In quadrilateral abcd below,
1 answer:
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Explanation:
Consider the attached diagram.
Circles A and B have mutual tangents CF and DE that intersect at point F.
We know that the two tangents to a circle from an external point are congruent, so ...
FE≅ FC
FD ≅ FC
By the transitive property of congruence, both segments congruent to FC are congruent to each other:
FD ≅ FE
Therefore, CF is a bisector of DE.
You can see the trapezoid in the attached picture.
The data given in the problem are:
KF = 10
LK // MF
A(KLMF) = A(FMN)
We know that KLMF is a parallelogram because:
LM // KF (bases of the trapezoid)
LK // MF (hypothesis).
The area of a parallelogram is given by the formula:
A(KLMF) = b × h
= KF <span>× h</span>
= 10 × h
The area of a triangle is given by the formula:
<span>A(FMN) = (b × h) / 2
= (FN </span>× h) / 2
The problem states that the two areas are congruent, therefore:
A(KLMF) = <span>A(FMN)
</span>10 × h = FN <span>× h / 2
10 = FN </span><span>/ 2
FN = 20
Therefore we can calculate:
KN = KF + FN = 10 + 20 = 30
Hence, KN is 30 units long.</span>
The data given in the problem are:
KF = 10
LK // MF
A(KLMF) = A(FMN)
We know that KLMF is a parallelogram because:
LM // KF (bases of the trapezoid)
LK // MF (hypothesis).
The area of a parallelogram is given by the formula:
A(KLMF) = b × h
= KF <span>× h</span>
= 10 × h
The area of a triangle is given by the formula:
<span>A(FMN) = (b × h) / 2
= (FN </span>× h) / 2
The problem states that the two areas are congruent, therefore:
A(KLMF) = <span>A(FMN)
</span>10 × h = FN <span>× h / 2
10 = FN </span><span>/ 2
FN = 20
Therefore we can calculate:
KN = KF + FN = 10 + 20 = 30
Hence, KN is 30 units long.</span>