(a)
In the diagram, \(\Delta\) ABD is right-angled at B. |AB| = 3 cm, |AD| = 5 cm, \(\stackrel\frown{ACB}\) = 61° and \(\stackrel\frown{DAC}\) = x°. Calculate, correct to one decimal place, the value of x.
(b)
In the diagram, OABCD is a pyramid with a square base of side 2cm and a slant height of 4 cm. Calculate, correct to three significant figures : (i) the vertical height of the pyramid ; (ii) the volume of the pyramid.
In the diagram, \(\Delta\) ABD is right-angled at B. |AB| = 3 cm, |AD| = 5 cm, \(\stackrel\frown{ACB}\) = 61° and \(\stackrel\frown{DAC}\) = x°. Calculate, correct to one decimal place, the value of x.
(b)
In the diagram, OABCD is a pyramid with a square base of side 2cm and a slant height of 4 cm. Calculate, correct to three significant figures : (i) the vertical height of the pyramid ; (ii) the volume of the pyramid.
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Correct Answer: Option
Explanation:
(a) In \(\Delta ACB, < CAB = 180° - (90° + 61°)\)
= \(180° - 151°\)
= \(29°\)
In \(\Delta ADB, < DAB = (29 + x)°\)
\(\cos (29 + x) = \frac{3}{5} = 0.6\)
\((29 + x)° = \cos^{-1} (0.6)\)
\((29 + x)° = 53.13°\)
\(x = 53.13° - 29°\)
= \(24.13°\)
\(\approxeq 24.1°\) (1 decimal place).
(b) (i)
\(DB = 2DN = 2NB\)
\(OA = OB = OC = OD = 4cm\)
\(AB = BC = CD = DA = 2cm\)
\(DB^{2} = BC^{2} + DC^{2}\)
= \(2^{2} + 2^{2}\)
= \(\sqrt{8}\)
= \(2\sqrt{2} cm\)
\(2DN = DB \implies DN = \frac{2\sqrt{2}}{2} = \sqrt{2} cm\)
In \(\Delta DON\),
\(OD^{2} = ON^{2} + DN^{2}\)
\(4^{2} = h^{2} + (\sqrt{2})^{2} \implies h^{2} = 16 - 2\)
\(h^{2} = 14 \implies h = \sqrt{14}\)
(ii) \(V = \frac{1}{3} \times \base area \times \perp distance\)
\(V = \frac{1}{3} \times 2^{2} \times \sqrt{14}\)
= \(\frac{4\sqrt{14}}{3}\)
= \(4.99 cm^{3}\)
(a) In \(\Delta ACB, < CAB = 180° - (90° + 61°)\)
= \(180° - 151°\)
= \(29°\)
In \(\Delta ADB, < DAB = (29 + x)°\)
\(\cos (29 + x) = \frac{3}{5} = 0.6\)
\((29 + x)° = \cos^{-1} (0.6)\)
\((29 + x)° = 53.13°\)
\(x = 53.13° - 29°\)
= \(24.13°\)
\(\approxeq 24.1°\) (1 decimal place).
(b) (i)
\(DB = 2DN = 2NB\)
\(OA = OB = OC = OD = 4cm\)
\(AB = BC = CD = DA = 2cm\)
\(DB^{2} = BC^{2} + DC^{2}\)
= \(2^{2} + 2^{2}\)
= \(\sqrt{8}\)
= \(2\sqrt{2} cm\)
\(2DN = DB \implies DN = \frac{2\sqrt{2}}{2} = \sqrt{2} cm\)
In \(\Delta DON\),
\(OD^{2} = ON^{2} + DN^{2}\)
\(4^{2} = h^{2} + (\sqrt{2})^{2} \implies h^{2} = 16 - 2\)
\(h^{2} = 14 \implies h = \sqrt{14}\)
(ii) \(V = \frac{1}{3} \times \base area \times \perp distance\)
\(V = \frac{1}{3} \times 2^{2} \times \sqrt{14}\)
= \(\frac{4\sqrt{14}}{3}\)
= \(4.99 cm^{3}\)