In a triangle PQR; PQ = 9 cm, QR= 12 cm and 30 RP = 15 cm. A perpendicular dropped from Q, meets the side RP at S. A circle of radius QS (with centre Q) is drawn. If the circle cuts PQ and QR at T and U respectively then RU is equal to
Answers
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
Learn More:
https://brainly.in/question/23950586
The length of perpendicular side of the right angle triangle is 7 cm ...
https://brainly.in/question/15400131
1)The perpendicular from the square corner ofa right triangle cuts ...
https://brainly.in/question/13095188
Answer:
Step-by-step explanation:
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