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

 Page 1


            
                                         
                                        
        
                 
      
        
 
   
     
      
     
 
         
             
 
   
              
     
     
 
   
    
             
       
                                                   
                                                  
        
Q u e s t i o n : 4 5
In a ?ABC it is given that base = 12 cm and height = 5 cm. Its area is
a 60 cm
2
b 30 cm
2
c
15v 3 cm
2
d 45 cm
2
S o l u t i o n :
b 30 cm
2
Area of triangle = 
1
2
×Base ×HeightArea of ? ABC =
1
2
×12 ×5 = 30 cm
2
Q u e s t i o n : 4 6
The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is
a 96 cm
2
b 120 cm
2
c 144 cm
2
d 160 cm
2
S o l u t i o n :
a 96 cm
2
Let: a = 20 cm, b = 16 cm and c = 12 cms = 
a+b+c
2
=
20+16+12
2
= 24 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 24(24 -20)(24 -16)(24 -12) = v 24 ×4 ×8 ×12 = v
Q u e s t i o n : 4 7
Each side of an equilateral triangle measure 8 cm. The area of the triangle is
a
8v 3 cm
2
b
16v 3 cm
2
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Q u e s t i o n : 4 5
In a ?ABC it is given that base = 12 cm and height = 5 cm. Its area is
a 60 cm
2
b 30 cm
2
c
15v 3 cm
2
d 45 cm
2
S o l u t i o n :
b 30 cm
2
Area of triangle = 
1
2
×Base ×HeightArea of ? ABC =
1
2
×12 ×5 = 30 cm
2
Q u e s t i o n : 4 6
The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is
a 96 cm
2
b 120 cm
2
c 144 cm
2
d 160 cm
2
S o l u t i o n :
a 96 cm
2
Let: a = 20 cm, b = 16 cm and c = 12 cms = 
a+b+c
2
=
20+16+12
2
= 24 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 24(24 -20)(24 -16)(24 -12) = v 24 ×4 ×8 ×12 = v
Q u e s t i o n : 4 7
Each side of an equilateral triangle measure 8 cm. The area of the triangle is
a
8v 3 cm
2
b
16v 3 cm
2
c
32v 3 cm
2
d 48 cm
2
S o l u t i o n :
b
16v 3 cm
2
Area of equilateral triangle = 
v
3
4
× Side)
2
=
v
3
4
× 8)
2
=
v
3
4
×64 = 16v 3 cm
2
Q u e s t i o n : 4 8
The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is
a
16v 5 cm
2
b
8v 5 cm
2
c
16v 3 cm
2
d
8v 3 cm
2
S o l u t i o n :
b
8v 5 cm
2
Area of isosceles triangle = 
b
4
v
4a
2
-b
2
Here, a = 6 cm and b = 8 cmThus, we have:
8
4
×
v
4 6)
2
-8
2
=
8
4
× v 144 -64 =
8
4
× v 80 =
8
4
×4v 5 = 8v 5 cm
2
Q u e s t i o n : 4 9
The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of the triangle is
a 8 cm
b
v 30 cm
c 4 cm
d
v 11 cm
S o l u t i o n :
c 4 cm
Height of isosceles triangle = 
1
2
v
4a
2
-b
2
=
1
2
v
4(5)
2
-6
2
       (a = 5 cm and b = 6 cm) =
1
2
× v 100 -36 =
1
2
× v 64 =
1
2
×8 = 4 cm
Q u e s t i o n : 5 0
Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is
