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SChand CLASS 9 Chapter 10 Rectilinear Figures Exercise 10(B)

 Exercise 10 B


(1) ABCD is a square, prove that AC2=2AB2

(2) In Fig. 10.26, AB=BC=CA=2a and segment AD side BC, Show that

(i) AD=a3, (ii) area of ΔABC=a23

Sol:1 (IMAGE TO BE ADDED)
 From, Right angle triangle , ABC
We get, AC2=AB2+BC2 (By Pythagoras Theorem)
=AB2+AB2 (ABCD is a square)
=2AB2

Sol:2  (IMAGE TO BE ADDED) 

(i) From right angle triangle ABC
 We get, 
(2a)2=a2+AD2 (by Pythagoras Theorem)
AD2 =4a2a2

(ii) Area of ABCa23

Question 3

 In Fig 10.27, prove that AB2AD2=CD2CB2,

(IMAGE TO BE ADDED) 

Sol : From , right angle triangle , ABC
We get, AC2=AB2+BC2(By Pythagoras theorem).......(i)

From right angle triangle 
ABC We get, 
AC2=AD2+DC2(By Pythagoras theorem)........(ii)

From (i) and (ii) We get 
AB2+BC2=AD2+DC2
AB2AD2=CD2CB2 Proved

Question 4
 In a ABC,ADBC, Prove that: AB2+CD2=AC2+BD2

(IMAGE TO BE ADDED) 

Sol:  From right angle triangle , 
ABD We get,
AB2=BD2+AD2(By Pythagoras theorem we get)

from ADC we get, 
AC2=AD2+DC2 .....(ii)

From (i) & (ii) We get, 
AB2BD2=AC2DC2
AB2+CD2= AC2+BD2 (Proved)


Question 5

In a quadrilateral ABCD, the diagonals AC,BD intersect at right angles. Prove that:
AB2+CD2=BC2+DA2

 (IMAGE TO BE ADDED) 

Sol: From right angle triangle , 
OAB We get, 
AB2=OA2+OB2..........(i)

OBC we get, 
BC2=OB2+OC2.........(ii)

OCD we get,
DC2=OC2+OD2............(iii)

OAD We get, 
AD2=OA2+OD2

From (i) and (ii) we get. 
AB2+CD2= OA2+OB2+OC2+OD2

From (iii) &(i) we get 

BC2+DA2= OA2+OB2+OC2+OD2
AB2+CD2= BC2+DA2

Question 6

 In ABC,B=90 and D is the mid point of BC, Prove that
(i) AC2=AD2+3CD2 (ii) BC2=4(AD2AB2)

 (IMAGE TO BE ADDED) 

From , right angle triangle ABC We get, 
AC2=AB2+BC2.........(i)

From right angle triangle
ABD We get, 
AD2=AB2 + B D^{2}.......(ii)

From (i) & (ii) We get, 
AC2=AD2BD2+BC2=AD2BD2+(2CD)2=AD2CD2+4CD2=AD2+3CD2 (proved) 

(ii)  (IMAGE TO BE ADDED) 
 AC2=AB2+BC2AD2=AB2+BD2










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