Cut a cuboid 16 cm long, 8 cm wide and 10 cm high into two cuboids of exactly the same shape and size. What is the maximum and minimum surface area of the cuboid?

Cut a cuboid 16 cm long, 8 cm wide and 10 cm high into two cuboids of exactly the same shape and size. What is the maximum and minimum surface area of the cuboid?


The maximum cross section is 16x10, the height is 4 and the surface area is (16x10 + 16x4 + 10x4) x2 = 528
The minimum cross section is 8x10, the height is 8, and the minimum surface area is (8x10 + 8x8 + 8x10) x2 = 448



A rectangular food box is 8 cm long, 6 cm wide and 10 cm high. If a circle of brand paper is pasted around the food box, how many square meters is the area of the brand paper at least?


(8 × 10 + 6 × 10) × 2 = (80 + 60) × 2 = 140 × 2 = 280 (square centimeter); 280 square centimeter = 0.028 square meter



The length of a cuboid is 10 cm, the width is 8 cm, and the height is 5 cm. After cutting it into two identical cuboids, the surface area will increase at least () square cm


The length of a cuboid is 10 cm, the width is 8 cm, and the height is 5 cm. After cutting it into two identical cuboids, the surface area will increase at least (80) square cm



As shown in the figure, in square ABCD, CE ⊥ Mn, if ∠ MCE = 35 °, then the degree of ∠ anm is______ .


If NP ⊥ BC is made to P through N, then NP = DC, ∫ MCE + ∠ NMC = 90 °, MNP + ∠ NMC = 90 ° and ∪ MCE = ∠ MNP, ∪ MNP = ∠ mcenp = CB ∠ NPM = ∠ CBE, ≌ BEC ≌ PMN, ∪ MCE = ∠ PNM and ∪ anm = 90 ° - MCE = 55 ° in △ MnP and △ ECB



6 / 7: simplify ratio and calculate ratio


It seems that I have to say something theoretical first: reduction ratio, as the name suggests, is to reduce a ratio to the simplest form, that is to say, the numbers on both sides of the comparison sign (colon) cannot be reduced, and the numbers on both sides are integers. Multiply two numbers by a number or divide them by a number at the same time, and the ratio will not change



One operation of quadratic radical
Sum of radical n (n + 1) (n + 2) (n + 3) + 1


N(N+1)(N+2)(N+3)+1
=[N(N+3)][(N+1)(N+2)]+1
=(N^2+3N)[(N^2+3N)+2]+1
=(N^2+3N)^2+2(N^2+3N)+1
=(N^2+3N+1)^2
So the root n (n + 1) (n + 2) (n + 3) + 1
=N^2+3N+1



The area of the circle in the figure is exactly 12 times that of the rectangle ABCD. The length of AB is 6.28cm, and the area of the shadow is______ Square centimeter


3.14 × (6.28 △ 2) 2 × 1.5 = 3.14 × 9.8596 × 1.5 = 30.959144 × 1.5 ≈ 46.44 (square centimeter); answer: the area of shadow part is 46 square centimeter. So the answer is: 46.44



Given that M & # 178; - 2m + 1 and | n-2 | are mutually opposite numbers, it is necessary to find the value of (m-n)


That is (m-1) &# 178; + | n-2 | = 0
So M-1 = n-2 = 0
m=1,n=2
So M-N = - 1



The first volume of the third grade is 20 ways of mathematical calculation, asking for help


1) 5.7-1.8= 2) 99×84= 3) 1.02×4= 4) 7.34-4= 5) 0.45÷0.15= 6) 800×0.03= 7) 3.27-1.27+4.9= 8) 11×25= 9) 0.14×50= 10) 0.81÷0.27= 11) 0.24×0.5= 12) 18.2+1.8= 13) 1.5×3= 14) 0.16×30= 15) 5.38+0.4...



If the length and width of a rectangle are increased by 3 cm, the area will be increased by 54 square cm. The perimeter of the original rectangle is______ Decimeter


As shown in the figure, (54-3 × 3) △ 3 × 2 = (54-9) △ 3 × 2 = 45 △ 3 × 2 = 30 (CM) = 3 (decimeter)