a
5v 10 cm
2
b 50 cm
2
c
10v 3 cm
2
d 75 cm
2
S o l u t i o n :
b 50 cm
2
Here, the base and height of the triangle are 10 cm and 10 cm, respectively.
Thus, we have:
Area of triangle = 
1
2
×Base ×Height =
1
2
×10 ×10 = 50 cm
2
Q u e s t i o n : 5 1
Each side of an equilateral triangle is 10 cm long. The height of the triangle is
a
10v 3 cm
b
5v 3 cm
c
10v 2 cm
d 5 cm
S o l u t i o n :
b
5v 3 cm
Height of equilateral triangle =
v
3
2
×Side =
v
3
2
×10 = 5v 3 cm
Q u e s t i o n : 5 2
The height of an equilateral triangle is 6 cm. Its area is
a
12v 3 cm
2
b
6v 3 cm
2
c
12v 2 cm
2
d 18 cm
2
S o l u t i o n :
a
12v 3 cm
2
Height of equilateral triangle = 
v
3
2
×Side ? 6 =
v
3
2
×Side ? Side =
12
v
3
×
v
3
v
3
=
12
3
× v 3 = 4v 3  cmNow, Area of equilateral triangle = 
v
3
4
× Side)
2
=
v
3
4
× 4v 3
2
=
v
3
4
×48 = 12v 3 cm
2
( (
(
(
( )
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Q u e s t i o n : 4 5
In a ?ABC it is given that base = 12 cm and height = 5 cm. Its area is
a 60 cm
2
b 30 cm
2
c
15v 3 cm
2
d 45 cm
2
S o l u t i o n :
b 30 cm
2
Area of triangle = 
1
2
×Base ×HeightArea of ? ABC =
1
2
×12 ×5 = 30 cm
2
Q u e s t i o n : 4 6
The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is
a 96 cm
2
b 120 cm
2
c 144 cm
2
d 160 cm
2
S o l u t i o n :
a 96 cm
2
Let: a = 20 cm, b = 16 cm and c = 12 cms = 
a+b+c
2
=
20+16+12
2
= 24 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 24(24 -20)(24 -16)(24 -12) = v 24 ×4 ×8 ×12 = v
Q u e s t i o n : 4 7
Each side of an equilateral triangle measure 8 cm. The area of the triangle is
a
8v 3 cm
2
b
16v 3 cm
2
c
32v 3 cm
2
d 48 cm
2
S o l u t i o n :
b
16v 3 cm
2
Area of equilateral triangle = 
v
3
4
× Side)
2
=
v
3
4
× 8)
2
=
v
3
4
×64 = 16v 3 cm
2
Q u e s t i o n : 4 8
The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is
a
16v 5 cm
2
b
8v 5 cm
2
c
16v 3 cm
2
d
8v 3 cm
2
S o l u t i o n :
b
8v 5 cm
2
Area of isosceles triangle = 
b
4
v
4a
2
-b
2
Here, a = 6 cm and b = 8 cmThus, we have:
8
4
×
v
4 6)
2
-8
2
=
8
4
× v 144 -64 =
8
4
× v 80 =
8
4
×4v 5 = 8v 5 cm
2
Q u e s t i o n : 4 9
The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of the triangle is
a 8 cm
b
v 30 cm
c 4 cm
d
v 11 cm
S o l u t i o n :
c 4 cm
Height of isosceles triangle = 
1
2
v
4a
2
-b
2
=
1
2
v
4(5)
2
-6
2
       (a = 5 cm and b = 6 cm) =
1
2
× v 100 -36 =
1
2
× v 64 =
1
2
×8 = 4 cm
Q u e s t i o n : 5 0
Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is
a
5v 10 cm
2
b 50 cm
2
c
10v 3 cm
2
d 75 cm
2
S o l u t i o n :
b 50 cm
2
Here, the base and height of the triangle are 10 cm and 10 cm, respectively.
Thus, we have:
Area of triangle = 
1
2
×Base ×Height =
1
2
×10 ×10 = 50 cm
2
Q u e s t i o n : 5 1
