find the area of the largest posible isosceles triangle with 2 sides equal to 6. thanks
h-hight , a-thirth side of triangle
al-angle ALPHA , A-area
sin(al)=(h/6) , h=6*sin(al)
cos(al)=(a/2)/6 =( a/12) , a=12*cos(al)
A=(1/2)a*h=(1/2)*6*sin(al)*12*cos(al)
=(1/2)*72*sin(al)*cos(al) =36*sin(al)*cos(al) =18*(2*sin(al)*cos(al))=18*sin(2al)
=dA/dal)=18*2*cos(2al=0 , cos(2al)=0
2al=90° al=45°
h=6*sin(45�‹)=6*(1/�ã2)=6/�ã2
a=12*12*cos(45)=12*(1/�ã2)=12/�ã2
A=(1/2)a*h=(1/2)6/�ã2*12/�ã2
=(1/2)*(1/�ã2)/*1/�ã2)*72=(1/2)*(1/)*72
=(1/4)*72= 18
Largest posible area= 18
�ã2 is square root
Area=(1/2)*(1/2)*72=18
thanks
To find the area of the largest possible isosceles triangle with two sides equal to 6, we need to determine the length of the base.
In an isosceles triangle, the two equal sides are also the legs, and the third side is the base. Since you have two sides equal to 6, let's call these two sides 6 units.
Let's assume that the length of the base is denoted by 'b'. Since it is an isosceles triangle, the base is the third side and is shorter than the two equal sides.
To find the largest possible base length, we can make an equilateral triangle by making the two equal sides coincide with the base. In an equilateral triangle, all three sides are equal.
Now, let's consider an equilateral triangle with side length 6.
Since it is an equilateral triangle, all angles are 60 degrees. We can divide the triangle into two congruent right-angled triangles, each with a leg of length 3 units (one-half of the base of the equilateral triangle) and a hypotenuse of length 6 units (the equal sides of the equilateral triangle).
Using Pythagoras' theorem, we can find the height of each right-angled triangle:
h^2 = 6^2 - 3^2
h^2 = 36 - 9
h^2 = 27
h ≈ √27 ≈ 5.196 units (rounded to three decimal places)
Since the height of the right-angled triangle is equal to half the height of the isosceles triangle, the height of the isosceles triangle is approximately 2 * 5.196 ≈ 10.392 units.
Now we can find the area of the isosceles triangle:
Area = (base * height) / 2
Area = (b * 10.392) / 2
Area = 5.196 * b
As we want to find the maximum area, we need to maximize the value of 'b'. In this case, the largest possible value occurs when 'b' is equal to the length of the equilateral triangle's side. Therefore,
b = 6 units
Substituting the value of 'b' into the area formula:
Area = 5.196 * 6
Area ≈ 31.176 square units
So, the largest possible area for the isosceles triangle with two sides equal to 6 units is approximately 31.176 square units.