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CBSE Questions for Class 9 Maths Heron'S Formula Quiz 4 - MCQExams.com
CBSE
Class 9 Maths
Heron'S Formula
Quiz 4
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are
15
c
m
,
14
c
m
,
13
c
m
, and the parallelogram stands on the base of
15
cm. Find the height of the parallelogram.
Report Question
0%
5.9
c
m
0%
6.8
c
m
0%
6.5
c
m
0%
5.6
c
m
Explanation
Area of triangle =
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
S
=
a
+
b
+
c
2
=
15
+
14
+
13
2
=
21
c
m
∴
area of
△
=
√
21
(
21
−
15
)
(
21
−
14
)
(
21
−
13
)
=
√
21
×
6
×
7
×
8
=
√
3
×
7
×
3
×
2
×
7
×
4
×
2
=
2
×
2
×
3
×
7
=
84
c
m
2
Now, area of
∥
gm = area of
△
=
b
×
h
=
84
c
m
2
⇒
15
×
h
=
84
⇒
h
=
84
15
=
5.6
c
m
∴
Height of
∥
gm =
5.6
c
m
What is the area of a triangle whose sides are
13
c
m
,
14
c
m
and
15
c
m
?
Report Question
0%
84
s
q
.
c
m
0%
64
s
q
.
c
m
0%
825
s
q
.
c
m
0%
105
s
q
.
c
m
Explanation
S
=
a
+
b
+
c
2
=
13
+
14
+
15
2
=
21
c
m
area of
△
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
=
√
21
(
21
−
13
)
(
21
−
14
)
(
21
−
15
)
=
√
21
×
8
×
7
×
6
=
√
3
×
7
×
2
×
2
×
2
×
7
×
2
×
3
=
2
×
2
×
3
×
7
=
84
c
m
2
Find the number of trees that can be planted in a triangular ground having sides
51
m
,
37
m
and
20
m
, when each tree occupies
6
m
2
of space:
Report Question
0%
54
0%
51
0%
82
0%
43
Explanation
Area of
△
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
,
s
=
a
+
b
+
c
2
=
51
+
37
+
20
2
=
54
m
∴
Area of triangle ground
=
√
54
(
54
−
51
)
(
54
−
37
)
(
54
−
20
)
=
√
54
×
3
×
17
×
34
=
306
m
2
Number of trees
=
306
6
=
51
The perimeter of a triangular field is
420
m and its sides are in the ratio
6
:
7
:
8
Find area of the triangular field.
Report Question
0%
2100
sq m
0%
2100
√
15
sq m
0%
2000
sq m
0%
2000
√
15
sq m
Explanation
Step - 1: Finding the sides
Given that the sides are in the ratio
6
:
7
:
8
⟹
a
=
6
x
,
b
=
7
x
,
c
=
8
x
Perimeter
=
a
+
b
+
c
⟹
420
=
6
x
+
7
x
+
8
x
=
21
x
⟹
x
=
20
⟹
a
=
120
,
b
=
140
,
c
=
160
Step - 2: Calculating area
s
=
a
+
b
+
c
2
=
210
⟹
Area
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
⟹
Area
=
√
210
×
90
×
70
×
50
⟹
Area
=
2100
√
15
Thus, the area of the given field is
2100
√
15
sq. m.
In
△
A
B
C
,
A
B
=
6
c
m
,
B
C
=
7
c
m
and
A
C
=
5
c
m
. Find the area of
△
A
B
C
Report Question
0%
6
√
6
c
m
2
0%
6
√
3
c
m
2
0%
6
√
2
c
m
2
0%
9
√
6
c
m
2
Explanation
Using Heron's formula,
Area of a triangle
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
where
s
=
semiperimeterof the triangle
a
=
length of side
B
C
b
=
length of side
A
C
c
=
length of side
A
B
In
△
A
B
C
,
a
=
7
,
b
=
5
,
c
=
6
Semiperemeter of
△
A
B
C
=
s
=
a
+
b
+
c
2
=
7
+
5
+
6
2
=
9
Hence, the area of
△
A
B
C
=
√
9
×
(
9
−
7
)
×
(
9
−
5
)
×
(
9
−
6
)
=
√
9
×
3
×
2
×
4
=
6
√
6
c
m
2
Find the area of the parallelogram with sides
11
c
m
and
13
c
m
and one of the diagonals as
16
c
m
.
