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

 Exercise 10 B

(1) $A B C D$ is a square, prove that $A C^{2}=2 A B^{2}$

(2) In Fig. 10.26, $A B=B C=C A=2 a$ and segment $A D \perp$ side $B C$, Show that

(i) $A D=a \sqrt{3}$, (ii) area of $\Delta A B C=a^{2} \sqrt{3}$

Sol:1 (IMAGE TO BE ADDED)

 From, Right angle triangle , $\triangle  ABC $

We get, $A C^{2}=A B^{2}+B C^{2}$ (By Pythagoras Theorem)

$=A B^{2}+A B^{2}$ (ABCD is a square)

$=2 A B^{2}$

Sol:2  (IMAGE TO BE ADDED) 

(i) From right angle triangle $\triangle  ABC $

 We get, 

$(2 a)^{2}=a^{2}+A D^{2}$ (by Pythagoras Theorem)

$A D^{2}$ =$4 a^{2}$ - $a^{2}$

(ii) Area of $\triangle A B C$ = $a^{2} \sqrt{3}$

Question 3

 In Fig 10.27, prove that $A B^{2}-A D^{2}=C D^{2}-C B^{2}$,

(IMAGE TO BE ADDED) 

Sol : From , right angle triangle , $\triangle A B C$

We get, $A C^{2}=A B^{2}+B C^{2}$(By Pythagoras theorem).......(i)

From right angle triangle 

$\triangle A B C$ We get, 

$A C^{2}=A D^{2}+D C^{2}$(By Pythagoras theorem)........(ii)

From (i) and (ii) We get 

$A B^{2}+B C^{2}=A D^{2}+D C^{2}$

$A B^{2}-A D^{2}=C D^{2}-C B^{2}$ Proved

Question 4

 In a $\triangle A B C, A D \perp B C$, Prove that: $A B^{2}+C D^{2}=A C^{2}+B D^{2}$

(IMAGE TO BE ADDED) 

Sol:  From right angle triangle , 

$\triangle A B D$ We get,

$A B^{2}$=$B D^{2}+A D^{2}$(By Pythagoras theorem we get)

from $\triangle A D C$ we get, 

$A C^{2}=A D^{2}+D C^{2}$ .....(ii)

From (i) & (ii) We get, 

$A B^{2}-B D^{2}=A C^{2}-D C^{2}$

$\therefore A B^{2}+C D^{2}=$ $A C^{2}+B D^{2}$ (Proved)

Question 5

In a quadrilateral $A B C D$, the diagonals $A C, B D$ intersect at right angles. Prove that:

$A B^{2}+C D^{2}=B C^{2}+D A^{2}$

 (IMAGE TO BE ADDED) 

Sol: From right angle triangle , 

$\triangle O A B$ We get, 

$A B^{2}=O A^{2}+O B^{2}$..........(i)

$\triangle $ OBC we get, 

$B C^{2}=O B^{2}+O C^{2}$.........(ii)

$\triangle O C D$ we get,

$D C^{2}=O C^{2}+OD^{2}$............(iii)

$\triangle O A D$ We get, 

$A D^{2}$=$O A^{2}+O D^{2}$

From (i) and (ii) we get. 

$A B^{2}+C D^{2}=$ $O A^{2}+O B^{2}+O C^{2}+O D^{2}$

From (iii) &(i) we get 

$B C^{2}+D A^{2}=$ $O A^{2}+O B^{2}+O C^{2}+O D^{2}$

$A B^{2}+C D^{2}=$ $B C^{2}+D A^{2}$

Question 6

 In $\triangle A B C, \angle B=90^{\circ}$ and $D$ is the mid point of $B C$, Prove that

(i) $A C^{2}=A D^{2}+3 C D^{2}$ (ii) $B C^{2}=4\left(A D^{2}-A B^{2}\right)$

 (IMAGE TO BE ADDED) 

From , right angle triangle $\triangle A B C$ We get, 

$A C^{2}=A B^{2}+B C^{2}$.........(i)

From right angle triangle

$\triangle A B D$ We get, 

$A D^{2}=A B^{2}$ + B D^{2}.......(ii)

From (i) & (ii) We get, 

$\begin{aligned} A C^{2} &=A D^{2}-B D^{2}+B C^{2} \\ &=A D^{2}-B D^{2}+(2 C D)^{2} \\ &=A D^{2}-C D^{2}+4 C D^{2} \\ &=A D^{2}+3 C D^{2} \text { (proved) } \end{aligned}$

(ii)  (IMAGE TO BE ADDED) 

 $\begin{aligned} A C^{2} &=A B^{2}+B C^{2} \\ A D^{2} &=A B^{2}+B D^{2} \end{aligned}$