RS Aggarwal MCQs: Areas of Parallelograms and Triangles | Mathematics (Maths) Class 9 PDF Download

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Question:39
Out of the following given figures which are on the same base but not between the same parallels?
Page 2


         
 
                              
   
 
 
         
           
 
                         
 
                  
 
 
  
  
              
 
 
              
   
  
                       
    
  
                        
                                      
  
                   
 
 
 
Question:39
Out of the following given figures which are on the same base but not between the same parallels?
Solution:
In this figure, both the triangles are on the same base (QR) but not on the same parallels.
Question:40
In which of the following figures, you find polynomials on the same base and between the same parallels?
Solution:
In this figure, the following polygons lie on the same base and between the same parallel lines:
a) Parallelogram ABCD
b) ???Parallelogram ABPQ
Question:41
The median of a triangle divides it into two
a
triangles of equal areas
b
congruent triangles
c
isosceles triangles
d
right triangles
Solution:
a
  triangles of equal areas
Question:42
The area of quadrilateral ABCD in the given figure is
a
57 cm
2
b
108 cm
2
c
114 cm
2
d
195 cm
2
Solution:
c
114 cm
2
ar (quad. ABCD) =  ar (? ABC)  +  ar (? ACD)
In right angle triangle ACD, we have:
AC  = 
v
17
2
 - 8
2
 = 
v
225 = 15 cm
In right angle triangle ABC, we have:
BC = 
v
15
2
 - 9
2
 = 
v
144 = 12 cm
Now, we have the following:
ar(?ABC) = 
1
2
( )
( )
Page 3


         
 
                              
   
 
 
         
           
 
                         
 
                  
 
 
  
  
              
 
 
              
   
  
                       
    
  
                        
                                      
  
                   
 
 
 
Question:39
Out of the following given figures which are on the same base but not between the same parallels?
Solution:
In this figure, both the triangles are on the same base (QR) but not on the same parallels.
Question:40
In which of the following figures, you find polynomials on the same base and between the same parallels?
Solution:
In this figure, the following polygons lie on the same base and between the same parallel lines:
a) Parallelogram ABCD
b) ???Parallelogram ABPQ
Question:41
The median of a triangle divides it into two
a
triangles of equal areas
b
congruent triangles
c
isosceles triangles
d
right triangles
Solution:
a
  triangles of equal areas
Question:42
The area of quadrilateral ABCD in the given figure is
a
57 cm
2
b
108 cm
2
c
114 cm
2
d
195 cm
2
Solution:
c
114 cm
2
ar (quad. ABCD) =  ar (? ABC)  +  ar (? ACD)
In right angle triangle ACD, we have:
AC  = 
v
17
2
 - 8
2
 = 
v
225 = 15 cm
In right angle triangle ABC, we have:
BC = 
v
15
2
 - 9
2
 = 
v
144 = 12 cm
Now, we have the following:
ar(?ABC) = 
1
2
( )
( )
 × 12 × 9 = 54 cm
2
ar(?ADC) = 
1
2
 × 15 × 8 = 60 cm
2
ar(quad. ABCD) = ? 54 + 60 = 114 cm
2
Question:43
The area of trapezium ABCD in the given figure is
a
62 cm
2
b
93 cm
2
c
124 cm
2
d
155 cm
2
Solution:
c
124 cm
2
In the right angle triangle BEC, we have:
EC  = 
v
17
2
-15
2
=
v
289 -225 =
v
64 = 8 cm
ar(trapez. ABCD) = 
1
2
× sum of parallel sides ×distance between them =
1
2
×31 ×8 = 124
cm
2
Question:44
In the given figure, ABCD is a || gm in which AB = CD = 5 cm and BD ? DC such that BD = 6.8 cm. Then, the area of || gm ABCD = ?
a
17 cm
2
b
25 cm
2
c
34 cm
2
d
68 cm
2
Solution:
c
34 cm
2
ar(parallelogram ABCD) = base × height = 5 ?× 6.8 =  34 cm
2
Question:45
In the given figure, ABCD is a || gm in which diagonals AC and BD intersect at O. If ar(||gm ABCD) is 52 cm
2
, then the ar(?OAB) = ?
a
26 cm
2
b
18.5 cm
2
c
39 cm
2
d
13 cm
2
( )
Page 4


         
 
                              
   
 
 
         
           
