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As shown in the figure, in △ ABC, ab = AC, BD bisect ∠ ABC, and BD = be, ∠ a = 100 °, then ∠ Dec=______ .
∵ in △ ABC, ab = AC, ∵ a = 100 °,
∴∠ABC=∠C=40°，
∵ BD bisection ∵ ABC,
∴∠DBE=1
2∠ABC=20°，
∴∠BDE=∠BED=80°，
∴∠DEC=100°．
So the answer is: 100 degrees
As shown in the figure, in △ ABC, ab = AC, BD bisect ∠ ABC, and BD = be, ∠ a = 100 °, then ∠ Dec=______ .
∵ in △ ABC, ab = AC, ∵ a = 100 °,
∴∠ABC=∠C=40°，
∵ BD bisection ∵ ABC,
∴∠DBE=1
2∠ABC=20°，
∴∠BDE=∠BED=80°，
∴∠DEC=100°．
So the answer is: 100 degrees
As shown in the figure, in △ ABC, ab = AC, BD bisect ∠ ABC, and BD = be, ∠ a = 100 °, then ∠ Dec=______ .
∵ in △ ABC, ab = AC, ∵ a = 100 °,
∴∠ABC=∠C=40°，
∵ BD bisection ∵ ABC,
∴∠DBE=1
2∠ABC=20°，
∴∠BDE=∠BED=80°，
∴∠DEC=100°．
So the answer is: 100 degrees
As shown in the figure, △ ABC is an isosceles right triangle, ∠ a = 90 ° E on BC, ab = AC = 1, ∠ a = 90 °, BDE = 90 °, and ∠ D is the surface of △ Dec at the midpoint of AC Δ ABC is an isosceles right triangle, ∠ a = 90 ° E on BC, ab = AC = 1, ∠ a = 90 °, BDE = 90 °, and ∠ D is the area of △ Dec at the midpoint of AC
If the vertical line of CD is intersected with F according to the crossing point E, then EF = FC. It is easy to know that ∠ EDF = ∠ DBA. There is a triangular EDF which is similar to a triangular DBA. Therefore, EF / ad = DF / AB, and DF = dc-fc = 1 / 2-ef, ad = 1 / 2, ab = 1, substituting into the above formula, we can get
2ef = 1 / 2-ef, EF = 1 / 6. Area s = 1 / 2dc * EF = 1 / 2 * 1 / 2 * 1 / 6 = 1 / 24
In the figure below, the area of the right triangle is 50cm. How many square centimeters is the area of the circle where the triangle is located? The formula is listed
This triangle belongs to a right triangle inscribed in a circle, so its hypotenuse is the diameter of the circle. The three sides of the right triangle are regarded as a, B, C. the height is C / 2
(a * b) / 2 = [(C / 2) * C] / 2 to obtain C, that is, r = C / 2
Area of circle = π r = 3.14 * (C / 2) squared = 157
A triangle sign, with an area of 2000cm2, a bottom length of 50cm and a height of?
2000×2÷50=80cm
The bottom of a triangular iron plate sign is 50cm, and the height is 40cm. What is the area of this signboard? How many pieces of this signboard can be made of 2M2
The bottom of a triangular iron plate sign is 50cm, and the height is 40cm. The area of this signboard is 1000 square centimeter, and there are five such signboards of 2M2
50cm*40cm/2=0.5m*0.4m/2=0.4m2
2m2/0.4m2=5
A cylindrical tin barrel with a diameter of 40cm and a height of 50cm. How many square decimeters of iron sheet should be used to make such a drum? What is the volume of this tin barrel?
Surface area: 3.14 × 40 × 50 + 3.14x20 × 20 × 2 = 8792 square centimeter = 87.92 square decimeter
Volume is volume; 3.14 × 20 × 20 × 50 = 62800 cubic centimeter = 62.8 cubic decimeter
The surface area of the cylinder = the perimeter of the bottom surface × height + the area of the bottom × 2 (it is not necessary to multiply by 2 if there is only one bottom)
Volume of cylinder = base area × height
A cylindrical metal fire bucket without cover, the bottom diameter is 40cm, the height is 50cm The diameter of the bottom surface is 40cm and the height is 50cm. The leftover material of such a fire bucket is exactly 5% of the actual area of the bucket. How much iron sheet is needed to make such a fire bucket?
Bottom diameter = 40, radius = 20,
Bottom area = 3.14x20x20 = 1256,
Bottom perimeter = 3.14x40 = 125.6
Side area = bottom perimeter x height = 125.6x50 = 6280
The iron sheet needed by the iron bucket is to find a bottom area + side area
=1256+6280=7536
The remaining 5% of the actual area is 7536x5% = 376.8
Therefore, to make a fire bucket, iron sheet = 7536 + 376.8 = 7912.80
Answer: make a fire bucket with iron sheet 7912.80 square centimeter
An open cuboid iron bucket with a length of 40cm, a width of 40cm and a height of 50cm. What is the volume of this bucket? How many square meters of iron sheet is needed to make this bucket?
Volume = 40 × 40 × 50 = 80000 cm3 = 80 liters
Iron sheet = 40 × 40 + 40 × 4 × 50 = 9600 square centimeter = 0.96 square meter
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