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The area of a trapezoid is 60 square centimeters, the bottom is 10 centimeters, and the height is 8 centimeters. How many centimeters is the top of the trapezoid
The area of trapezoid = (upper bottom + lower bottom) * height / 2. The solution of the equation assumes that the upper bottom of trapezoid is xcm (x + 10) * 8 / 2 = 60 (x + 10) * 4 = 60, x + 10 = 60 / 4, x + 10 = 15
For a right angle trapezoid, the length of the upper bottom is 4 / 7 of that of the lower bottom. If the upper bottom is increased by 7 meters and the lower bottom is increased by 1 meter, the trapezoid becomes a square. What is the area of the trapezoid?
4x+7=7x+1
4x+6=7x
3x=6
x=2
4*2=8m
7*2=14m
4*2+7=15m
(8+14)*15/2=165
The top and bottom of a right angled trapezoid are 5 decimeters long. If its top and bottom are increased by 3 decimeters, the area will be increased by 9 square meters, and it will just become a rectangle. How many square meters is the area of the original trapezoid?
Hello
Height: 9 × 2 △ 3 = 6 decimeters
Original area: (5 + 5 + 3) × 6 △ 2 = 39 square decimeters
A right angle trapezoid has a bottom of 40 meters. If the upper bottom is increased by 32 meters, it will become a square. How many square meters is the original trapezoid area?
Top and bottom = 40-32 = 8M
Original area = (40 + 8) × 40 △ 2 = 960 square meters
As shown in the figure, in the trapezoidal ABCD, we know ad ‖ BC, BC = BD, ad = AB = 4cm, ∠ a = 120 ° to find the area of the trapezoidal ABCD
As shown in the figure, make AE ⊥ BC in E, DF ⊥ BC in F, ∧ AE ∥ DF and ∧ ad ∥ BC, and ∧ a = 120 °, ∧ ABC = 60 °, AE = DF, ∧ AB = ad = 4, ∧ abd = ∧ ADB = ∧ DBC = 30 ° in RT △ Abe, AE = ab · cos30 ° = 4 × 32 = 23; in RT △ BDF, BD = 2DF = 2ae = 43 ∧ BC = BD = 43 ∧ s trapezoidal ABCD = 12 (AD + BC) · AE = (12 + 43) cm2
If EF = 10, high ah = 6, then AD + BC =? Area?
AD + BC = 20, area 60
As shown in the right figure, in the trapezoidal ABCD, ad: BC = 2:3, the area of the triangle AOB is 6 square centimeters, what is the area of the trapezoidal ABCD
Isosceles trapezoid Oh, the middle is the letter o
Make a straight line perpendicular to BC through point O, intersect ad, BC to e, F, OE, EF respectively, which is the height of △ AOD and △ abd. Because △ AOD ∽ BOC, ∽ OE: of = 2:3, OE: EF = 2:5 ∽ s ∽ AOD:S The results show that △ abd = 2:5 ∵ s △ AOB = 6, s △ abd = s △ AOD + s △ AOB ∵ s △ AOD: (s △ AOD + 6) = 2:5
As shown in the figure, square ABCD and isosceles right triangle are on the same straight line. Now the triangle does not move, and the square moves to the right at a speed of 2 cm per second along the straight line at a constant speed. What is the overlapping area of square and triangle in the following periods? (1) In the 10th second, the overlap area is______ (2) in the 11th second, the area of the overlap is______ (3) in the 13th second, the area of the overlap is______ (4) in the 16th second, the area of the overlap is______ (5) in the 18th second, the area of the overlap is______ cm2．
(1) (2) 6 × 6 / 2 = 18 (cm2); (3) 6 × 6-4 × 4 / 2 = 36-8 = 28 (cm2); (4) 6 × 6-2 × 2 / 2 = 36-2 = 34 (cm2); (4) at the 18th second, the overlap area is half of the square: 6 × 6 / 2 = 36 / 2 = 18 (cm2). At the 10th second, the overlap area is (8) square centimeter; at the 11th second, the overlap area is (18) square centimeter; (3) at the 10th second, the overlap area is (8) square centimeter ）In the 13th second, the overlap area is (28) square centimeter; (4) in the 16th second, the overlap area is (34) square centimeter; (5) in the 18th second, the overlap area is (18) square centimeter
The length of the square ABCD is 10 cm, the length of the hypotenuse EF of the isosceles right triangle EFG is 16 cm, and the interval between the points B.E. of AB EF on a straight line is 30 cm
The square ABCD moves 5 cm to the right along the square ABCD and 3 cm to the left along the straight line. When moving for 5 seconds, what is the local area of the square stacked with the isosceles right triangle?
Square ABCD moves 5cm / s to the right along a straight line, and isosceles right triangle EFG moves 3cm / s to the left along a straight line,
Taking e as the reference point, the velocity of point B relative to point E is v = V1 + V2 = 5 + 3 = 8 cm / s
Relative displacement BB '= VT = 8 × 5 = 40 cm
EB '= be-bb' = 30-40 = 10 cm
Let the intersection of b'c 'and GF be m, then the stacking part is egmb'
GE=GF=EF/√2=16/√2
MB'=B'F=EF-EB'=16-10=6
Stacking area = s △ gef-s △ mb'f = 1 / 2 * 16 / √ 2 * 16 / √ 2 - 1 / 2 * 6 * 6 = 64-18 = 46 square cm
, cut off four equal isosceles right triangles on the four corners of a square piece of paper ABCD with a side length of 2cm to get an octagon and find its length
Get a regular octagon efghijkl, find the length of the octagon
x=√2*(2-x)/2;
x=2√2-2;
It grows to 2 √ 2-2 cm
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