In a right triangle ABC right angled at C. P and Q are points on sides AC and CB respectively which divide these sides in the ratio of 2 : 1.
Prove that (i) 9AQ² = 9AC² + 4BC²
(ii) 9BP² = 9BC² + 4AC²
(iii) 9(AQ² + BP²) = 13AB²
Answers
P , Q are on AC , CB respectively
i ) Given ,
AQ dividing CB in the ratio 2 : 1
then CQ : QB = 2 : 1 ,
CQ : BC = 2 : 3
CQ = 2BC/3
In ∆ACQ , <C = 90°
AQ² = AC² + CQ²
[ By Phythogarian theorem ]
AQ² = AC² + ( 2BC/3 )²
= AC² + 4BC²/9
= ( 9AC² + 4BC² )/9
9AC² = 9AC² + 4BC² ----( 1 )
__________________________
ii ) Given ,
P divides CA in ratio 2 : 1
CP : PA = 2 : 1
=> CP : CA = 2 : 3
=> CP = 2AC/3
In ∆PCB ,
BP² = BC² + PC²
[ By Phythogarian theorem ]
BP² = BC² + ( 2AC/3 )²
= BC² + 4AC²/9
= ( 9BC² + 4AC² )/9
9BP² = 9BC² + 4AC²------( 2 )
_________________________
iii ) By adding ( 1 ) and ( 2 ) , we get
9( AQ² + BP² ) = 13AC² + 13BC²----( 3 )
But in ∆ACB ,
AB² = AC² + BC² ---( 4 )
[ By Phythogarian theorem ]
substitute ( 4 ) in ( 3 ) , we get
9( AQ² + BP² ) = 13AB²
Hence proved.
I hope this helps you.
: )
Answer:
Step-by-step explanation:
In ∆ACB , <C = 90°
P , Q are on AC , CB respectively
i ) Given ,
AQ dividing CB in the ratio 2 : 1
then CQ : QB = 2 : 1 ,
CQ : BC = 2 : 3
CQ = 2BC/3
In ∆ACQ , <C = 90°
AQ² = AC² + CQ²
[ By Phythogarian theorem ]
AQ² = AC² + ( 2BC/3 )²
= AC² + 4BC²/9
= ( 9AC² + 4BC² )/9
9AC² = 9AC² + 4BC² ----( 1 )
__________________________
ii ) Given ,
P divides CA in ratio 2 : 1
CP : PA = 2 : 1
=> CP : CA = 2 : 3
=> CP = 2AC/3
In ∆PCB ,
BP² = BC² + PC²
[ By Phythogarian theorem ]
BP² = BC² + ( 2AC/3 )²
= BC² + 4AC²/9
= ( 9BC² + 4AC² )/9
9BP² = 9BC² + 4AC²------( 2 )
_________________________
iii ) By adding ( 1 ) and ( 2 ) , we get
9( AQ² + BP² ) = 13AC² + 13BC²----( 3 )
But in ∆ACB ,
AB² = AC² + BC² ---( 4 )
[ By Phythogarian theorem ]
substitute ( 4 ) in ( 3 ) , we get
9( AQ² + BP² ) = 13AB²
Hence proved.
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