Each side of an equilateral triangle is 10 cm long. The height of the triangle is
a
10v 3 cm
b
5v 3 cm
c
10v 2 cm
d 5 cm
S o l u t i o n :
b
5v 3 cm
Height of equilateral triangle =
v
3
2
×Side =
v
3
2
×10 = 5v 3 cm
Q u e s t i o n : 5 2
The height of an equilateral triangle is 6 cm. Its area is
a
12v 3 cm
2
b
6v 3 cm
2
c
12v 2 cm
2
d 18 cm
2
S o l u t i o n :
a
12v 3 cm
2
Height of equilateral triangle = 
v
3
2
×Side ? 6 =
v
3
2
×Side ? Side =
12
v
3
×
v
3
v
3
=
12
3
× v 3 = 4v 3  cmNow, Area of equilateral triangle = 
v
3
4
× Side)
2
=
v
3
4
× 4v 3
2
=
v
3
4
×48 = 12v 3 cm
2
( (
(
(
( )
Q u e s t i o n : 5 3
The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is
a
480 m
2
b
320 m
2
c
384 m
2
d
360 m
2
S o l u t i o n :
c
384 m
2
Let: a = 40 m, b = 24 m and c = 32 ms = 
a+b+c
2
=
40+24+32
2
= 48 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 48(48 -40)(48 -24)(48 -32) = v 48 ×8 ×24 ×16 = v 24 ×
Q u e s t i o n : 5 4
The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 cm. The area of the triangle is
a 375 cm
2
b 750 cm
2
c 250 cm
2
d 500 cm
2
S o l u t i o n :
b 750 cm
2
Let the sides of the triangle be 5x cm, 12x cm and 13x cm.
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, the sides of the triangle are 5 ×
5 cm, 12 ×
5 cm and 13 ×
5 cm, i.e., 25 cm, 60 cm and 65 cm.
Now,
Let: a = 25 cm, b = 60 cm and c = 65 cms = 
150
2
= 75 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 75(75 -25)(75 -60)(75 -65) = v 75 ×50 ×15 ×10 = v 15 ×5 ×5
Q u e s t i o n : 5 5
The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of the altitude of the triangle corresponding to the smallest side is
a 24 cm
b 18 cm
c 30 cm
d 12 cm
S o l u t i o n :
a 24 cm
Let: a = 30 cm, b = 24 cm and c = 18 cms = 
a+b+c
2
=
30+24+18
2
= 36 cmOn applying Heron's formula, we get: Area of triangle = v s(s -a)(s -b)(s -c) = v 36(36 -30)(36 -24)(36 -18) = v 36 ×6 ×12
The smallest side is 18 cm.
Hence, the altitude of the triangle corresponding to 18 cm is given by:
Area of triangle = 216 cm
2
?
1
2
×Base ×Height = 216 ? Height = 
216×2
18
= 24 cm
Q u e s t i o n : 5 6
The base of an isosceles triangle is 16 cm and its area is 48 cm
2
. The perimeter of the triangle is
a 41 cm
b 36 cm
c 48 cm
d 324 cm
S o l u t i o n :
b 36 cm
Let ?
PQR be an isosceles triangle and PX
? QR.
Now,
Area of triangle = 48 cm
2
 ?
1
2
×QR ×PX = 48 ? h =
96
16
= 6 cmAlso, QX = 
1
2
×24 = 12 cm and PX = 12 cm
PQ =
v
QX
2
+PX
2
a =
v
8
2
+6
2
= v 64 +36 = v 100 = 10 cm
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Q u e s t i o n : 4 5
In a ?ABC it is given that base = 12 cm and height = 5 cm. Its area is
a 60 cm
2
b 30 cm
2
c
15v 3 cm
2
d 45 cm
2
S o l u t i o n :
b 30 cm
2
Area of triangle = 
1
2
×Base ×HeightArea of ? ABC =
