The perimeter of a triangle is 8 cm. One of the sides is 3 cm. Find the other two sides such that
the triangle has maximum area.
Answers
Step-by-step explanation:
The perimeter of a triangle is 8cm and one of its side is 3cm. The area of the triangle is maximum . Then the other two sides are?
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My answer requires knowledge of conic sections. These are the properties I use
a) in an ellipse, the sum of the distances of a point P on the curves to the foci is a constant
b) The length of perpendicular from P to major axis is max when the perpendicular is minor axis
c) Let is take an ellipse with foci be S and S’, and SS’= 3 cm. Extend SS’ either side to intersect the ellipse at A and A’ (AA’ is major axis =2a). Let C be center of the ellipse and BC be semi-minor axis (b)
d) As the point P moves from A to B along the ellipse, the length of perpendicular from P to AA’ increases reaching maximum at B (the length of the perpendicular here = max = b). Then it decreases as P moves from B to A’
Now area of triangle SS’P is = (1/2)(SS’)*h = 3h/2, where h is height of perpendicular from P to AA’. h reaches a max value of b. So the area = 3b/2
Now we have to find b. That is simple. Triangle BSS’ is isosceles (BS=BS’). Since its perimeter = 8 and SS’=3, we have BS+BS’= 5 => BS=BS’=2.5
So in right angle triangle BSC (rt. angle at C) we have BS = 2.5, SC = 3/2 = 1.5 => BC =2 (apply pythagoras).
So max area of the required triangle = (1/2)(3)(2) = 3 sq units
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Let other two sides are x cm and y cm.
z=3 cm , x+y+z= 8 cm. , x+y = 5 cm. or y=5-x. , s=8/2=4 cm.
Area (A) of the triangle=√{s.(s-x).(s-y).(s-z)}
A=√{4.(4-x).(4–5+x).(4–3)}
A^2 =4.(4-x).(x-1).1 = 4(-4+5x-x^2)
or. p= -16+20x-4x^2
dp/dx=20–8x. , for maximum or minimum putting dp/dx=0
0=20–8x
or. x=20/8=5/2=2.5
d2p/dx^2=-8 (negative ), there exist maximum at x=2.5
But y=5-x =5–2.5 =2.5 cm.
Thus ,other two sides are 2.5 cm. long each. Answer