Question

Let P be a point inside an equilateral triangle with each sides of length 'l'. If h1, h2, h3 are the
perpendicular distances from P to the three sides of the triangle. Then the value of h1 + h2+ h3 is​

Answer 1

The value of h1 + h2+ h3 is $\frac{\sqrt{3}}{2} * l$.

Step-by-step explanation:

It is given that,

(as shown in the figure attached below)

P is a point inside the equilateral triangle ABC  

Each side AB = BC = CA = “l”

Join each of the vertices A, B and C with point P to form three triangles APB, PBC & APC.

Let “h1” be the perpendicular height on AB, “h2” be the perpendicular height on BC and “h3” be the perpendicular height on AC.

Now, we know that the area of an equilateral triangle with side “a” is given as,

Area = $\frac{\sqrt{3}}{2}$ * a²

From the figure attached below, we can see that

Area of the equilateral triangle ABC = [Area of ∆APB] + [Area of ∆PBC] + [Area of ∆APC]  

⇒ $\frac{\sqrt{3}}{4}$ * (l)² = [½ * h1 * l] + [½ * h2 * l] + [½ * h3 * l]

⇒ $\frac{\sqrt{3}}{4}$ * (l)² = ½ * l * [h1 + h2 + h3]  

⇒ h1 + h2 + h3 = 2 * $\frac{\sqrt{3}}{4}$ * (l)

⇒ h1 + h2 + h3 = $\frac{\sqrt{3}}{2}$ * (l)

Thus, the value of h1 + h2 + h3 is [ $\frac{\sqrt{3}}{2}$ * (l)].

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Also View:

From point in the interior of an equilateral triangle , perpendicular are drawn on each of its three sides.the length of the perpendiculars are 14cm,10cm and 6cm.find the area of the triangle ?

https://brainly.in/question/5602151

Answer 2

Area of the equilateral triangle having length of each side 1 is given by √34⋅12=√34squnit

Again it's area= sum of areas of three triangles obtained by joining P with the three vertices.

12⋅1⋅h1+12⋅1⋅h2+12⋅1⋅h3

So12(h1+h2+h3)=√34

⇒h1+h2+h3=√32

Step-by-step explanation:

Explanation:

For any equilateral #triangle ABC of side 'a',

the area of the triangle=12(√32a)(a)=√34a2

= sum of the areas of △s PBC,PCAandPAB

=12(h1+h2+h3)a, and so,

(h1+h2+h3)=(√3/2)a

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