If the perimeter of a right triangle is 48 cm and the sum of the sides of two right angles is 28 cm, the area of the triangle can be calculated?

If the perimeter of a right triangle is 48 cm and the sum of the sides of two right angles is 28 cm, the area of the triangle can be calculated?

Upstairs, why are you so troublesome
In fact, this problem can be easily calculated without solving the problem of a and B. here is my method:
Let two right angles be a and B respectively, and the hypotenuse be c
From the meaning of the title, a + B = 28, a + B + C = 48, we can get: C = 20
Here's the essence. Take a closer look~
Because this triangle is a right triangle, so a ^ 2 + B ^ 2 = C ^ 2, and the area of the triangle should be 1 / 2Ab
(a+b)^2-c^2=a^2+b^2-c^2+2ab=2ab=28^2-20^2=384
So AB = 192, so the area of the triangle is 1 / 2Ab = 96cm ^ 2
The advantage of this method is that there is no need to solve the equation,

A rectangle is (2a + 3) cm long and (2a-3) cm wide. What is the area of the rectangle

Area is length times width, so
s=（2a+3)*(2a-3)
In this case, the square difference formula is used to get the result
2A * 2a-3 * 3 is 4A * 2-9
(I can't express this square by computer. I should change it to understand it.)

The length to width ratio of the rectangle is 7 to 5, and the area of the new rectangle is 112 CM & sup2 larger than that of the original rectangle

Suppose that the length of the original rectangle is 7x and the width is 5x
4·（5X）+4·（7X+4） =112
The solution is x = 2
So: 7x = 14, 5x = 10
Therefore, the area of the original rectangle is 14x10 = 140

A 40 cm long iron wire is used to form a rectangle. If the length of the rectangle is 4 cm less than 3 times of the width, the area of the rectangle is - CM & # 178;

If the width is x, the length is 3x-4, and the wire length is 40cm, that is, the perimeter is 40cm
2×（x+3x-4）=40
When x = 6, the width is 6cm and the length is 14cm
The area is 6 × 14 = 84 square centimeters