In the figure shown, show that $∠ 1 + ∠ 3 + ∠ 8 + ∠ 6 = 360^{0}$ . [2 MARKS]

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Solution

Steps : 1 Mark
Proof : 1 Mark

$∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 = 360^{0}$

$∠ 4 = ∠ 8$ [Pair of corresponding angles]

and $∠ 2 = ∠ 6$ [Pair of corresponding angles]

$\Rightarrow ∠ 1 + ∠ 3 + ∠ 8 + ∠ 6 = 360^{0}$

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Similar questions

In the given figure, show that $∠ 1 + ∠ 3 + ∠ 8 + ∠ 6 = 360^{0}$ . [2 MARKS]

Show that four points (0, – 1), (6, 7), (–2, 3) and (8, 3) are the vertices of a rectangle. Also, find its area. [3 MARKS]

\frac{1}{(3 - \sqrt{8})} - \frac{1}{(\sqrt{8} - \sqrt{7})} + \frac{1}{(\sqrt{7} - \sqrt{6})} - \frac{1}{(\sqrt{6} - \sqrt{5})} + \frac{1}{(\sqrt{5} - 2)} = 5

\frac{1}{(3 - \sqrt{8})} - \frac{1}{(\sqrt{8} - \sqrt{7})} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - 2} = 5

\frac{4}{1 +} \frac{6}{2 +} \frac{8}{3 +} \dots \frac{2 n + 2}{n +} \dots = \frac{2 (e^{2} - 1)}{e^{2} + 1}

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