a rectangle is to be inscribed in a isosceles triangle of height 8 and base 10. Find the greatest area of such rectangle.
Did you make a sketch?
Draw triangle ABC so that AB = AC
Draw the height AD, D on BC
Let P be the point where the rectangle touches AC
Draw PE perpendicular to BC, E on BC
let DE = x, then EC = 5-x
let PE = y
by similar triangles:
y/(5-x) = 8/5
y = (40-8x)/5
Area of rectangle = 2xy
= 2x(40-8x)/5
= (80x - 16x^2)/5
d(Area)/dx = (80 - 32x)/5 = 0 for max of area
32x = 80
x = 80/32 = 5/2
y = (40 - 8(2.5))/5 = 4
greatest area = 2xy = 2(5/2)(4) = 20
Well, if a rectangle is inscribed in an isosceles triangle, it can be quite the "tri"-cky situation! But fear not, I'm here to help with a little humor.
Let's start by visualizing the problem. We have an isosceles triangle with a height of 8 and a base of 10. To maximize the area of the inscribed rectangle, we want to make it as big as possible without exceeding the boundaries of the triangle.
Now, let's try to fit that rectangle in there. One clever way to maximize its area is to place its base along the base of the triangle. This way, the height of the rectangle will coincide with the height of the triangle, which is 8.
So, we know that the base of the rectangle is 10 (same as the base of the triangle) and the height is 8 (same as the height of the triangle). Multiplying these two dimensions together gives us an area of 10 times 8, which is... drumroll please... 80!
Therefore, the greatest area for the inscribed rectangle in this isosceles triangle is 80 square units.
I hope this "tri"-cky little explanation brings a smile to your face!
To find the greatest area of a rectangle inscribed in an isosceles triangle, we need to consider that the rectangle's base will be along the base of the triangle, and its height will be along the height of the triangle.
Let's denote the base of the rectangle as "b" and the height as "h."
Given that the isosceles triangle has a base of 10 and a height of 8, we can determine the relationship between the base, height, and hypotenuse of the triangle using the Pythagorean theorem.
The hypotenuse of the triangle can be calculated as follows:
hypotenuse = sqrt((base/2)^2 + height^2)
= sqrt((10/2)^2 + 8^2)
= sqrt(25 + 64)
= sqrt(89)
Since the base of the rectangle will be along the base of the triangle, the length of the base of the rectangle (b) will be equal to the base of the triangle (10).
The height of the rectangle (h) will be equal to the height of the triangle (8).
Therefore, the area of the rectangle can be calculated as follows:
Area = base * height
= 10 * 8
= 80
Thus, the greatest area of a rectangle inscribed in the given isosceles triangle is 80 square units.
To find the greatest area of a rectangle inscribed in an isosceles triangle, we need to consider the dimensions and proportions.
Let's draw the isosceles triangle and inscribe a rectangle inside it.
First, draw the isosceles triangle with a height of 8 units and a base of 10 units. Label the top vertex as A, and the two bottom vertices are B and C.
Now, draw a line segment from vertex A perpendicular to the base BC. Let's label the point where the line segment intersects with the base as D.
We can see that the line segment AD splits the base BC into two equal parts, so BD = DC = 5 units.
Now, let's consider the rectangle that is inscribed in the triangle. Label the vertices of the rectangle as E, F, G, and H, starting from the top left corner and going clockwise.
By looking at the diagram, we can see that the sides of the rectangle EF and GH are parallel to the base BC of the triangle. So, the length of the rectangle will be equal to the length of the base, which is 10 units.
The sides FG and EH of the rectangle are parallel to the height of the triangle, so the width of the rectangle will be equal to the height, which is 8 units.
Now, let's calculate the area of this rectangle. The area of a rectangle is given by the formula: Area = length × width.
In this case, the length of the rectangle is 10 units and the width is 8 units.
So, the area of the rectangle is 10 × 8 = 80 square units.
Therefore, the greatest area of a rectangle inscribed in this isosceles triangle is 80 square units.