there is an isosceles triangle ABC in which line PQis drawn parallel to BC passing through point A. AB=AC.there is a square that has side equal to double the distance between the parallel line. find the area of the circle whose circumference touches all the vertices of the square if the equal sides of the triangle measure 5 cm and the third side of the triangle measures 6cm.
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Given:
AB || PR, BC || QR, AC || PQ
To Find:
Perimeter of triangle PQR : Perimeter of triangle ABC.
Solution:
AC || PQ (Given)
AC || BQ
Similarly, BC || AQ
Therefore, AQBC is a parallelogram.
Similarly, ARCB & ABPC are parallelograms.
AQ = BC___1
AR = BC___2
(opposite sides of parallelogram are equal)
Adding equations 1 and 2,
BC + BC = AQ + AR
2BC = QR
Similarly, 2AB = PR
2AC = PQ
Perimeter of triangle ABC
= AB + BC + AC
Perimeter of triangle PQR
= PQ + QR + PR
= 2AC + 2BC + 2AB
= 2(AB + BC + AC)
Ratio of perimeters of triangle PQR & triangle ABC
Ratio of perimeters = 2:1
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UnknownUsername:
i didn't ask that. see the question carefully. find the circle's area. well, the question is silly, but i have figured out the question. don't worry. try again
where did the parallelogram come from? you have to know the height of the triangle by pythagoras theorem and then find the diagonal of the square. the height of the triangle will be equal to the length between the parallel lines. if you want more hints, comment.
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