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· started a discussion

· 1 Months ago

yy (Question: 184943.0 )

what is solution....

Question:
The length of the perpendiculars drawn from a point in the interior of an equilateral triangle are 6 cm, 7cm, and 8 cm. What is the length of the side of triangle ? 
Options:
A) 7 cm
B) 10.5 cm
C) \(14\sqrt{3}\)  cm
D) \(\cfrac{14\sqrt{3}}{3}\) cm
Solution:

Let ∆ABC be an equilateral triangle.

 Length of each side be ‘a’ cm.

 O is a point in the interior of the triangle. OX⊥BC, OY⊥CA and OZ⊥AB

 Let OZ = 6 cm, OX = 7 cm and OY = 8 cm

Area of ∆ABC = Area of ∆AOB + Area of ∆BOC + Area of ∆COA

 √3/4 *a^2 = (1/2 x OZ x AB) + (1/2 x OX x BC) + (1/2 x OY x AC)

 √3/4 * a^2 = (1/2 x 6 x a) + (1/2 x 7 x a) + (1/2 x 8 x a)

 √3/4 * a^2 = 3a + 7/2 a + 4a

 √3/4 *a^2 =7a+  7a/2 = 21a/2

 a = (21 x 4) / 2√3

 a = 42 / √3 cm = 14*3/√3 cm = 14√3 cm

 Therefore, the length of the side = 14√3 cm

Knowledge Expert

· commented

· 1 Months ago

Let ∆ABC be an equilateral triangle.
Length of each side be ‘a’ cm.
O is a point in the interior of the triangle. OX⊥BC, OY⊥CA and OZ⊥AB
Let OZ = 6 cm, OX = 7 cm and OY = 8 cm
Area of ∆ABC = Area of ∆AOB + Area of ∆BOC + Area of ∆COA
√3/4 *a^2 = (1/2 x OZ x AB) + (1/2 x OX x BC) + (1/2 x OY x AC)
√3/4 * a^2 = (1/2 x 6 x a) + (1/2 x 7 x a) + (1/2 x 8 x a)
√3/4 * a^2 = 3a + 7/2 a + 4a
√3/4 *a^2 =7a+ 7a/2 = 21a/2
a = (21 x 4) / 2√3
a = 42 / √3 cm = 14*3/√3 cm = 14√3 cm
Therefore, the length of the side = 14√3 cm
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