Question

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Q. ABCD is a parallelogram, M is the midpoint of DC. If AP = 65 and PM = 30, then the largest possible integral value of AB is:

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Answer 1

Concept:-Value of AB<AP+PB as sum of sides of two triangle is always greater than third side.

Solution:-Use the concept of similarty of triangle

angleAPB=angleMPC (vertically opposite)

angle PAB=anglePCD (alternate interior Angle)

so, ∆APB~∆MPC (AA criteria)

using property of similarty we can write

AP/PC=PB/MP=AB/MC....i)

it is said that

MC=DC/2

so,

MC=AB/2=>2MC=AB

put this value in i)

PB/MP=2MC/MC

PB=2MP

PB=30×2=60

Now as we get AP=60,,,PB=65(given)

so,

AB<AP+PB

AB<60+65

AB<125

The maximum value of AP can be 124.9...

but as it is said only integral value and we know

Maximum integral value less than 125 is 124

so, maximum value of AB=124cm

Answer 2

Answer:

△ABP∼△CMP

AB/MC=BD/MP=AD/CP=2/1

BP=2×Mp=2×30=60

In a triangle, sum of two sides is always greater than the third side

∴AB<AP+BP=65+60=125

So, the largest integral value of of AB is 124.

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