In a right triangle ABC right angled at B. P and Q are the points on the
sides CA and CB respectively, which divide these sides in the ratio 2:1.
Prove that
i) 9 AQ'2 = 9 AC'2 + 4BC'2
9 BP'2= 9BC'2 +4AC'2
ii) 9 (AQ? + BP'2 ) = 13AB'2
{and ('2) means square.
For eg:AQ square}
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7th
Maths
The Triangle and Its Properties
Right Angled Triangles and Pythagoras Property
In a right angled triangle ...
MATHS
In a right angled triangle ABC, right angled at C, P and Q are the points on the side CA and C respectively which divide these sides in the ratio 2 : 1. Prove that
(i) 9AQ
2
=9AC
2
+4
2
(ii) 9B
2
=9B
2
+4AC
2
(iii) 9(AQ
2
+BP
2
)=13AB
2
EASY
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ANSWERANSWER
In ACQ
(1AQ
2
=AC
2
+QC
2
QB
CQ
=
2
1
&
CA
CP
=
1
2
QB
CQ
=
1
2
⇒
BC−CQ
CQ
=2
⇒3C=2BC
⇒C=
3
2
BC
By (1)
AQ
2
=AC
2
+(
3
2
BC)
2
2)9AQ
2
=9AC
2
+4BC
2
−−−−−(A)
In BCP
CA−CP
CP
=2
PB
2
=BC
2
+PC
2
−−C=
3
2
CA
PB
2
=BC
2
+(
3
2
CA)
2
⇒9PB
2
=9BC
2
+4AC
2
−−−−−(B)
(3) Adding (A&(B)
9AQ
2
=9BC
2
=13(BC
2
+AC
2
)
⇒9(AQ
2
+BP
2
)=13AB
2