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Question Number 210608 by peter frank last updated on 13/Aug/24

Commented by peter frank last updated on 13/Aug/24

$help (b)$
	Answered by mr W last updated on 13/Aug/24

	Commented by mr W last updated on 13/Aug/24

	$t h e a r e a o f t h e p a r a l l e l o g r a m w i t h$ $d i a g o n a l s a a n d b i s$ $A = \frac{∣ a ∣ \cdot ∣ b ∣ \sin θ}{2}$ $∣ p ∣=∣ q ∣= 1$ $∣ a ∣^{2} =∣ b ∣^{2} = 1^{2} + 2^{2} - 2 \times 1 \times 2 \cos 150 ° = 5 + 2 \sqrt{3}$ $\cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ $= \frac{(p + 2 q) \cdot (2 p + q)}{∣ a ∣∣ b ∣}$ $= \frac{2 p \cdot p + 2 q \cdot q + 5 p \cdot q}{∣ a ∣∣ b ∣}$ $= \frac{2 + 2 + 5 \cos 30 °}{(5 + 2 \sqrt{3})}$ $= \frac{8 + 5 \sqrt{3}}{2 (5 + 2 \sqrt{3})} = \frac{9 \sqrt{3} + 10}{26}$ $\sin θ = \frac{3 \sqrt{37 - 20 \sqrt{3}}}{26}$ $\Rightarrow A = \frac{(5 + 2 \sqrt{3}) \times 3 \sqrt{37 - 20 \sqrt{3}}}{52} = 0.75 ✓$
	Commented by ajfour last updated on 14/Aug/24

	$\frac{1}{2} ∣ (p + 2 q) \times (q + 2 p) ∣$ $= \frac{1}{2} {- \frac{1}{2} + 4 (\frac{1}{2})} = \frac{3}{4}$
	Commented by mr W last updated on 14/Aug/24
	��
	Commented by peter frank last updated on 14/Aug/24

	$thank you$
	Answered by mr W last updated on 14/Aug/24

	Commented by mr W last updated on 16/Aug/24

	$t h e a r e a o f t h e p a r a l l e l o g r a m w i t h a$ $a n d b a s d i a g o n a l s i s h a l f o f t h e a r e a$ $o f t h e p a r a l l e l o g r a m w i t h a a n d b$ $a s s i d e s .$ $A = \frac{∣ a ∣∣ b ∣ \sin θ}{2} = \frac{∣ a \times b ∣}{2}$ $= \frac{∣ (p + 2 q) \times (2 p + q) ∣}{2}$ $= \frac{∣ 2 p \times p + 4 q \times p + p \times q + 2 q \times q ∣}{2}$ $= \frac{3 ∣ q \times p ∣}{2} = \frac{3 \times 1 \times 1 \times \sin 30 °}{2} = \frac{3}{4} ✓$
	Commented by peter frank last updated on 14/Aug/24

	$thank you .$
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