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find the area of a triangle where the
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base equals 2 and the angles at the base
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45° as usual before watching pause the
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video and try to solve the problem
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yourself let's label the triangle's
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vertices as a b and c and drop a
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perpendicular from vertex C to the base
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AB calling the height CD we can then
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find the area of triangle ABC as half
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the product of Base ab and CD the sum of
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the acute angles in the right triangle
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BCD equals 90° meaning the other acute
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angle of right triangle BCD is
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45° hence triangle BCD is isoceles and
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its two legs BD and CD are equal let's
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denote the height CD of triangle BCD as
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X thus BD which is part of the base AB
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will also equal x now note that in the
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right triangle ACD leg CD lies opposite
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the 30° angle meaning it will be half
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the hypotenuse AC therefore hypotenuse
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AC will equal 2 * X by the Pythagorean
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theorem leg a d will equal the square
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root of the difference between the
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squares of hypotenuse AC and leg CD
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substituting the values of AC and CD 2X
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and X respectively we find that a d
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equal x multiplied the square < TK of 3
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given that the base AB equals the sum of
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a d and BD we substitute the values
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giving us x * the < TK 3 + x since the
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base a is 2 we solve for x x * the < TK
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3 + x = 2 solving this we find x = 2 /
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the < TK 3 + 1 to find the area we
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substitute ab = 2 and CD = X and divide
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by 2 simplifying we get 2 / the < TK of
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3 + 1 to remove the irrationality in the
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denominator we multiply by the conjugate
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< TK 3 - 1 in both the numerator and
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denominator the denominator becomes 3 -
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1 simplifying the expression we get the
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square < TK of 3 - 1 thus the area of
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the triangle is equal to the square < TK
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of 3 - 1 square units if you understand
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the solution please like leave a comment
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and don't forget to subscribe to the
