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A triangle is formed inside a square by joining corner C with the mid points of sides AD and AB. If a point in the square is chosen randomly, what is the probability that the point will be inside the triangle as well?

bou

The answer is 37.5%. However, I am unsure of how to arrive at that answer. I calculated the following areas for the unshaded triangles but I don't know what steps to take afterwards:

area of 2 big triangles = $[(s/2)^2 * (1/2)]$× 2 big triangles→ $s^2/4 $

area of 1 small triangle = $[(s/2)^2 * (1/2)]$ ×1 small triangle→ $s^2/8$

Total area of unshaded triangles = $s^2/4+s^2/8=(3s^2)/8$

orangebull
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1 Answers1

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So the first thing to note here is that your area for the two big triangles (lets call them $A_1, A_2$) and the small triangle (call it $A_3$) is wrong...
The area for the a triangle is simple base multiplied by height multiplied by 0.5.
If you look carefully, one of the sides (for the big triangle) is $\frac{s}{2}$ while the other is $s$, hence the area is: $$ A_1 = A_2 = \frac{s}{2}\cdot s\cdot \frac{1}{2} = \frac{s^2}{4} $$ The base and height of the small triangle is $\frac{s}{2}$ each. Hence, its area is: $$ A_3 = \frac{s}{2}\cdot\frac{s}{2}\cdot \frac{1}{2} = \frac{s^2}{8} $$ Hence total area covered by non-shaded area is: $$ A_1 + A_2 + A_3 = \frac{s^2}{4} + \frac{s^2}{4}+ \frac{s^2}{8} = \frac{5s^2}{8} $$ we know area covered by shaded part = Area of square - area covered by non-shaded part: $$ s^2 - \frac{5s^2}{8} = \frac{3s^2}{8} = 0.375\cdot s^2 $$ Probability of landing in shaded part = $\frac{\text{area of shaded part}}{\text{total area}}$ $$ \therefore P = \frac{0.375s^2}{s^2} = 0.375 = 37.5\% $$

Hiten
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