In the figure △ A B C is a right triangle with legs A B = 6 and A C = 8 . A square is drawn as shown, with a side along B C and corners on A B and A C . Find the length of the side of the square.
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Let ∠ A C B = α . Then tan α = 4 3 .
Let the length of each side of the square be a . Then
( 3 4 + 1 + 4 3 ) a = 6 2 + 8 2 = 1 0
⟹ a = 3 7 1 2 0 .
Nice Mr. Alak
I have a derived formula about problems like this. Let x be the side length of the square. Then,
x = b + h b h where: b is the side of the triangle where the side of the square coincides and h is the perpendicular distance from b to the opposite vertex.
Note that △ A B C is the popular pythagorean triple, 3 − 4 − 5 right triangle. So the length of B C is 1 0 . Now we need to find h ,
sin ∠ A B C = B C A C = A B h
1 0 8 = 6 h
h = 5 2 4
Now, substitute.
x = b + h b h = 1 0 + 5 2 4 1 0 × 5 2 4 = 3 7 1 2 0 a n s w e r
B C = 1 0 (pythagorean theorem) l e t H G = x = H I = G J △ A B C ∼ △ H B I ⇒ H B A B = H I A C ⇒ H B = 4 3 x △ A B C ∼ △ G C J ⇒ G C A B = G J A C ⇒ G C = 3 4 x B C = B H + H G + G C ⇒ 1 0 = 1 2 3 7 x ⇒ x = 3 7 1 2 0
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We note the right △ A B C is the popular right triangle with integer side lengths of ratio 3 : 4 : 5 or 0 . 6 : 0 . 8 : 1 . We also note that △ A I J and △ B I H are similar to △ A B C .
Let the side length of the square be a , then A I = 0 . 6 a and I B = 0 . 8 a . Since:
A I + I B 0 . 6 a + 0 . 8 a 0 . 8 0 . 4 8 a + a ⟹ a = A B = 6 = 6 = 1 . 4 8 4 . 8 0 = 3 7 1 2 0