(a) AB is a chord of a circle centre O. If |AB| = 24.2 cm and the perimeter of \(\Delta\) AOB is 52.2 cm, calculate < AOB, correct to the nearest degree.
(b) A rectangular tank 60cm by 80cm by 100cm is half filled with water. How many litres of water is it holding?
(b) A rectangular tank 60cm by 80cm by 100cm is half filled with water. How many litres of water is it holding?
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Correct Answer: Option n
Explanation:
(a)
\(2r + 24.2 = 52.2\)
\(2r = 52.2 - 24.2 = 28\)
\(r = 14 cm\)
\(|AB|^{2} = OA^{2} + OB^{2} - 2(OA)(OB) \cos <AOB\)
\(24.2^{2} = 14^{2} + 14^{2} - 2(14)(14) \cos < AOB\)
\(\cos < AOB = \frac{14^{2} + 14^{2} - 24.2^{2}}{2(14)(14)}\)
\(\cos < AOB = \frac{392 - 585.64}{392}\)
= \(\frac{-193.64}{392}\)
\(\cos < AOB = - 0.4940\)
\(< AOB = \cos^{-1} (0.4940)\)
= \(119.6° \approxeq 120°\)
(b) Volume of water in the tank = \(\frac{1}{2} \times l \times b \times h\)
= \(\frac{1}{2} \times 60 \times 80 \times 100\)
= \(240000 cm^{3}\)
Number of litres of water = \(\frac{240000}{1000}\)
= \(240 litres\)
(a)
\(2r + 24.2 = 52.2\)
\(2r = 52.2 - 24.2 = 28\)
\(r = 14 cm\)
\(|AB|^{2} = OA^{2} + OB^{2} - 2(OA)(OB) \cos <AOB\)
\(24.2^{2} = 14^{2} + 14^{2} - 2(14)(14) \cos < AOB\)
\(\cos < AOB = \frac{14^{2} + 14^{2} - 24.2^{2}}{2(14)(14)}\)
\(\cos < AOB = \frac{392 - 585.64}{392}\)
= \(\frac{-193.64}{392}\)
\(\cos < AOB = - 0.4940\)
\(< AOB = \cos^{-1} (0.4940)\)
= \(119.6° \approxeq 120°\)
(b) Volume of water in the tank = \(\frac{1}{2} \times l \times b \times h\)
= \(\frac{1}{2} \times 60 \times 80 \times 100\)
= \(240000 cm^{3}\)
Number of litres of water = \(\frac{240000}{1000}\)
= \(240 litres\)