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anonymous

  • one year ago

Quesrion 1 The vertices of a quadrilateral ABCD are A(-3, 4), B(-4, 1), C(-7, 2), and D(-7, 6). The vertices of another quadrilateral EFCD are E(-11, 4), F(-10, 1), C(-7, 2), and D(-7, 6). Which conclusion is true about the quadrilaterals? Their corresponding diagonals are equal. The measures of their corresponding angles are not identical. The lengths of their corresponding sides are unequal. Their shapes and sizes are not identical. Question 2 Use ΔABC shown below to answer the question that follows: Triangle ABC with segment AD drawn from vertex A and intersecting side CB. What statement is needed to prove that ΔABC is similar to ΔDBA? Segment BC is a hypotenuse. Angle B is congruent to itself. Segment BA is shorter than segment BC. Segment BC is intersected by segment AD.

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  1. anonymous
    • one year ago
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    For option A, find length AC and BD using distance formula. This will give you the length of diagonals for ABCD. Do the same for EFCD. Find length of diagonal EC and FD. If the length of the diagonals of both quadrilaterals are equal, then option A is correct. It turns out option A is correct. I have done the computation. Do it yourself and for an example the distance AC = sqrt((-3+7)^2 + (2-4)^2).

  2. anonymous
    • one year ago
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    oh ok i see what u did so is that for question 1 and for question 2 it be be B?

  3. anonymous
    • one year ago
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    @lokeshh are u there??

  4. anonymous
    • one year ago
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    I havn't done the second question yet.

  5. anonymous
    • one year ago
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    Please provide the figure for second question.

  6. anonymous
    • one year ago
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    here and sorry for the long wait (:

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  7. anonymous
    • one year ago
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    For second question, notice that BAC is a right angle. What angle do we need in case of triangle DBA so that both triangles are similar? Notice how BAC is formed by interchanging the letters of ABC. Similarly use the same interchanging of letters to know which letters to interchange in triangle DBA to find the angle which should be right angle. It turns out its BDA. You get the idea? The idea is to derive the angle from triangle DBA which corresponds to angle BAC in triangle ABC.

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