Report Question
0%
23
√
32
sq. cm
0%
25
√
35
sq. cm
0%
24
√
35
sq. cm
0%
None of these
Explanation
ar
∥
gm ABCD =
2
×
a
r
△
A
B
C
ar
△
ABC =
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
,
s
=
a
+
b
+
c
2
=
16
+
13
+
11
2
=
20
c
m
∴
ar
△
ABC =
√
20
(
20
−
16
)
(
20
−
13
)
(
20
−
11
)
=
√
20
×
4
×
7
×
9
=
√
5
×
4
×
4
×
7
×
9
=
4
×
3
√
35
=
12
√
35
c
m
2
∴
ar
△
ABCD =
2
×
12
√
35
=
24
√
35
c
m
2
Find the area of the quadrilateral whose sides are
9
m
,
40
m
,
28
m
,
15
m
. The angle between first two sides is
90
o
.
Report Question
0%
348
sq. m
0%
315
sq. m
0%
306
sq. m
0%
None of these
Explanation
Area of quadrilateral
A
B
C
D
= ar
[
△
A
B
C
]
+ ar
[
△
A
C
D
]
In
△
A
B
C
:
a
r
[
△
A
B
C
]
=
1
2
×
B
C
×
A
B
=
1
2
×
40
m
×
9
m
=
180
m
2
Using pythagoras theorem,
A
C
=
√
A
B
2
+
B
C
2
=
√
(
9
)
2
+
(
40
)
2
=
√
1681
=
41
m
Using Heron's formula,
a
r
[
△
A
C
D
]
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
s
=
a
+
b
+
c
2
=
41
+
28
+
15
2
=
84
2
=
42
m
a
r
[
△
A
C
D
]
=
√
42
(
42
−
41
)
(
42
−
28
)
(
42
−
15
)
=
√
42
×
1
×
14
×
27
=
126
m
2
Therefore, area of quadrilateral
A
B
C
D
=
180
+
126
=
306
m
2
Find the area of the parallelogram with sides
9
c
m
and
11
c
m
and one of the diagonals as
14
c
m
.
Report Question
0%
22
√
15
0%
24
√
17
0%
21
√
17
0%
None of these
Explanation
Area of
∥
gm ABCD = 2
×
area
△
ABC
ar
△
ABC =
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
,
&
s
=
a
+
b
+
c
2
=
14
+
11
+
19
2
=
17
c
m
∴
ar
△
ABC =
√
17
(
17
−
14
)
(
17
−
11
)
(
17
−
9
)
=
√
17
×
3
×
6
×
8
=
12
√
17
c
m
2
Therefore, ar
∥
gm ABCD = 2
×
12
√
17
=
24
√
17
c
m
2
The parallel sides of a parallelogram are
60
m and
25
m and one of the diagonals is
65
m. Find the area of the parallelogram,
Report Question
0%
1430
sq m
0%
1576
sq m
0%
1500
sq m
0%
None of these
Explanation
The area of parallelogram
A
B
C
D
=
2
×
A
r
△
A
B
C
Now,
A
r
△
A
B
C
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
s
=
a
+
b
+
c
2
=
65
+
60
+
25
2
=
75
m
Thus,
∴
A
r
△
A
B
C
=
√
75
(
75
−
65
)
(
75
−
60
)
(
75
−
25
)
=
√
75
×
10
×
15
×
50
=
750
m
2
Thus, area of parallelogram
=
2
×
750
m
2
=
1500
m
2
Find the area of an equilateral triangle each of whose side is 4 cm long?
Report Question
0%
2
√
3
0%
3
√
3
0%
5
√
3
0%
4
√
3
Explanation
solution:
A
r
e
a
=
√
3
4
r
2
A
r
e
a
=
√
3
4
4
2
A
r
e
a
=
√
3
1
4
hence the correct opt: D
In the given figure, the area of the
△
A
B
C
is
Report Question
0%
13.24
c
m
2
0%
12.28
c
m
2
0%
11.32
c
m
2
0%
15.37
c
m
2
Explanation
We know that the area of the triangle using heron's formula is given by
Area
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
So,
S
=
8
+
4
+
11
2
=
23
2
=
11.5
Area
=
√
11.5
(
11.5
−
11
)
(
11.5
−
8
)
(
11.5
−
4
)
=
√
11.5
×
0.5
×
3.5
×
7.5
=
12.28
cm
2
If the sides of a triangle are
y
z
+
z
x
,
z
x
+
x
y
and
x
y
+
y
z
then its area is
Report Question
0%
xyz
0%
√
x
y
+
y
z
+
z
x
0%
√
x
y
z
0%
x
2
y
2
z
2
2
A triangular park in a city has dimensions 100m X 90m X110m.A contract is given to a company for planting grass in the park at the rate of Rs.4000 per hectare.Find the amount to be paid to the company.(Take
√
2
=
1.414
)
Report Question
0%
Rs.
