinant of $\mathbf{BA}$ .

Worked solution (try it first)

(a)

The curved surface of a cone is $\pi r l$ .
The first cone has $\pi \times 5 \times 12 = 60\pi$ .
For the second, $\pi \times 6 \times l = 60\pi$ , so its slant height is $l = 10$ cm.
Height of the second cone: $h = \sqrt{10^2 - 6^2} = \sqrt{64} = 8$ cm.

(b)(i)

Multiply rows of $\mathbf B$ by columns of $\mathbf A$ : $\mathbf{BA} = \begin{pmatrix} 3(2) + (-2)(-1) & 3(5) + (-2)(-3) \\ 4(2) + 1(-1) & 4(5) + 1(-3) \end{pmatrix}$
$= \begin{pmatrix} 8 & 21 \\ 7 & 17 \end{pmatrix}$ .

(ii)

$|\mathbf{BA}| = 8 \times 17 - 21 \times 7$
$= 136 - 147$
$= -11$ .

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