Question

Find the sum of squares of the medians MP and OQ drawn from the two acute angled vertices of a right angled triangle MNO. The longest side of Δ MNO is 20cm .

Answer 1

Answer:

In △ABC,

AE and CD are the medians of the triangle.

So, AE=6 and CD=8

Hence, BE=EC=21BC and AD=DB=21AB

In △ABE

AE2=AB2+BE2

∴36=AB2+41BC2           ....(1)

In △BCD

CD2=DB2+BC2

∴64=41AB2+BC2           ....(2)

Adding (1) and (2), we get

100=45(AB2+BC2)

∴AC2=80

∴AC=45 cm

Answer 2

Concept

The Pythagorean theorem states, "In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides." The sides of this triangle were called the perpendicular, the base and the hypotenuse. Here, the hypotenuse is on the opposite side of the 90° angle, so it is the longest side.

Given

We have been given that two median  MP and OQ  is drawn from  two acute angled vertices of a right angled triangle . and the longest side of Δ MNO is 20cm .

Find

We are asked to determine the sum of squares of the medians .

Solution

As P is a median of length NO then, P divides the NO in equal length which means that Let OP = x then NP will also = x

Similarly ,  As Q is a median of length MN then, Q divides the MN in equal length which means that Let MQ = y then QN will also = y

In Δ MNO

According to the Pythagorean theorem

$(MO)^2=(MN)^2+(NO)^2\\\\(20)^2=(2y)^2+(2x)^2\\\\400=4y^2+4x^2$  

On dividing by 4 , we get

$x^2+y^2=100$    ...(1)

In Δ MNP

According to the Pythagorean theorem

$(MP)^2=(MN)^2+(NP)^2\\\\(MP)^2=(2y)^2+(x)^2\\\\(MP)^2=4y^2+x^2$....(2)

In  Δ QNO

According to the Pythagorean theorem

$(OQ)^2=(QN)^2+(NO)^2\\\\(OQ)^2=(y)^2+(2x)^2\\\\(OQ)^2=y^2+4x^2$....(3)

On adding equation (2) and (3) , we get

$(MP)^2+(OQ)^2=4y^2+x^2+4x^2+y^2\\\\ (MP)^2+(OQ)^2=5x^2+5y^2\\(MP)^2+(OQ)^2=5(x^2+y^2)$

Putting the value from equation (1) in above equation , we get

$(MP)^2+(OQ)^2=5(100)\\(MP)^2+(OQ)^2=500cm$

Therefore ,  the sum of squares of the medians MP and OQ  is 500 cm.

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