CBSE Class 9 SA 2 Solved 3 marks Questions for Maths

Linear Equation in two variables:

1. Represent 5x + y = 6 by a graph. Write the coordinates of the point where it meets :
    (a) x-axis   (b) y-axis

Solution:
For such question we will create a table for values of x and the corresponding values of y.

for the equation the table thus formed is :

x	1	0	-1
y	1	6	11

Therefore the three set of points thus obtained are (1,1),(0,6),(-1,11)
Plot these points on the graph paper to get:

for the coordinates of the point where it meets :
(a) x-axis

Put y=0  in the equation 5x + y = 6 we get x=6/5
Therefore, the line 5x + y = 6 meets the x axis at (1.2 , 0)

 (b) y-axis
Put x=0  in the equation 5x + y = 6 we get y=-6
Therefore, the line 5x + y = 6 meets the y axis at (0,6 )

2. The cost of a toy horse is same as that of 3 balls. Express this statement as a linear equation in two variables. Also draw its graph.
Solution:
Let the cost of one toy horse be x and the cost of one ball be y.
therefore according to the conditions given in the question , we have
                  x= 3y   and this is the required equation for the linear equation.

For drawing the graph of this equation we will create the table of corresponding values of x and y for the equation  x= 3y

x	0	3	-3
y	0	1	-1

 
 Now, plotting above points on the graph paper to get the graph as

3.Money donated by two friends Saif and Joseph in a charitable trust is 1200. Represent this situation in the form of a linear equation in two variables and draw its graph.

Solution:

Let the money donated by Saif and Joseph be x and y respectively.
Then it is given that they donated in total a amount of Rs. 1200.
The equation thus formed is      x + y = 1200.

Now we create a table for the corresponding values of x and y by putting different values of x and finding the corresponding value of y.

x	400	500	600
y	800	700	500

 

 Plotting these points on the graph to get the required one.

4. In 2x + y = 13 , express y in terms of x. Also find three solutions of the above equation and draw its graph.
Solution:
Expressing y in terms of x.
y = 13 - 2x
Now, for finding three solution of this equation, we will put different values of x and then solve to find the required value of y.
The required three values are given below

x	0	3	4
y	13	7	5

And the graph thus obtained is : 
 .

Quadrilateral:

Q1. Prove that the diagonals of a rectangle are equal in length,

Solution: 

Given : ABCD is a rectangle where AC and BD are the two diagonals. 

To Prove:   AC =BD

Proof:

              Taking the two triangles inside the rectangle ABCD , ΔABC and ΔBAD
                          In  ΔABC and ΔBAD
                                    BC = AD           (opposite sides of a rectangle are equal)
                              ∠DAB = ∠CBA     (both 90°)
                                    AB = AB           (common to both )
          Therefore,     ΔABC ≡  ΔBAD    (SAS congruence )

Hence,                          AC= BD           (Congruent sides of congruent triangle)            

                              

Q2. ABCD is a parallelogram with with diagonals AC = BD equal to each other. Prove that ABCD is a rectangle.

Solution:
Given: ABCD is a parallelogram in which AC =BD
To Prove : ABCD is a rectangle.

Proof : 
            In ΔABC and ΔBAD, we have
                             BC = AD    (opposite sides of a parallelogram are equal)
                             AC = BD    (given)
                             AB = BA     (common)
     Therefore,       ΔABC ≡  ΔBAD             (by SSS congruence condition)
and by CPCT     ∠ABC = ∠BAD             ------(i)
Again,   ∠ABC +  ∠BAD  =180 (angles on same side of the transversal add up to 180)
But       ∠ABC = ∠BAD  from (i)
therefore,     2∠ABC = 180
              =>     ∠ABC = 90
This also means ∠BAD = 90.
Using the same method above, we have∠ADC = ∠DCB = 90.

Q3. In the adjoining figure, ABCD is a rhombus. Find the values of x and y. Also, find all the angles of the rhombus.

Solution:
ABCD is a rhombus
So,        ∠AOD = ∠AOB = 90°     (Diagonal of a rhombus bisect each other)                                 
 Now, from    ΔAOD,
                      x + 90 + 50 = 180
     =>                             x =  40

Similarly, from  ΔAOB = we have,
                     ∠BAO = ∠DAO = 50
So, y + 90 +  50 = 180
=>                  y = 180 - 140 = 40
Now,             ∠ABC = 2y = 2 x 40 = 80
                     ∠ADC = 2x = 2 x 40 = 80
                      ∠BAD = 2 x 50 = 100
                       ∠BCD = ∠BD = 100
Thus, angles of the rhombus are 80, 100, 80 and 100.            


Q4.      ABCD is a parallelogram. If the bisectors DP and CP of the angles D and C respectively meet at P on side AB  , then show that P is the mid-point of side AB.

Solution:
                            

Given:  DP and CP are the bisectors of angle D and C.
              therefore,   ∠1 = ∠2
                        and  ∠3 = ∠4

To prove : AP =PB

Now,
          ∠DPA = ∠1               (alternate interior angles,. DC || AB and DP is the transverse  )
  
 But    ∠1 = ∠2             (given )
   =>  ∠2 = ∠DPA
This means  in ΔDAP,   side DA = AP   (sides opposite to equal angles are also equal)  ----(i)

Again,   ∠CPB =  ∠3      (alternate interior angles,. DC || AB and CP is the transverse  )        
But      ∠3 = ∠4             (given)  
=>        ∠CPB =∠4
This means  in ΔCBP, side CB = PB  (sides opposite to equal angles are also equal)  ----(ii)

Again, given that ABCD is a parallelogram it means DA = CB
From point (i) and (ii) mentioned above we get AP = PB
Hence proved that P is the mid-point of AB.                                            

Q5 In the adjoining figure, AB= AC, CD||AB and AD is the bisector of ∠CAE. Show that ABCD is a parallelogram.

Solution:

Given : CD||AB , AB =AC and ∠CAD = ∠DAE
To Prove : ABCD is a parallelogram
Proof:
Given that  CD||AB  we just have to prove that AD||BC. (as a quadrilateral having parallel opposite sides is a parallelogram)
Here,  
Since    ∠CAD = ∠DAE         (given)    -------------(i)
And      ∠CAD = ∠ACB        (alternate opposite angles for CD||AB)     -------------(ii)
Therefore from (i) and (ii),  ∠ACB =  ∠DAE          -------------(iii)

Again, in ΔABC,
                              AB=AC  (given)
Therefore,       ∠ACB = ∠ABC   (angles opposite to equal sides are equal)-------------(iv)


But from (i)  above, ∠ACB =  ∠DAE
Using (iii)  and  and (iv) we have
                      ∠ABC = ∠DAE
                          AB || DC  ( if corresponding angles are equal then the lines are parallel)