 
                         
 
                  
 
 
  
  
              
 
 
              
   
  
                       
    
  
                        
                                      
  
                   
 
 
 
Question:39
Out of the following given figures which are on the same base but not between the same parallels?
Solution:
In this figure, both the triangles are on the same base (QR) but not on the same parallels.
Question:40
In which of the following figures, you find polynomials on the same base and between the same parallels?
Solution:
In this figure, the following polygons lie on the same base and between the same parallel lines:
a) Parallelogram ABCD
b) ???Parallelogram ABPQ
Question:41
The median of a triangle divides it into two
a
triangles of equal areas
b
congruent triangles
c
isosceles triangles
d
right triangles
Solution:
a
  triangles of equal areas
Question:42
The area of quadrilateral ABCD in the given figure is
a
57 cm
2
b
108 cm
2
c
114 cm
2
d
195 cm
2
Solution:
c
114 cm
2
ar (quad. ABCD) =  ar (? ABC)  +  ar (? ACD)
In right angle triangle ACD, we have:
AC  = 
v
17
2
 - 8
2
 = 
v
225 = 15 cm
In right angle triangle ABC, we have:
BC = 
v
15
2
 - 9
2
 = 
v
144 = 12 cm
Now, we have the following:
ar(?ABC) = 
1
2
( )
( )
 × 12 × 9 = 54 cm
2
ar(?ADC) = 
1
2
 × 15 × 8 = 60 cm
2
ar(quad. ABCD) = ? 54 + 60 = 114 cm
2
Question:43
The area of trapezium ABCD in the given figure is
a
62 cm
2
b
93 cm
2
c
124 cm
2
d
155 cm
2
Solution:
c
124 cm
2
In the right angle triangle BEC, we have:
EC  = 
v
17
2
-15
2
=
v
289 -225 =
v
64 = 8 cm
ar(trapez. ABCD) = 
1
2
× sum of parallel sides ×distance between them =
1
2
×31 ×8 = 124
cm
2
Question:44
In the given figure, ABCD is a || gm in which AB = CD = 5 cm and BD ? DC such that BD = 6.8 cm. Then, the area of || gm ABCD = ?
a
17 cm
2
b
25 cm
2
c
34 cm
2
d
68 cm
2
Solution:
c
34 cm
2
ar(parallelogram ABCD) = base × height = 5 ?× 6.8 =  34 cm
2
Question:45
In the given figure, ABCD is a || gm in which diagonals AC and BD intersect at O. If ar(||gm ABCD) is 52 cm
2
, then the ar(?OAB) = ?
a
26 cm
2
b
18.5 cm
2
c
39 cm
2
d
13 cm
2
( )
Solution:
d
13 cm
2
The diagonals of a parallelogram divides it into four triangles of equal areas.
? Area of ?OAB = 
1
4
? ar(||gm ABCD)
? ar(?OAB) = ?? 1
4
? ? 52 = 13 cm
2
Question:46
In the given figure, ABCD is a || gm in which DL ? AB. If AB = 10 cm and DL = 4 cm, then the ar(||gm ABCD) = ?
a 40 cm
2
b 80 cm
2
c 20 cm
2
d 196 cm
2
Solution:
a
40 cm
2
ar(||gm ABCD) = base × height =  10 ?× 4 =  40 cm
2
Question:47
The area of ||gm ABCD is
a
AB × BM
b
BC × BN
c
DC × DL
d
AD × DL
Solution:
Area of a parallelogram is base into height.
Height = DL = NB
Base = AB = CD
So, area of parallelogram ABCD = DC ×
 DL
Hence, the correct answer is option c
. 
Question:48
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
a
1 : 2
b
1 : 1
c
2 : 1
d
3 : 1
Page 5


         
 
                              
   
 
 
         
           
 
                         
 
                  
 
 
  
  
              
 
 
              
   
  
                       
    
  
                        
                                      
  
                   
 
 
 
