Question on: JAMB Mathematics - 2010
A
7cm
B
8cm
C
5cm
D
6cm
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Correct Option: D
Let A denote the area of \(\bigtriangleup\)PQR, then A = \(\frac{1}{2} bh\)
Using Sin 60o = \(\frac{h}{q}\)
h = q sin 60o
So A = \(\frac{1}{2}b(q \sin 60^o)\)
12\(\sqrt{3} = \frac{1}{2} \times 8 \times q \times \frac{\sqrt{3}}{3}\)
12\(\sqrt{3}\) - 2q\(\sqrt{3}\)
q = \(\frac{12}{2} = 6\)cm
Using Sin 60o = \(\frac{h}{q}\)
h = q sin 60o
So A = \(\frac{1}{2}b(q \sin 60^o)\)
12\(\sqrt{3} = \frac{1}{2} \times 8 \times q \times \frac{\sqrt{3}}{3}\)
12\(\sqrt{3}\) - 2q\(\sqrt{3}\)
q = \(\frac{12}{2} = 6\)cm
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