The sides of an isosceles triangle are 10 cm, 10 cm, and 12 cm. A rectangle is inscribed in the triangle with one side on the triangle's base. The other two vertices of the rectangle are on the triangle's legs. Find the Maximum Area of the Rectangle.

To find the maximum area of the rectangle inscribed in the given isosceles triangle, we can start by understanding the properties of the triangle and the rectangle.

1. Let's label the isosceles triangle ABC, where AC and BC are the congruent sides and AB is the base.
2. Assume that the rectangle is inscribed such that one side lies on the base AB, and the other two vertices lie on the congruent sides AC and BC.
3. Let the rectangle be PQRS, where PQ and RS are parallel to the base AB, and QR and PS are parallel to the congruent sides AC and BC.

Now, let's proceed with finding the maximum area of the rectangle.

Step 1: Identify the dimensions of the rectangle:
Since one side of the rectangle, PQ, lies on the base AB, its length will be equal to the length of AB, which is 12 cm.

Step 2: Determine the width of the rectangle:
The width of the rectangle can be found by considering the relationship between similar triangles. We can observe that:

- Triangle ABC is similar to triangle SPQ (by the AA similarity criterion) since the angles at points P and Q are congruent to the corresponding angles at points A and B.
- Additionally, triangles ABC and SPQ share a common angle (angle ACB = angle PQS) since the sides of the rectangle are parallel to the respective sides of the triangle.

Therefore, we can set up the following proportion:

AC (side of the triangle) / PQ (side of the rectangle) = BC (side of the triangle) / PS (side of the rectangle)

Using the given values, we have:

10 cm / PQ = 10 cm / PS

Simplifying, we get:

PQ / PS = 1

Since PQ = 12 (as determined in step 1), we can conclude that PS, the width of the rectangle, is also 12 cm.

Step 3: Calculate the maximum area of the rectangle:
The area of the rectangle is given by the formula:

Area = length × width

Substituting the values we found in steps 1 and 2, we get:

Area = 12 cm × 12 cm = 144 cm²

Therefore, the maximum area of the rectangle inscribed in the given isosceles triangle is 144 cm².

To find the maximum area of the rectangle inscribed in the isosceles triangle, we can use the following steps:

Step 1: Draw the isosceles triangle and mark the lengths of its sides. Let's call the two equal sides of the triangle 'a' and the base 'b'. In this case, 'a' is 10 cm and 'b' is 12 cm.

Step 2: Draw a rectangle inside the triangle with one side on the base of the triangle. Let's call the length of the rectangle 'x' and the width 'y'.

Step 3: Since the rectangle is inscribed in the triangle, the length of the rectangle 'x' will be equal to the length of the base 'b'. Therefore, 'x' = 12 cm.

Step 4: Determine the height of the rectangle 'y' using the properties of the isosceles triangle. The height 'y' is perpendicular to the base and divides the triangle into two congruent right triangles. We can use the Pythagorean theorem to find 'y'.

Applying the Pythagorean theorem:
a^2 = b^2 - (y^2/4)
10^2 = 12^2 - (y^2/4)
100 = 144 - (y^2/4)
y^2/4 = 144 - 100
y^2/4 = 44
y^2 = 4 * 44
y^2 = 176
y = √176
y ≈ 13.23 cm

Step 5: Calculate the area of the rectangle using the length 'x' and the width 'y':
Area of Rectangle = Length * Width
Area of Rectangle = x * y
Area of Rectangle = 12 cm * 13.23 cm
Area of Rectangle ≈ 158.76 cm^2

Therefore, the maximum area of the rectangle inscribed in the isosceles triangle with side lengths 10 cm, 10 cm, and 12 cm is approximately 158.76 square cm.