Question:39
Out of the following given figures which are on the same base but not between the same parallels?
Solution:
In this figure, both the triangles are on the same base (QR) but not on the same parallels.
Question:40
In which of the following figures, you find polynomials on the same base and between the same parallels?
Solution:
In this figure, the following polygons lie on the same base and between the same parallel lines:
a) Parallelogram ABCD
b) ???Parallelogram ABPQ
Question:41
The median of a triangle divides it into two
a
triangles of equal areas
b
congruent triangles
c
isosceles triangles
d
right triangles
Solution:
a
  triangles of equal areas
Question:42
The area of quadrilateral ABCD in the given figure is
a
57 cm
2
b
108 cm
2
c
114 cm
2
d
195 cm
2
Solution:
c
114 cm
2
ar (quad. ABCD) =  ar (? ABC)  +  ar (? ACD)
In right angle triangle ACD, we have:
AC  = 
v
17
2
 - 8
2
 = 
v
225 = 15 cm
In right angle triangle ABC, we have:
BC = 
v
15
2
 - 9
2
 = 
v
144 = 12 cm
Now, we have the following:
ar(?ABC) = 
1
2
( )
( )
 × 12 × 9 = 54 cm
2
ar(?ADC) = 
1
2
 × 15 × 8 = 60 cm
2
ar(quad. ABCD) = ? 54 + 60 = 114 cm
2
Question:43
The area of trapezium ABCD in the given figure is
a
62 cm
2
b
93 cm
2
c
124 cm
2
d
155 cm
2
Solution:
c
124 cm
2
In the right angle triangle BEC, we have:
EC  = 
v
17
2
-15
2
=
v
289 -225 =
v
64 = 8 cm
ar(trapez. ABCD) = 
1
2
× sum of parallel sides ×distance between them =
1
2
×31 ×8 = 124
cm
2
Question:44
In the given figure, ABCD is a || gm in which AB = CD = 5 cm and BD ? DC such that BD = 6.8 cm. Then, the area of || gm ABCD = ?
a
17 cm
2
b
25 cm
2
c
34 cm
2
d
68 cm
2
Solution:
c
34 cm
2
ar(parallelogram ABCD) = base × height = 5 ?× 6.8 =  34 cm
2
Question:45
In the given figure, ABCD is a || gm in which diagonals AC and BD intersect at O. If ar(||gm ABCD) is 52 cm
2
, then the ar(?OAB) = ?
a
26 cm
2
b
18.5 cm
2
c
39 cm
2
d
13 cm
2
( )
Solution:
d
13 cm
2
The diagonals of a parallelogram divides it into four triangles of equal areas.
? Area of ?OAB = 
1
4
? ar(||gm ABCD)
? ar(?OAB) = ?? 1
4
? ? 52 = 13 cm
2
Question:46
In the given figure, ABCD is a || gm in which DL ? AB. If AB = 10 cm and DL = 4 cm, then the ar(||gm ABCD) = ?
a 40 cm
2
b 80 cm
2
c 20 cm
2
d 196 cm
2
Solution:
a
40 cm
2
ar(||gm ABCD) = base × height =  10 ?× 4 =  40 cm
2
Question:47
The area of ||gm ABCD is
a
AB × BM
b
BC × BN
c
DC × DL
d
AD × DL
Solution:
Area of a parallelogram is base into height.
Height = DL = NB
Base = AB = CD
So, area of parallelogram ABCD = DC ×
 DL
Hence, the correct answer is option c
. 
Question:48
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
a
1 : 2
b
1 : 1
c
2 : 1
d
3 : 1
Solution:
Parallelograms on the same base and between the same parallels are equal in area.
So, the ratio of their areas will be 1 : 1.  
Hence, the correct answer is option b
. 
Question:49
In the given figure, ABCD and ABPQ are two parallelograms and M is a point on AQ and BMP is a triangle.
Then, ar(?BMP) = 
1
2
ar(||gm ABCD) is
a
true
b
false
Solution:
We know parallelogram on the same base and between the same parallels are equal in area. 
Here, AB is the common base and AB || PD
Hence, arABCD
= arABPQ
                             .....1
Also, when a triangle and a parallelogram are on the same base and between the same parallels then the
area of triangle is half the area of the parallelogram. 
Here, for the ?BMP and parallelogrm ABPQ, BP is the common base and they are between the common parallels BP and AQ
So, ar? BMP
= 
1
2
 ar| | gmABPQ
                      .....2
From 1
and 2
we have
ar? BMP
= 
1
2
 ar| | gmABCD
 
Thus, the given statement is true.
Hence, the correct answer is option a
.    
 
Question:50
The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to
a
1
2
(ar? ABC)
b
1
3
(ar? ABC)
c
1
4
(ar? ABC)
Solution: