IIT JEE , AIEEE Forum - Mathematics , Algebra

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Let P be a point inside an equilateral triangle of side 10 cm. If the distances of P from two sides are 2cm and 3cm, find the distance of P from the third side.

joyfrancis

Let the equilateral traingle have co-ordinates A(-5,0), B(5,0) and C(0,5

3)

Take point P as (a,b)

Let it be at a distance of 3cm from AB and 2cm from BC

.:b=3

point p is (a,3)

It's distance from BC=2cm

Eqn of BC

=>x

3+y-5

3=0

it's distnce from(a,3) is

(a

3+3-5

3)/2

since this distance=2 we get

a=(1+5

3)/

3

.: P is [(1+5

3)/

3 , 3]

Eqn of AC

=>y-x

3-5

3=0

It's distance from P

=(5+5

3-3+5

3)/2

=53-5------>ans

gagsdon

This Question can be solved by Simply comparing areas os the Independent triangles

Let ABC be an equilateral triangle and P be any point in the interior of it.

Let Pl , Pm and Pn perpendicular to the sides AB , BC and CA.

Now area of Triangle ABC = ^{[ ](}

3)/4 a²= 25^{[ ]}

3

Now ar triangle PAB = 1/2 * base * ht

=10

and ar triangle PBC = 15

Therefore ar Triangle PAC = 25(^{[ ]}

3 - 1)

Hence 1/2 * 10 * Pn = 25(^{[ ]}

3 - 1)

hence Pn = 5(^{[ ]}

3 - 1)

hence my answer is 5(^{[ ]}

3 - 1)

Plzzzz Rate me. um new in this field

puneet

hey

the solution of gagsdon is easy and best .. this is wat comes to my mind straight .. try this one ..

cheers

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