What is the area of the largest triangle that can be fitted into a rectangle of length 10 m and width 9 m ?

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Let OA=10cm along X axis and OB be 9 cm along Y axis. Let OACB be the rectangle.

The triangle fitted in OACB has to have two of its vertices on any two adjacent sides, say at X On OA and Y on OB and the third vertex on any of the other 2 sides. But if you make the other vertex of triangle coincide with C, then the triangle XYC fitted in the rectangle OACB will be maximum for a given position of X on OA and Y on OB.

Now the area of this triangle XYC is (by coordinate geometry) = (1/2){OA*OY+OB*OX- OX*OY}

=(1/2){l*y+bx-xy}, where l = OA and b = OB, OX =x and OY =y

Therefore area of XYC will be maximum when partial derivatives wrt x and Y are zero and d2/dxdy of area A(x,y) of triangle XYC is -ve. So,

d/dx A(x,y) = 0= d/dy A(x,y) and for these x and y ,d2/dxdy A(x,y) is -ve.

d/dx A(x,y) = 0 gives b-y = 0 and d/dy A = 0 gives l-x = 0

So x =l and y=b and

Obviously (d2/dxdy A) at x=l, y=b is -1.

Therefore, the maximum area of the triangle XYC = 1/2{lb+lb-lb} = (1/2)lb

=(1/2)10*9 = 45 sq units Or half the area of the rectangle.

The last answer was very accurate...but simplify it by using the concept of triangles. A rectangle is nothing but two identical triangles put together...

Find the area of the rectangle and then cut it in half...

10(9) = 90 that is the area of the rectangle. Now divide by 2.

90 /2 is 45. Your answer is 45.

The formula for the area of a rect. is A = LW and for a triangle it is 1/2 times LW = A

Good luck...

simple

lets call the triangle points A B C

so on a triangle the sum of two sides are greater the the third so.

A+B>C

A+C>B

B+C>A

so 10m+9m=19m

but 10m-9m=1m

so the last side will be 18m-2m but I think you would like 18m.

hope this helps ;P

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