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.
Solution:
Consider a equilateral triangle ABC
P, Q and R are the perpendicular drawn on the three sides
The length of the perpendiculars are 14 cm, 10 cm and 6 cm
We have to find the area of the triangle.
Let the sides of an equilateral triangle be a cm
Area of triangle = 1/2 × base × height
Area of triangle OAB = 1/2 × AB × OP
= 1/2 × a × 14
= 7a cm²
Area of triangle OBC = 1/2 × BC × OQ
= 1/2 × a × 10
= 5a cm²
Area of triangle OAC = 1/2 × AC × OR
= 1/2 × a × 6
= 3a cm²
Area of triangle ABC = area of triangle (OAB + OBC + OAC)
= 7a + 5a + 3a
= 15a cm² ----------------------- (1)
By Heron’s formula,
Area of triangle = √s(s - a)(s - b)(s - c)
Where s = semiperimeter
s = (a + b + c)/2
So, s = (a + a + a)/2
= 3a/2
s = 1.5a
Area = √1.5a(1.5a - a)(1.5a - a)(1.5a - a)
= √1.5a(0.5a)³
= √1.5(0.125)a⁴
= a²√0.1875
= 0.4330a² ------------------------ (2)
Comparing (1) and (2),
15a = 0.4330 a²
0.433a = 15
a = 15/0.433
a = 34.64 cm
Area of equilateral triangle = √3/4 (side)²
= √3/4 (34.64)²
= √3/4 (1199.99)
= √3 (299.99)
= 300√3 cm²
Therefore, the area of triangle is 300√3 cm²
✦ Try This: From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 16 cm, 10 cm and 8 cm. Find the area of the triangle.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 12
NCERT Exemplar Class 9 Maths Exercise 12.3 Problem 3
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.
Summary:
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. The area of the triangle is 300√3 cm²
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