4532.90
0%
Rs.
4242
0%
Rs.
1696.80
0%
Rs.
1000
Explanation
H
e
r
o
n
′
s
F
o
r
m
u
l
a
:
Area of a triangle =
Δ
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
where
s
is the semi-perimeter and
a
,
b
and
c
are the sides of the triangle.
s
=
a
+
b
+
c
2
Here, for the triangular park,
a
=
100
m
,
b
=
90
m
,
c
=
110
m
and
s
=
100
+
90
+
110
2
=
150
m
Area of the triangular park =
Δ
=
√
150
×
50
×
60
×
40
=
3000
√
2
m
2
=
0.3
√
2
h
a
=
0.4242
h
a
Rate of planting grass =
R
s
.
4000
/
h
a
So, the amount to be paid =
4000
×
0.4242
=
R
s
.
1696.80
.
[
C
]
Find the area of a triangle whose two sides are
8
cm and
11
cm and the perimeter is
32
cm.
Report Question
0%
8
c
m
2
0%
8
√
30
c
m
2
0%
8
√
30
c
m
2
0%
24
c
m
2
Explanation
Let third side be
x
Perimeter
=
8
+
11
+
x
19
+
x
=
32
x
=
13
s
=
32
2
=
16
Heron's formula
A
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
=
√
16
(
16
−
8
)
(
16
−
11
)
(
16
−
13
)
=
√
16
×
8
×
5
×
3
=
8
√
30
c
m
2
One side of an equilateral triangle measures
8
cm. Find its area using Heron's formula. What is its altitude
Report Question
0%
12
√
3
cm
2
,
3
√
3
cm
0%
16
√
3
cm
2
,
4
√
3
cm
0%
14
√
3
cm
2
,
7
2
√
3
cm
0%
20
√
3
cm
2
,
5
√
3
cm
Explanation
According to the Heron's formula, Area (A) of the triangle having sides
a
,
b
,
c
units is
A
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
where
s
=
a
+
b
+
c
2
For the given triangle,
a
=
b
=
c
=
8
c
m
s
=
8
+
8
+
8
2
=
12
A
=
√
12
(
12
−
8
)
(
12
−
8
)
(
12
−
8
)
A
=
√
3
×
4
×
4
3
=
4
×
4
√
3
=
16
√
3
Area of the triangle =
1
2
×
b
a
s
e
×
a
l
t
i
t
u
d
e
=
16
√
3
1
2
×
8
×
altitude
=
16
√
3
Altitude
=
16
√
3
4
=
4
√
3
The perimeter of a triangle is
300
m
. If its sides are in the ratio
3
:
5
:
7
. Find the area of the triangle.
Report Question
0%
1500
√
5
m
2
0%
1500
m
2
0%
1500
√
3
m
2
0%
1100
√
3
m
2
Explanation
Let, 1 side be =
3
x
2 side be =
5
x
3 side be =
7
x
According to question,
3
x
+
5
x
+
7
x
=
300
⟹
x
=
20
So, sides are
60
,
100
,
140
Area =
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
s
=
s
e
m
i
−
p
e
r
i
m
e
t
e
r
s
=
150
,
a
=
60
,
b
=
100
,
c
=
140
Area=
√
150
(
150
−
60
)
(
150
−
100
)
(
150
−
140
)
⟹
√
150
(
90
)
(
50
)
(
10
)
⟹
1500
√
3
m
2
Find the area of a triangle whose sides are 9 cm, 12 cm, and 15 cm.
Report Question
0%
34
c
m
2
0%
44
c
m
2
0%
54
c
m
2
0%
55
c
m
2
Explanation
Given sides of the triangle
a
=
9
cm
b
=
12
cm
c
=
15
cm
The given triangle is a right angle triangle
because
9
2
+
12
2
=
15
2
∴
Area
=
1
2
×
9
×
12
=
54
sq cm
Area
=
54
sq cm.
