in a triangle pqr PQ equal to 9 centimetre QR equal to 12 cm and RP equal to 15 centimetres a perpendicular dropped from Q meets the side RP at is a circle of radius QS(with center Q) is drawn. if the circle cuts pq and qr at t and u respectively then PT :RU is equal to
Answers
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Given : Triangle PQR PQ = 9 cm , QR = 12 cm , RP = 15 cm a perpendicular dropped from Q meets the side RP at S
a circle of radius QS(with center Q) is drawn. circle cuts PQ and QR at T and u respectively
To Find : PT :RU
Solution:
PQ = 9 cm
QR = 12 cm
RP = 15 cm
9² + 12 ² = 81 + 144 = 225 = 15²
=> PQ² + QR² = PR²
Using converse of Pythagoras theorem
PQR is a right angle triangle at Q
Area of Of Δ PQR = (1/2) * PQ * QR
Area of Of Δ PQR = (1/2) * PR * QS
=> (1/2) * PQ * QR = (1/2) * PR * QS
=> 9 * 12 = 15 * QS
=> QS = 108/15
=> QS = 7.2 cm
QS = QT = QU = 7.2 cm Radius
PT = PQ - QT = 9 - 7.2 = 1.8 cm
RU = QR - QU = 12 - 7.2 = 4.8 cm
PT : RU = 1.8 : 4.8
= 3 : 8
PT :RU = 3 : 8
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