After two cubes are put together into a cuboid, the surface area is reduced by 162 square centimeters, and the surface area and volume of the cuboid are calculated

Put two cubes together to form a cuboid. If the area of each square is 162 / 2 = 81 (square centimeter), the side length of the square is 9 cm. At the same time, the surface area of the original two cubes is 81 * 6 + 81 * 6 = 972 (square centimeter), and the surface product of the cuboid is 972-162 = 810 (square centimeter)

Five identical cubes form a cuboid. The surface area of the cuboid is 198 square centimeters. What are the surface area and volume of each cube?

Five identical cubes form a cuboid. The surface area of the cuboid is the area of 22 squares,
198 divided by 22 = 9 square centimeter is the area of a face. So the edge length is 3cm
Surface area: 9x6 = 54 square centimeter
Volume: 3x3x3 = 27 CC

A cuboid of wood, cut from the lower and upper parts of the cuboid 3 cm and 2 cm high, will become a cube. Surface area reduced by 120 square percent
What is the bottom area of the original cuboid,
Isn't it necessary to increase the area

120 △ 3 + 2 △ 4 = 6cm
6 × 6 = 36 square centimeter

The side length of square ABCD is 2, AE = EB, Mn = 1, the two ends of line Mn slide on CB and CD, and the triangle AED is similar to the triangle MNC

Because ⊿ AED is similar to ⊿ MNC, AD / NC = AE / MC = de / Mn or ad / MC = AE / NC = de / Mn (1) if AD / NC = AE / MC = de / Mn, then because the side length of square ABCD is 2, AE = EB, so de = √ 5, so 1 / MC = √

When cm = (), the △ AED is similar to the triangle whose vertices are n, m and C

Let cm be X. according to Pythagorean theorem, in triangle ade, ed = √ 5. Then ed / Mn = ad / cm = √ 5 / 1 = 2 / x, the solution is x = 2 √ 5 / 5. Or AE / cm = ed / Mn, bring in the number, and get x = √ 5 / 5. When cm = 2 * √ 5 / 5 or cm = √ 5 / 5, △ AED is similar to the triangle with m, N, C as vertex

As shown in the figure, e is a point on the edge BC of rectangle ABCD, EF ⊥ AE, EF intersects AC respectively, CD intersects at points m, F, BG ⊥ AC, perpendicular foot is g, BG intersects AE at point h. (1) prove: △ Abe ⊥ ECF; (2) find a triangle similar to △ ABH, and prove it; (3) if e is the midpoint of BC, BC = 2Ab, ab = 2, find the length of EM

(1) It is proved that: ∵ quadrilateral ABCD is a rectangle, ∵ Abe = ∠ ECF = 90 °. ∵ AE ⊥ EF, ∵ AEB + ∠ FEC = 90 °. ∵ AEB + ∠ BAE = 90 °, ∵ BAE = ∠ CEF, ∽ Abe ∽ ECF; (2) △ ABH ∽ ECM In RT △ EMR, EM = mrsin45 ° is 223

As shown in the figure, in square ABCD, e is the midpoint of BC, AE and BD intersect at F, proving: CF ⊥ De
Such as the title

F is a point on BD,
Square ABCD
∴△ABF△CBF
∴∠BCF = ∠BAF
∵ e is the midpoint of BC
∴DE = AE
∴∠EDA = ∠EAD
∴∠EAB = ∠EDC
∴∠BCF = ∠EDC
∵∠DCF + ∠FCB = 90°
∴∠EDC + ∠ECF = 90°
∴CF⊥DE

As shown in the figure, the perimeter of rectangle ABCD is 20cm. Take AB and ad as the sides, make square abef and square adgh outwards. If the sum of square abef and adgh area is 68cm2, then the area of rectangle ABCD is ()
A. 21cm2B. 16cm2C. 24cm2D. 9cm2

Let AB = xcm, ad = (10-x) cm, then the area of square abef is x2cm2, the area of square adgh is (10-x) 2cm2. According to the title, we get x2 + (10-x) 2 = 68, we get x2-10x + 16 = 0, and the solution is X1 = 2, X2 = 8, so AB = 2cm, ad = 8cm or AB = 8cm, ad = 2cm

As shown in the figure, the circumference of the rectangle ABCD is 20 cm. Take AB and ad as sides and make square abef and adgh outward. If the sum of square abef and adgh area is 68 square cm, calculate the area of rectangle ABCD

The circumference of rectangle ABCD is 20cm. Take AB and ad as sides, make square abef and square adgh outwards. If the difference between the circumference of square abef and adgh is 24, then what is the area of rectangle ABCD? Use binary linear equation

Let AB = a, ad = B, abef = 4A, adgh = 4B
∴ 4b-4a=24
2a+2b=20
We can get two numbers AB, but there should be two groups of results, because we can't know who is big or small of a or B

After two cubes are put together into a cuboid, the surface area is reduced by 162 square centimeters, and the surface area and volume of the cuboid are calculated