Question

Let M be a point inside the rectangle ABCD. Suppose that MA=3cm, MB=2cmMA=3cm,MB=2cm and MC=1cmMC=1cm. Find the value of MD^2MD 2 .

Answer 1

Solution:

Construction:

Draw a line XY passsing through M and

parallel to BC.

Then MY is perpendicular to CD.

MX is perpendicular to AB

Pythagoras theorem,

In a right angled triangle square on the hypootenuse is equal to sum of the squares of the other two sides.

$In triangle MDY, MD^2=MY^2+YD^2\\In triangle MYC, MC^2= MY^2+YC^2\\In triangle MXA, MA^2= MX^2+XA^2\\In triangle MXB, MB^2=MX^2+XB^2$

$Now,\\MD^2=MY^2+YD^2\\MD^2=(MC^2-YC^2)+AX^2\\MD^2=(1-YC^2)+(MA^2-MX^2)\\MD^2=(1-YC^2)+(9-MX^2)\\MD^2=10-YC^2-MX^2\\MD^2=10-YC^2-(MB^2-XB^2)\\MD^2=10-YC^2-(4-YC^2)\\MD^2=10-YC^2-4+YC^2\\MD^2=6$