Question

in triangle abc in which the medians from b and c are perpendicular then b^2+c^2=​

Answer 1

Answer:

b² + c² = 5a²

Step-by-step explanation:

Given data:

In ∆ ABC, medians through B and C are perpendicular to each other.

To find: The value of (b² + c²)

As shown in the figure below., Let BE be the median of length “y” from B to AC and CF  be the median of length “x” from C to AB. Also, the medians are intersecting each other perpendicularly at point G called its centroid.

We know that a centroid divide the median in the ratio of 2:1. Therefore,  

BG : GE = 2y : y

CG : GF = 2x : x

And,

Let the length of  

Side BC be “a”

Side AC be “b”

Side AB be “c”

Since the medians divide the adjacent side of the triangle into equal half. Therefore,

AE = EC = b/2

CF = FA = c/2  

In ∆ BGF, by pythagoras theorem we get

c²/4 =  x² + 4y²

or, c² = 4x² + 16y² ….. (i)

In ∆ CEG, by pythagoras theorem we get

b²/4 =  y² + 4x²

or, b² = 4y² + 16x² ….. (ii)

In ∆ BCG, by pythagoras theorem we get

a² =  4x² + 4y² …… (iii)

From (i), (ii) & (iii) we will calculate the value for b²+c²

∴ b² + c²

= 4y² + 16x² + 4x² + 16y²

= 20x² + 20y²

= 5 [4x² + 4y²]

= 5a²

Answer 2

Answer:

5a^2

Step-by-step explanation:

in ∆ ABC, BC= a

BE & CD Medians intersect at O & Perpendicular

DE//BC ; DE = BC/2 ;DE = a/2 (MPT)

EC = b/2 : BD = c/2

in ∆ DOE, O^ = 90° by pythog th...

OE^2 + OD^2 = DE^2

OE^2 + OD^2 = (BC/2)^2

OE^2 + OD^2 = a^2/4.....(1)

in ∆ BOC, O^ = 90° by pythog th...

OB^2 + OC^2 = BC^2

OB^2 + OC^2 = a^2. .......(2)

Now in ∆ COE, O^ = 90° by pythog..

OE^2 + OC^2 = EC^2

OE^2 + OC^2 = b^2/4 .....(3)

in ∆ BOD, O^ = 90° by pythog..

OB^2 + OD^2 = BD^2

OB^2 + OD^2 = c^2/4......(4)

now (3) + (4)....

(OE^2 + OC^2 ) + (OB^2 + OD^2) = (b^2/4) + (c^2/4)

(OE^2 + OD^2) + (OC^2 + OB^2) = (b^2 + c^2) / 4

using (1) & (2) ....

[(a^2)/4 + a^2]= (b^2 + c^2) / 4

5(a^2)/4 = (b^2 + c^2) /4

hence (b^2 + c^2) = 5a^2