From a point in the interior of a equilateral triangle, perpendiculars are drawn to its sides. If the perpendiculars are 14cm, 10cm and 6cm. Find the area of the triangle.
Answers
Step-by-step explanation:
Let ABC be an equilateral triangle, O be the interior point and AQ, BR and CP are the perpendicular drawn from point O.
Let the each side of an equilateral triangle be m.
A
r
e
a
o
f
Δ
O
A
B
=
1
2
×
A
B
×
O
P
[
∵
a
r
e
a
o
f
a
t
r
i
a
n
g
l
e
=
1
2
(
b
a
s
e
×
h
e
i
g
h
t
)
]
=
1
2
×
a
×
14
=
7
a
c
m
2
⋯
(
i
)
A
r
e
a
o
f
Δ
O
B
C
=
1
2
×
B
C
×
O
Q
=
1
2
×
a
×
10
=
5
a
c
m
2
⋯
(
i
i
)
A
r
e
a
o
f
Δ
O
A
C
=
1
2
×
A
C
×
O
R
=
1
2
×
a
×
6
=
3
a
c
m
2
⋯
(
i
i
i
)
∴
Area of an equilateral
Δ
A
B
C
,
=
A
r
e
a
o
f
(
Δ
O
A
B
+
Δ
O
B
A
+
Δ
O
A
C
)
=
(
7
a
+
5
a
+
3
a
)
=
15
a
c
m
2
⋯
(
i
v
)
We have, semi-perimeter
s
=
a
+
a
+
a
2
⇒
s
=
3
a
2
c
m
∴
Area of an equilateral
Δ
A
B
C
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
[by Heron’s formula]
=
√
3
a
2
(
3
a
2
−
a
)
(
3
a
2
−
a
)
(
3
a
2
−
a
)
=
√
3
a
2
×
a
2
×
a
2
×
a
2
=
√
3
4
a
2
⋯
(
v
)
From Equations (iv) and (v),
√
3
4
a
2
=
15
a
⇒
a
=
15
×
4
√
3
×
√
3
√
3
=
60
√
3
3
=
20
√
3
c
m
putting
a
=
20
√
3
in Equation (v), we get
A
r
e
a
o
f
Δ
A
B
C
=
√
3
4
(
20
√
3
)
2
=
√
3
4
×
400
×
3
=
300
√
3
c
m
2
Hence, the area of an equilateral triangle is
300
√
3
c
m
2
MathematicsNCERT ExemplarStandard IX
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SIMILAR QUESTIONS
Q. From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are
14
c
m
,
10
c
m
and
6
c
m
. Find the area of the triangle.
Mathematics
Q. From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are
14
cm,
10
cm and
6
cm. Find the area of the triangle.