The perimeter of a triangle field is
450
m
and its sides are in the ratio
25
:
17
:
12
. Find the area of the triangle.
Report Question
0%
2250
m
2
0%
6250
m
2
0%
5250
m
2
0%
6050
m
2
Explanation
Let, 1 side be =
25
x
2 side be =
17
x
3 side be =
12
x
According to question,
25
x
+
17
x
+
12
x
=
450
⟹
x
=
450
54
⟹
x
=
25
3
So, sides are
625
3
,
425
3
,
100
1
Area =
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
s
=
s
e
m
i
−
p
e
r
i
m
e
t
e
r
s
=
225
,
a
=
625
3
,
b
=
425
3
,
c
=
100
1
Area=
√
225
(
225
−
625
3
)
(
225
−
425
3
)
(
225
−
100
)
⟹
√
225
(
50
3
)
(
250
3
)
(
125
)
⟹
6250
m
2
Find the area of triangle of sides
6
c
m
,
8
c
m
,
10
c
m
Report Question
0%
24
c
m
2
0%
12
c
m
2
0%
30
c
m
2
0%
n
o
n
e
Explanation
The sides are
6
c
m
,
8
c
m
,
10
c
m
s
=
6
+
8
+
10
2
=
24
2
=
12
The area is given as
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
√
12
(
6
)
(
4
)
(
2
)
√
576
=
24
c
m
2
Find the area of a triangle whose sides are respectively
150
c
m
,
120
c
m
, and
200
c
m
.
Report Question
0%
8545.77
sq. cm
0%
8966.57
sq. cm
0%
9257.66
sq. cm
0%
5627.66
sq. cm
Explanation
Let
a
=
150
c
m
,
b
=
120
c
m
and
c
=
200
c
m
.
s
=
a
+
b
+
c
2
⇒
s
=
150
+
120
+
200
2
⇒
s
=
235
c
m
Now, by using the heron's formula,
Area
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
=
√
235
(
235
−
150
)
(
235
−
120
)
(
235
−
200
)
=
√
235
(
85
)
(
115
)
(
35
)
=
√
80399375
=
8966.57
sq. cm
If every side of a triangle is doubled, the area of the new triangle is
K
times the area of the old one. Find the value of
K
.
Report Question
0%
√
2
0%
2
0%
3
0%
4
Explanation
Let, the sides of the original triangle be
a
,
b
and
c
so, the semi-perimeter,
s
=
a
+
b
+
c
2
Original area
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
Now every side is doubled
So, the new sides are
2
a
,
2
b
and
2
c
new semi-perimeter,
s
′
=
2
a
+
2
b
+
2
c
2
=
2
(
a
+
b
+
c
2
)
=
2
s
New area
=
√
s
′
(
s
′
−
2
a
)
(
s
′
−
2
b
)
(
s
′
−
2
c
)
=
√
(
2
s
)
(
2
s
−
2
a
)
(
2
s
−
2
b
)
(
2
s
−
2
c
)
=
√
2.2.2.2
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
=
4
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
=
4
×
O
r
i
g
i
n
a
l
A
r
e
a
Hence, the value of
K
is
4
.
Use Heron's formula to find the area of a triangle of lengths
5
,
7
and
8
.
Report Question
0%
10
√
3
0%
20
√
3
0%
2
√
6
0%
20
√
6
Three sides of a triangular field are 20m, 21m, and 29m long, respectively. The area of the field is :
Report Question
0%
215
m
2
0%
230
m
2
0%
210
m
2
0%
None of these
Explanation
Given: Sides of the triangular field;
a
=
20
m
,
b
=
21
m
,
c
=
39
m
Now, Semiperimeter
S
=
a
+
b
+
c
2
=
20
+
21
+
29
2
=
35
m
A
r
e
a
=
√
s
×
(
s
−
a
)
×
(
s
−
b
)
×
(
s
−
c
)
=
√
35
×
(
35
−
20
)
×
(
35
−
21
)
×
(
35
−
29
)
=
√
35
×
15
×
14
×
6
=
210
s
q
.
m
The area of a triangle whose sides are 15 m, 16 m and 17 m is :
Report Question
0%
24
√
4
m
2
0%
24
√
3
m
2
0%
24
√
21
m
2
0%
None of these
Explanation
The sides of the triangle are
a
=
15
m
,
b
=
16
m
,
c
=
17
m
Area of the triangle is given by
S
=
a
+
b
+
c
2
=
15
+
16
+
17
2
=
48
2
=
24
m
Area
=
√
S
(
S
−
a
)
(
S
−
b
)
(
S
−
c
)
=
√
24
(
25
−
15
)
(
24
−
16
)
(
24
−
17
)
=
√
24
×
9
×
8
×
7
=
√
8
_
×
3
×
3
_
×
3
×
8
_
×
7
=
24
√
21
sq. m.
