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Use three square with 13 cm circumference to form a rectangle, and find the circumference of the rectangle
As shown in the figure: 13 × 3 - 13, = 39 - 13, = 26 (CM); answer: the circumference of this rectangle is 26 cm
After sawing a 3 meter wide strip from a 36 square meter rectangular board, the remaining area is 30 square meters?
The width of the rectangle is (36-30) / 3 = 2m
The length of the rectangle is: 36 / 2 = 18 meters
The perimeter of the rectangle is: (18 + 2) * 2 = 40 meters
A rectangle is 19.2 meters in circumference and 7.2 meters in length. How many meters is the rectangle wide?
We have to do it with equations, we have to solve it, and we have equations
Set the width as X meters
2 * (length + width) = perimeter
2*（7.2+X）=19.2
7.2+X=9.6
X=9.6-7.2
X=2.4
It's 2.4 meters wide
Cut a 20cm long iron wire into two sections, and make a square with the length of each section as the perimeter. (1) to make the sum of the two squares equal to 17cm2, what are the lengths of the two sections? (2) Can the sum of the areas of two squares be equal to 12 cm2? If yes, find out the length of two pieces of wire; if not, please explain the reason
(1) Let the side length of one square be xcm, then the side length of the other square be (5-x) cm. According to the equation, we can get x2 + (5-x) 2 = 17, and we can get x2-5x + 4 = 0, (x-4) (x-1) = 0. Solving the equation, we can get X1 = 1, X2 = 4, 1 × 4 = 4cm, 20-4 = 16cm; or 4 × 4 = 16cm, 20-16 = 4cm
Cut a 20 cm long iron wire into two sections and make a square with the length of each section as the perimeter. Then the sum of the two areas is the minimum
emergency
Let the side length of a square be x, s = x ^ 2 + ((20-4x) / 4) ^ 2, so the minimum is 12.5
Cut a 20 cm long iron wire into two sections, and make a square with the length of each section as the perimeter. To make the sum of the two squares equal to 17 square centimeters, what are the lengths of the two sections?
Let the side length of one square be xcm, then the side length of the other square be (5-x) cm. According to the equation, we can get x2 + (5-x) 2 = 17, and we can get x2-5x + 4 = 0, (x-4) (x-1) = 0. Solving the equation, we can get X1 = 1, X2 = 4, 1 × 4 = 4cm, 20-4 = 16cm; or 4 × 4 = 16cm, 20-16 = 4cm
Cut a 20cm long iron wire into two sections, and make a square with the length of each section as the perimeter. (1) to make the sum of the two squares equal to 17cm2, what are the lengths of the two sections? (2) Can the sum of the areas of two squares be equal to 12 cm2? If yes, find out the length of two pieces of wire; if not, please explain the reason
(1) Let the side length of one square be xcm, then the side length of the other square be (5-x) cm. According to the equation, we can get x2 + (5-x) 2 = 17, and we can get x2-5x + 4 = 0, (x-4) (x-1) = 0. Solving the equation, we can get X1 = 1, X2 = 4, 1 × 4 = 4cm, 20-4 = 16cm; or 4 × 4 = 16cm, 20-16 = 4cm
Cut a 20cm long iron wire into two sections, and make a square with the length of each section as the perimeter. (1) to make the sum of the two squares equal to 17cm2, what are the lengths of the two sections? (2) Can the sum of the areas of two squares be equal to 12 cm2? If yes, find out the length of two pieces of wire; if not, please explain the reason
(1) Let the side length of one square be xcm, then the side length of the other square be (5-x) cm. According to the equation, we can get x2 + (5-x) 2 = 17, and we can get: x2-5x + 4 = 0, (x-4) (x-1) = 0. Solving the equation, we can get X1 = 1, X2 = 4, 1 × 4 = 4cm, 20-4 = 16cm; or 4 × 4 = 16cm, 20-16 = 4cm Reason: let the sum of the areas of two squares be y, then y = x2 + (5-x) 2 = 2 (X-52) 2 + 252, ∵ a = 2 ＞ 0, ∵ when x = 52, the minimum value of y = 12.5 ＞ 12, ∵ the sum of the areas of two squares can not be equal to 12cm2; (in addition, from (1) we can know that x2 + (5-x) 2 = 12, after simplification, we get 2x2-10x + 13 = 0, ∵ △ = (- 10) 2-4 × 2 × 13 = - 4 ＜ 0, ∵ the equation has no relation So the sum of the areas of two squares can't be equal to 12cm2
Cut a 20cm long iron wire into two sections, and make a square with the length of each section of iron wire as the perimeter, then the minimum value of the sum of the areas of the two squares is______ cm2．
Let the length of one section of iron wire be x and the length of the other section be (20-x), then the side lengths are 14x and 14 (20-x) respectively, then s = 116x2 + 116 (20-x) (20-x) = 18 (X-10) 2 + 12.5. By the function, when x = 10cm, s is the smallest, which is 12.5cm2
Cut a 20cm long iron wire into two sections, and make a square with the length of each section of iron wire as the perimeter, then the minimum value of the sum of the areas of the two squares is______ cm2．
Let the length of one section of iron wire be x and the length of the other section be (20-x), then the side lengths are 14x and 14 (20-x) respectively, then s = 116x2 + 116 (20-x) (20-x) = 18 (X-10) 2 + 12.5. By the function, when x = 10cm, s is the smallest, which is 12.5cm2
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