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the legs of a right triangle are equal
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to 3 and five find the bis sector drawn
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from the right angle as usual before
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viewing try to solve this problem
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yourself according to the condition the
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bis sector divides the right triangle
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into two parts meaning two smaller
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triangles this means the area of the
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larger right triangle is equal to the
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sum of the areas of its two parts that
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is the sum of the areas of the two
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smaller triangles the area of the
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original right triangle is found as half
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the product of the legs which equals 12
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* 3 and multiplied 5 now to find the
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areas of the smaller triangles let's
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denote the bis sector we are looking for
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as X the area of the first small
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triangle can be found as half the
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product of its two sides 3 * X and the
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sign of the angle between them that is
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the sign of 45° similarly the area of
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the second small triangle is half the
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product of its two sides 5 * X and the
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sign of 45° on the left side we get 15
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halves and on the right side we combine
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terms it's convenient to factor out X
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45° in parentheses we get 3 halves + 5
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which equals 8 halves or
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four thus on the right side we get 4
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multiplied X and the sign of
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45° the sign of 45° is the square < TK
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we cancel out the twos and are left with
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15 halves on the left side and 2 * the <
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TK of 2 * X on the right solving for x
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we get X is equal to 15 / 4un of 2 the
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final answer is that the bis sector is
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equal to 15 roots of 2 / 8 the problem
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is solved if you understood the solution
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give it a like leave a comment and
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