Question

Висота трапеції в 2 рази більша за середню лінію. Знайдіть основи трапеції , якщо їх довжини відносяться як 3:5, а площа трапеції дорівнює 128 см квадратних

Answer 1

Середня лінія дорівнює сумі основ,поділених на 2.тобто(3х+5х)/2,отже висота дорівнює 8х.Площа     S  =  (a + b) /2*h=(3x+5x)/2*8x=4x*8x=32x:  x=128/32=4. Основи 3*4=12см і 5*4=20 сс

Answer 2

Problem Analysis

We are given that the height of a trapezoid is twice the length of the average line. We need to find the lengths of the bases of the trapezoid, given that their lengths are in a ratio of 3:5 and the area of the trapezoid is 128 square cm.

Solution

Let's assume the lengths of the bases of the trapezoid are 3x and 5x, where x is a common factor.

The formula for the area of a trapezoid is (1/2) * (sum of the lengths of the bases) * (height).

We are given that the height is twice the length of the average line, so the height is 2 * (3x + 5x) = 16x.

We can substitute the values into the formula to find the area of the trapezoid:

(1/2) * (3x + 5x) * 16x = 128

Simplifying the equation:

(8x) * (8x) = 128

64x^2 = 128

Dividing both sides by 64:

x^2 = 2

Taking the square root of both sides:

x = √2

Now we can find the lengths of the bases:

The length of the smaller base is 3x = 3 * √2.

The length of the larger base is 5x = 5 * √2.

Therefore, the lengths of the bases of the trapezoid are 3 * √2 and 5 * √2.

Answer

The lengths of the bases of the trapezoid are 3 * √2 and 5 * √2.