1
2
×12 ×5 = 30 cm
2
Q u e s t i o n : 4 6
The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is
a 96 cm
2
b 120 cm
2
c 144 cm
2
d 160 cm
2
S o l u t i o n :
a 96 cm
2
Let: a = 20 cm, b = 16 cm and c = 12 cms = 
a+b+c
2
=
20+16+12
2
= 24 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 24(24 -20)(24 -16)(24 -12) = v 24 ×4 ×8 ×12 = v
Q u e s t i o n : 4 7
Each side of an equilateral triangle measure 8 cm. The area of the triangle is
a
8v 3 cm
2
b
16v 3 cm
2
c
32v 3 cm
2
d 48 cm
2
S o l u t i o n :
b
16v 3 cm
2
Area of equilateral triangle = 
v
3
4
× Side)
2
=
v
3
4
× 8)
2
=
v
3
4
×64 = 16v 3 cm
2
Q u e s t i o n : 4 8
The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is
a
16v 5 cm
2
b
8v 5 cm
2
c
16v 3 cm
2
d
8v 3 cm
2
S o l u t i o n :
b
8v 5 cm
2
Area of isosceles triangle = 
b
4
v
4a
2
-b
2
Here, a = 6 cm and b = 8 cmThus, we have:
8
4
×
v
4 6)
2
-8
2
=
8
4
× v 144 -64 =
8
4
× v 80 =
8
4
×4v 5 = 8v 5 cm
2
Q u e s t i o n : 4 9
The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of the triangle is
a 8 cm
b
v 30 cm
c 4 cm
d
v 11 cm
S o l u t i o n :
c 4 cm
Height of isosceles triangle = 
1
2
v
4a
2
-b
2
=
1
2
v
4(5)
2
-6
2
       (a = 5 cm and b = 6 cm) =
1
2
× v 100 -36 =
1
2
× v 64 =
1
2
×8 = 4 cm
Q u e s t i o n : 5 0
Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is
a
5v 10 cm
2
b 50 cm
2
c
10v 3 cm
2
d 75 cm
2
S o l u t i o n :
b 50 cm
2
Here, the base and height of the triangle are 10 cm and 10 cm, respectively.
Thus, we have:
Area of triangle = 
1
2
×Base ×Height =
1
2
×10 ×10 = 50 cm
2
Q u e s t i o n : 5 1
Each side of an equilateral triangle is 10 cm long. The height of the triangle is
a
10v 3 cm
b
5v 3 cm
c
10v 2 cm
d 5 cm
S o l u t i o n :
b
5v 3 cm
Height of equilateral triangle =
v
3
2
×Side =
v
3
2
×10 = 5v 3 cm
Q u e s t i o n : 5 2
The height of an equilateral triangle is 6 cm. Its area is
a
12v 3 cm
2
b
6v 3 cm
2
c
12v 2 cm
2
d 18 cm
2
S o l u t i o n :
a
12v 3 cm
2
Height of equilateral triangle = 
v
3
2
×Side ? 6 =
v
3
2
×Side ? Side =
12
v
3
×
v
3
v
3
=
12
3
× v 3 = 4v 3  cmNow, Area of equilateral triangle = 
v
3
4
× Side)
2
=
v
3
4
× 4v 3
2
=
v
3
4
×48 = 12v 3 cm
2
( (
(
(
( )
Q u e s t i o n : 5 3
The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is
a
480 m
2
b
320 m
2
c
384 m
2
d
360 m
2
S o l u t i o n :
c
384 m
2
Let: a = 40 m, b = 24 m and c = 32 ms = 
a+b+c
2
=
40+24+32
2
= 48 mBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 48(48 -40)(48 -24)(48 -32) = v 48 ×8 ×24 ×16 = v 24 ×
Q u e s t i o n : 5 4
The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 cm. The area of the triangle is
a 375 cm
2
b 750 cm
2
c 250 cm
2
d 500 cm
2
S o l u t i o n :
b 750 cm
2
