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Exercise 6.5 · Q13

Q.If the equation of the base, opposite to the vertex (2,3)(2, 3), of an equilateral triangle is x+y=2x+y=2, then the length of a side is

(1) 32\sqrt{\dfrac32}
(2) 66
(3) 6\sqrt6
(4) 323\sqrt2
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The perpendicular distance from vertex (2,3)(2,3) to the base x+y=2x+y=2 is the triangle's height hh; use h=32sh=\dfrac{\sqrt3}{2}s for an equilateral triangle to find the side ss.

Step 1. Find the perpendicular distance from the vertex (2,3)(2,3) to the base line x+y−2=0x+y-2=0.

h=∣2+3−2∣12+12=32.h = \frac{|2+3-2|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt2}.

This distance is exactly the triangle's altitude from the opposite vertex to the base.

Step 2. Relate the altitude to the side length of an equilateral triangle. For an equilateral triangle of side ss, the altitude is

h=32s.h = \frac{\sqrt3}{2}s.

Step 3. Solve for ss.

s=2h3=23⋅32=66=6.s = \frac{2h}{\sqrt3} = \frac{2}{\sqrt3}\cdot\frac{3}{\sqrt2} = \frac{6}{\sqrt6} = \sqrt6. …

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