Given a semicircle that has a diameter XY and the point A is any point on the arc XY. The point A can move but it is required that XA+XY=25. How would one find the maximum area of the triangle XAY? Thanks!!!
-
You don't really need to apply differentiation here. Have the following concepts in mind before you attempt this problem
The area of the triangle = (1/2) base height
Now, take the base of the triangle as the diameter XY and the height as a perpendiular AO from the point A. The area of the triangle will be maximum when the height will be maximum
By symmetry, the height will be maximum when arc XA = arc AY
Remember that since angle in a semi circle is 90 degrees, in such a case , by congruency of triangles XAO and YAO , angle XAO = angle YAO = 45 deg
take the radius of the traingle as r .
=> r = x sin 45 ........ (1) from triang XAO
& r = (25-x) sin 45 ...... (2) from triang AOY
divide both the equations, and u get x = 25/2 and r = 25/(2 root 2)
The area of the triangle in such a case is (1/2 ) .(2 r) . r ( as AO perpendicular to XY also forms the radius )
=> area = r^2
= 625/8
= 78.1 square units
The area of the triangle = (1/2) base height
Now, take the base of the triangle as the diameter XY and the height as a perpendiular AO from the point A. The area of the triangle will be maximum when the height will be maximum
By symmetry, the height will be maximum when arc XA = arc AY
Remember that since angle in a semi circle is 90 degrees, in such a case , by congruency of triangles XAO and YAO , angle XAO = angle YAO = 45 deg
take the radius of the traingle as r .
=> r = x sin 45 ........ (1) from triang XAO
& r = (25-x) sin 45 ...... (2) from triang AOY
divide both the equations, and u get x = 25/2 and r = 25/(2 root 2)
The area of the triangle in such a case is (1/2 ) .(2 r) . r ( as AO perpendicular to XY also forms the radius )
=> area = r^2
= 625/8
= 78.1 square units
-
In this such problem the angle A (XAY) play big roll
angle A=90 this is proven
Now let see what we got here
XA is related to the circle diameter XY that put a condition in the circle size, meaning that not any diameter for the circle will give such relation XA+XY=25.
XA+XY=25
Find the side XA
XA=25-XY
Let start with the area of the right angled triangle XAY
Triangle Area= 1/2 base * height
Area triangle XAY = 1/2 XA * YA notice I can use one of the sides as height
Area triangle XAY = 1/2 (25-XY) * YA
we need now to find YA
since
XY² = XA² + YA²
then YA² = XY² - XA²
YA =sqrt( XY² - XA²)
Let put that in the area formula
Area triangle XAY = 1/2 (25-XY) * sqrt( XY² - XA²)
well we want this formula only have one variable XY (because it relate the triangle area to the diameter of the circle)
since XA=25-XY let plug it in the area formula to become
Area triangle XAY = 1/2 (25-XY) * sqrt( XY² - (25-XY)²)
now we got the triangle area formula in terms of circle diameter
to find the maximum area we need to do the derivative
Let make XY=x
A = 1/2 (25-x) * sqrt( x² - (25-x)²)
d/dx(1/2 (25-x) sqrt(x^2-(25-x)^2)) = -(5 (3 x-50))/(2 sqrt(2 x-25))
let it equal zero to get x
-(5 (3 x-50))/(2 sqrt(2 x-25)) =0
x=50/3=16.67
Diameter =XY = x= 16.67
so the area
Area triangle XAY = 1/2 (25-16.67) * sqrt( 16.67² - (25-16.67)²) = 60.14
angle A=90 this is proven
Now let see what we got here
XA is related to the circle diameter XY that put a condition in the circle size, meaning that not any diameter for the circle will give such relation XA+XY=25.
XA+XY=25
Find the side XA
XA=25-XY
Let start with the area of the right angled triangle XAY
Triangle Area= 1/2 base * height
Area triangle XAY = 1/2 XA * YA notice I can use one of the sides as height
Area triangle XAY = 1/2 (25-XY) * YA
we need now to find YA
since
XY² = XA² + YA²
then YA² = XY² - XA²
YA =sqrt( XY² - XA²)
Let put that in the area formula
Area triangle XAY = 1/2 (25-XY) * sqrt( XY² - XA²)
well we want this formula only have one variable XY (because it relate the triangle area to the diameter of the circle)
since XA=25-XY let plug it in the area formula to become
Area triangle XAY = 1/2 (25-XY) * sqrt( XY² - (25-XY)²)
now we got the triangle area formula in terms of circle diameter
to find the maximum area we need to do the derivative
Let make XY=x
A = 1/2 (25-x) * sqrt( x² - (25-x)²)
d/dx(1/2 (25-x) sqrt(x^2-(25-x)^2)) = -(5 (3 x-50))/(2 sqrt(2 x-25))
let it equal zero to get x
-(5 (3 x-50))/(2 sqrt(2 x-25)) =0
x=50/3=16.67
Diameter =XY = x= 16.67
so the area
Area triangle XAY = 1/2 (25-16.67) * sqrt( 16.67² - (25-16.67)²) = 60.14