Let the sides of the triangle be 5x cm, 12x cm and 13x cm.
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, the sides of the triangle are 5 ×
5 cm, 12 ×
5 cm and 13 ×
5 cm, i.e., 25 cm, 60 cm and 65 cm.
Now,
Let: a = 25 cm, b = 60 cm and c = 65 cms = 
150
2
= 75 cmBy Heron's formula, we have: Area of triangle = v s(s -a)(s -b)(s -c) = v 75(75 -25)(75 -60)(75 -65) = v 75 ×50 ×15 ×10 = v 15 ×5 ×5
Q u e s t i o n : 5 5
The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of the altitude of the triangle corresponding to the smallest side is
a 24 cm
b 18 cm
c 30 cm
d 12 cm
S o l u t i o n :
a 24 cm
Let: a = 30 cm, b = 24 cm and c = 18 cms = 
a+b+c
2
=
30+24+18
2
= 36 cmOn applying Heron's formula, we get: Area of triangle = v s(s -a)(s -b)(s -c) = v 36(36 -30)(36 -24)(36 -18) = v 36 ×6 ×12
The smallest side is 18 cm.
Hence, the altitude of the triangle corresponding to 18 cm is given by:
Area of triangle = 216 cm
2
?
1
2
×Base ×Height = 216 ? Height = 
216×2
18
= 24 cm
Q u e s t i o n : 5 6
The base of an isosceles triangle is 16 cm and its area is 48 cm
2
. The perimeter of the triangle is
a 41 cm
b 36 cm
c 48 cm
d 324 cm
S o l u t i o n :
b 36 cm
Let ?
PQR be an isosceles triangle and PX
? QR.
Now,
Area of triangle = 48 cm
2
 ?
1
2
×QR ×PX = 48 ? h =
96
16
= 6 cmAlso, QX = 
1
2
×24 = 12 cm and PX = 12 cm
PQ =
v
QX
2
+PX
2
a =
v
8
2
+6
2
= v 64 +36 = v 100 = 10 cm
? Perimeter = 10 +10 +16
cm = 36 cm
Q u e s t i o n : 5 7
The area of an equilateral triangle is 36v 3 cm
2
. Its perimeter is
a 36 cm
b
12v 3 cm
c 24 cm
d 30 cm
S o l u t i o n :
a 36 cm
Area of equilateral triangle = 
v
3
4
× Side)
2
?
v
3
4
× Side)
2
 = 36v 3
? Side)
2
= 144 ? Side = 12 cm
Now,
Perimeter = 3 × Side = 3 × 12 = 36 cm
Q u e s t i o n : 5 8
Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is
a 156 cm
2
b 78 cm
2
c 60 cm
2
d 120 cm
2
S o l u t i o n :
c 60 cm
2
Area of isosceles triangle = 
b
4
v
4a
2
-b
2
Here, a = 13 cm and b = 24 cmThus, we have:
24
4
×
v
4 13)
2
-24
2
= 6 × v 676 -576 = 6 × v 100 = 6 ×10 = 60 cm
2
Q u e s t i o n : 5 9
The base of a right triangle. is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is
a 168 cm
2
b 252 cm
2
c 336 cm
2
d 504 cm
2
S o l u t i o n :
c 336 cm
2
Let ?
PQR be a right-angled triangle and PQ ?
QR.
Now,
PQ =
v
PR
2
-QR
2
=
v
50
2
-48
2
= v 2500 -2304 = v 196 = 14 cm
? Area of triangle =
1
2
×QR ×PQ =
1
2
×48 ×14 = 336 cm
2
Q u e s t i o n : 6 0
The area of an equilateral triangle is 81v 3 cm
2
. Its height is
a
9v 3 cm
b
6v 3 cm
c
18v 3 cm
d 9 cm
S o l u t i o n :
a
9v 3 cm
Area of equilateral triangle = 81v 3 cm
2
?
v
3
4
× Side)
2
= 81v 3 ? Side)
2
= 81 ×4 ? Side)
2
= 324 ? Side = 18 cmNow, Height = 
v
3
2
×Side =
v
3
2
×18 = 9v 3 cm
( (
(
(
( ( (