proof of area theorem.
Answers
Area Theorem : The ratio of areas of two similar triangles is equal to the squares of the ratio of their corresponding sides.
Given, △ABD∼△PQR
∴∠A=∠P, ∠B=∠Q, and ∠C=∠R
PQ AB = QRBC = PR CA .......(1)
The ratio of area of △ABD and △PQR is given as,
ar(△PQR)
ar(△ABC) = 21 ×QR×PS2 1 ×BC×AD
ar(△PQR) ar(△ABC) = QR BC × PSAD ........(2)
In △ABD and △PQS
∠B=∠Q [Given]
∠ADB=∠PSQ [Each 90o]
∴△ABD∼△PQS [AA similarity]
⇒ PS AD = PQ AB = QS BD .......(3) [Corresponding sides are proportional]
From (1) and (3), we get
PQ AB = QR BC = PR CA = PS AD
QRBC = PS AD
From (2) and (4), we get
ar(△PQR) ar(△ABC) = QRBC × QR BC = QR2 BC2
∴ ar(△PQR) ar(△ABC) = PQ 2 AB2 = QR2 BC2 = PR2 CA2 [Using (1)]
Answer:
hello mate..
Step-by-step explanation:
Area Theorem : The ratio of areas of two similar triangles is equal to the squares of the ratio of their corresponding sides.
Given, △ABD∼△PQR
∴∠A=∠P, ∠B=∠Q, and ∠C=∠R
PQ
AB
=
QR
BC
=
PR
CA
.......(1)
The ratio of area of △ABD and △PQR is given as,
ar(△PQR)
ar(△ABC)
=
2
1
×QR×PS
2
1
×BC×AD
ar(△PQR)
ar(△ABC)
=
QR
BC
×
PS
AD
........(2)
In △ABD and △PQS
∠B=∠Q [Given]
∠ADB=∠PSQ [Each 90
o
]
∴△ABD∼△PQS [AA similarity]
⇒
PS
AD
=
PQ
AB
=
QS
BD
.......(3) [Corresponding sides are proportional]
From (1) and (3), we get
PQ
AB
=
QR
BC
=
PR
CA
=
PS
AD
QR
BC
=
PS
AD
From (2) and (4), we get
ar(△PQR)
ar(△ABC)
=
QR
BC
×
QR
BC
=
QR
2
BC
2
∴
ar(△PQR)
ar(△ABC)
=
PQ
2
AB
2
hope you get your answer