Question-178947

Question Number 178947 by HeferH last updated on 23/Oct/22

Answered by mr W last updated on 23/Oct/22

Commented by mr W last updated on 23/Oct/22

\frac{\sin β}{4 k} = \frac{\sin (\frac{π}{4} + β)}{G B}

\frac{\sin α}{3 k} = \frac{\sin (\frac{π}{4} + α)}{G B}

\frac{3 \sin β}{4 \sin α} = \frac{\sin (\frac{π}{4} + β)}{\sin (\frac{π}{4} + α)} = \frac{\cos β + \sin β}{\cos α + \sin α}

α + β = \frac{π}{2}

\frac{3 \cos α}{4 \sin α} = 1

\Rightarrow \tan α = \frac{3}{4}

\frac{4 k}{\sin β} = \frac{G B}{\sin (\frac{π}{4} + β)}

\Rightarrow G B = \frac{4 k (\cos β + \sin β)}{\sqrt{2} \sin β} = \frac{7 \sqrt{2} k}{2}

G H = G B - H B = \frac{7 \sqrt{2} k}{2} - \frac{4 \sqrt{2} k}{2} = \frac{3 \sqrt{2} k}{2}

H E = G H \tan α = \frac{3 \sqrt{2} k}{2} \times \frac{3}{4} = \frac{9 \sqrt{2} k}{8}

m a g e n t a a r e a = \frac{G B \times H E}{2}

= \frac{1}{2} \times \frac{7 \sqrt{2} k}{2} \times \frac{9 \sqrt{2} k}{8} = \frac{63 k^{2}}{16}

a r e a o f s q u a r e = {(4 k)}^{2} = 16 k^{2}

y e l l o w a r e a = 16 k^{2} - \frac{63 k^{2}}{16} = \frac{193 k^{2}}{16}

\frac{m a g e n t a}{y e l l o w} = \frac{63}{196} ✓

Commented by Tawa11 last updated on 23/Oct/22

Great sir .

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