A rectangle is inscribed in an isosceles triangle. If the sides of the triangle are 10, 10 and 12, what are the dimensions of the rectangle having the greatest area? The diagram shows the triangle on a set of axes with base 12 on the x axis and the top vertices on the y axis.
To find the dimensions of the rectangle inscribed in the isosceles triangle with sides 10, 10, and 12, we need to maximize its area.
First, let's draw the diagram to get a better understanding of the problem. The base of the triangle is on the x-axis, and the top vertices are on the y-axis.
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^ y-axis
|
|
| /\
| / \
| / \
| /______\
|_________\
x-axis
```
Let's label the triangle vertices: A, B, and C. A is the vertex on the y-axis, B is the vertex on the x-axis, and C is the vertex opposite to the base.
Now, let's find the dimensions of the rectangle.
Step 1: Find the height of the triangle (AC):
Since it is an isosceles triangle, the height is also the median, and it bisects the base.
Let's label the midpoint of BC as D.
Since AD is a median, BD = DC = half of the base = 12/2 = 6.
Step 2: Find the height of the triangle using the Pythagorean theorem:
The length of BC is 10, and half of it (BD) is 6.
Using the Pythagorean theorem, we can find the height (AC):
AC^2 = BC^2 - BD^2
AC^2 = 10^2 - 6^2
AC^2 = 100 - 36
AC^2 = 64
AC = √64
AC = 8
Step 3: Find the dimensions of the rectangle:
Since the rectangle is inscribed in the triangle, its width is equal to AC.
The length of the rectangle is equal to half of the base, which is 12/2 = 6.
Therefore, the dimensions of the rectangle with the greatest area inscribed in the isosceles triangle are:
Width: 8
Length: 6
To find the dimensions of the rectangle with the greatest area inscribed in the given isosceles triangle, we can use the following steps:
1. Start by drawing the isosceles triangle on a set of axes according to the given information. The base of the triangle is 12 units on the x-axis, and the top vertices are on the y-axis.
2. Label the two congruent sides of the isosceles triangle as "10" units each. This will give us an idea of the maximum height of the rectangle that can be inscribed in the triangle.
3. Now, consider that the rectangle inscribed in the triangle will have its diagonal aligned with the base of the triangle. This means the longer side of the rectangle will have a length equal to the base of the triangle.
4. In this case, the longer side of the rectangle will be 12 units, which is the same as the base of the triangle. This will ensure that the diagonal of the rectangle lies along the base of the triangle.
5. Now, we have to find the height of the rectangle. Since the rectangle is inscribed in the triangle, its height will be equal to the height of the triangle at that particular point on the x-axis.
6. To find the height of the triangle at that point, we can use similar triangles. The given isosceles triangle is symmetrical, so we can draw a line from the top vertex perpendicular to the base.
7. This line will divide the isosceles triangle into two congruent right-angled triangles. Therefore, the height of the triangle is equal to the hypotenuse of one of these right-angled triangles.
8. Applying the Pythagorean theorem, we can find the height of the triangle. Let's call it "h". Considering one of the right-angled triangles with a base of 6 units (half of the base of the isosceles triangle), the height "h" can be found using the equation:
h^2 + 6^2 = 10^2
Solving this equation, we get:
h^2 + 36 = 100
h^2 = 100 - 36
h^2 = 64
h = √64
h = 8
9. Therefore, the height of the rectangle (which is also the height of the triangle at that point) is 8 units.
10. Now, we have the dimensions of the rectangle with the greatest area inscribed in the given isosceles triangle. The longer side of the rectangle is equal to the base of the triangle, which is 12 units. The height of the rectangle is equal to the height of the triangle at that point, which is 8 units.
Hence, the dimensions of the rectangle with the greatest area are 12 units (base) and 8 units (height).