Calculate the surface area of the half cylinder in centimetres High: 40 Diameter: 10

Calculate the surface area of the half cylinder in centimetres High: 40 Diameter: 10

Radius: 10 / 2 = 5cm
5 * 5 * 3.14 = 39.25 square centimeter
10 * 3.14 * 40 / 2 = 628 square centimeter
10 * 40 = 400 square centimeter
39.25 + 628 + 400 = 1067.25 square centimeter

In △ ABC, ∠ C = 90 °, AC = 4cm, BC = 5cm, point D is on BC, and CD = 3cm. There are two moving points P and Q starting from point a and point B at the same time, where point P moves along AC to terminal C at the speed of 1cm / s, and point Q moves along BC to terminal C at the speed of 1.25cm/s. Passing point P, PE ‖ BC is connected to point E, and EQ is connected (1) The length of AE and De is expressed by the algebraic expression containing x; (2) when the point Q moves on BD (excluding points B and D), let the area of △ EDQ be y (cm2), find the functional relationship between Y and time x, and write out the value range of the independent variable x; (3) when the value of X is, the △ EDQ is a right angle triangle

(1) In RT △ ADC, AC = 4, CD = 3, ∧ ad = 5, ∧ EP ∥ DC, ∧ AEP ∧ ADC, ∧ EAAD = APAC, that is ea5 = x4, ∧ EA = 54x, de = 5-54x When point Q moves on BD for x seconds, DQ = 2-1.25x, then y = 12 × DQ × CP = 12 (4 − x) (2 − 1.25x) = 58 & nbsp; x2 − 72x + 4 (6) the analytic expression of Y and X is y = 58X2 − 72x + 4, in which the value range of independent variable is 0 ＜ x ＜ 1.6. (3) the discussion is divided into two cases: (1) when ∠ eqd = 90 °, it is obvious that EQ = PC = 4-x, and ∵ EQ ∥ AC, ∥ EDQ ∥ ADC ∥ eqac = dqdc, DQ = 1.25x-2, that is, 4 − X4 = 1.25x − 23 The solution is x = 2.5 (9 points) ② when ∠ QED = 90 degree, ∵ CDA = ∠ EDQ, ∵ QED = ∠ C = 90 °∫ EDQ ∫ CDA ∫ dqda = high on the hypotenuse of RT △ EDQ, high on the hypotenuse of RT △ EDQ: 4-x, high on the hypotenuse of RT △ CDA: 125. ∫ 1.25x − 25 = 5 (4 − x) 12, the solution is x = 3.1 (12 points)

In the triangle ABC, angle c = RT angle, AC = 4cm, BC = 5cm, point D is on BC, and CD = 3cm. There are two moving points P and Q starting from point a and point B at the same time,

∵PE//BC ∴PE⊥AC
Also ∵ △ ape ∽ ACD
AP=X CD=3 AC=4
∴AP/AC=PE/CD
∴PE=AP*CD/AC=3X/4
In RT △ Ape:
AE=√(AP^2+PE^2)
=5/4X
Because:
AD=√(AC^2+CD^2)=5
So:
DE=AD-AE=5-5/4X
(2) Only when QE / / AC, i.e. QE ⊥ BC is satisfied, △ EDQ is a right triangle
∴△DEQ∽△DAC
∵QB=X
∴DQ=QB-DB=X-2
From DQ / DC = de / DA, it can be concluded that:
That is: X-2 / 3 = 5 - (5 / 4x) / 5
The solution is x = 20 / 7
So when x = 20 / 7, △ DEQ is a right triangle

A is the midpoint of de side of triangle CDE, BC = CD / 3, triangle ABC area is 5, calculate the area of triangle abd and triangle ace

The area of the triangle abd is 10
The area of the triangle ace is 15
Analysis: triangle ABC and triangle abd are based on BC and BD respectively, with the same height, while 2BC = BD, so the area of triangle abd is twice that of triangle ABC, equal to 10
The area of triangle ADC is 10 + 5 = 15
Triangle ace and triangle ADC have the same base and height, and the same area, which is also equal to 15