Question

AD is a median of the triangle ABC.if AB^2+AC^2=P.(BD^2+AD^2) then options are (a)p=2,(b)p=1/2,(c)p=1/4 and (d)p=3

Answer 1

Question:

AD is a median of the ∆ABC.

If AB² + AC² = p•( BD² +AD² ) , then find the value of p .

Options:-

(a) p = 2

(b) p = 1/2

(c) p = 1/4

(d) p = 3

Answer:

option(a)

p = 2

Note:

• The line segment which joins a vertex and the mid point of its opposite side of a triangle is called Median .

• A triangle has three medians corresponding to each side.

• Pythagoras theorem:- In a right angled triangle , the square of the hypotenuse is equal to the sum of squares of other two sides.

Solution:

Given:-

AD is the median of the ∆ABC , thus ;

BD = CD ---------(1)

Construction:-

Let's draw an altitude AE from the vertex A to the side BC .

Now,

Applying Pythagoras theorem in right angled

∆ABE , ∆ADE and ∆ACE , we have ;

AB² = AE² + BE² --------(2)

AC² = AE² + CE² --------(3)

AD² = AE² + DE² --------(4)

Now ,

Adding eq-(2) and eq-(3) , we get;

=> AB² + AC² = AE² + BE² + AE² + CE²

=> AB² + AC² = 2AE² + BE² + CE²

=> AB² + AC² = 2(AD²- DE²) + BE² + CE²

{ using eq-(4) }

=> AB² + AC² = 2AD²- 2DE² + (BD-DE)² + (CD+DE)²

{ since, BE = BD - DE and CE = CD + DE }

=> AB² + AC² = 2AD²- 2DE² + BD² + DE² - 2BD•DE

+ CD² + DE² + 2CD•DE

=> AB² + AC² = 2AD²- 2DE² + 2DE² + BD² + CD²

- 2BD•DE + 2CD•DE

=> AB² + AC² = 2AD² + BD² + CD² - 2BD•DE

+ 2CD•DE

=> AB² + AC² = 2AD² + BD² + BD² - 2BD•DE

+ 2BD•DE

{ using eq-(1), CD = BD }

=> AB² + AC² = 2AD² + 2BD²

=> AB² + AC² = 2(AD² + BD²) ------------(5)

Now,

Comparing eq-(5) with the given condition ,

AB² + AC² = p(AD² + BD²) , we get ;

p = 2.

Hence,

The required value of p is 2 .