The sides of a triangle are
7
cm,
9
cm and
14
cm. Its area is
Report Question
0%
12
√
5
cm
2
0%
12
√
3
cm
2
0%
24
√
5
cm
2
0%
63
cm
2
Explanation
Area of triangle =
√
S
(
S
−
a
)
(
S
−
b
)
(
S
−
c
)
S
=
a
+
b
+
c
2
Given
a
=
7
cm
b
=
9
cm
c
=
14
cm
S
=
7
+
9
+
14
2
=
30
2
S
=
15
c
m
∴
Area of triangle =
√
S
(
S
−
a
)
(
S
−
b
)
(
S
−
c
)
=
√
15
(
15
−
7
)
(
15
−
9
)
(
15
−
14
)
=
√
15.8.6.1
=
√
720
=
√
5
×
144
=
√
12
×
12
×
5
=
12
√
5
c
m
2
∴
Area of triangle =
12.
√
5
c
m
2
So, the answer is A.
12.
√
5
c
m
2
The sides of a triangle are in the ratio
3
:
5
:
7.
If the perimeter of the triangle is
60
cm, then its area
Report Question
0%
60
√
3
cm
2
0%
30
√
3
cm
2
0%
15
√
3
cm
2
0%
120
cm
2
Explanation
Given sides in the ratio
3
:
5
:
7
perimeter of the triangle
=
60
c
m
3
x
+
5
x
+
7
x
=
60
⇒
15
x
=
60
⇒
x
=
60
/
15
⇒
x
=
4
3
x
=
3
×
4
=
12
cm
5
x
=
5
×
4
=
20
cm
7
x
=
7
×
4
=
28
cm
Let
S
be the semiperimeter of given triangle.
S
=
1
2
(
a
+
b
+
c
)
=
1
2
⋅
60
=
30
Now use formula foe area of triangle
Area
=
√
30
(
30
−
12
)
(
30
−
20
)
(
30
−
28
)
=
√
30
⋅
18
⋅
10
⋅
2
=
√
60
⋅
180
=
√
10800
=
√
3600
⋅
3
=
60
√
3
c
m
2
∴
Area of triangle
=
60
√
3
c
m
2
.
The percentage increase in the area of a triangle if its each side is doubled, is
Report Question
0%
200
%
0%
400
%
0%
300
%
0%
500
%
Explanation
If the sides of triangle are 4, 5 and 6 cm. Then the area (in sq cm) of triangle is
Report Question
0%
π
4
0%
π
4
√
7
0%
4
15
0%
4
15
√
7
0%
15
4
√
7
Explanation
Given, triangle of sides
a
,
b
,
c
=
4
,
5
,
6
c
m
∴
S
=
a
+
b
+
c
2
=
4
+
5
+
6
2
=
15
2
Then, area of triangle
A
=
√
S
(
S
−
a
)
(
S
−
b
)
(
S
−
c
)
=
√
15
2
(
15
2
−
4
)
(
15
2
−
5
)
(
15
2
−
6
)
=
√
15
2
.
(
7
2
)
(
5
2
)
(
3
2
)
=
15
4
√
7
Find the area of equilateral triangle each of whose sides measures 20 cm.
Report Question
0%
173
c
m
2
0%
150
c
m
2
0%
120
c
m
2
0%
153
c
m
2
Explanation
Area of Equilateral Triangle =
√
3
b
2
4
where b is the Side of the Triangle.
Area=
√
3
×
20
2
4
Area=
173.2
c
m
2
Find the area of equilateral triangle (in
c
m
2
)each of whose sides measure 18 cm.
Report Question
0%
150
0%
140
0%
180
0%
190
Explanation
Area of Equilateral Triangle =
√
3
a
2
4
where
a
is the Side of the Triangle.
Given
a
=
18
c
m
Area
=
√
3
×
18
2
4
Area
=
140.29
c
m
2
0:0:3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
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Practice Class 9 Maths Quiz Questions and Answers
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