ABCD is a square,AB=a,AC intersects BC at O. M and N are points on the line segments BC anh DC respectively satisfying that MON=45.Then BMxDN=....a^2
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ABCD is a square,AB=a,ac intersects BC at O. M and N are points on the line segments BC anh DC respectively satisfying that MON=45.Then BMxDN=....a^2
ABCD is asquare,AB=a,ac intersects BC at O. M and N are points on the line segments BC anh DC respectively satisfying that MON=45.Then BMxDN=....a^2
Hình vuông ABCD, AB=a, giao điểm của AC và BD tại O. M và N là 2 điểm thuộc BC và DC thỏa mãn \(\widehat{MON}\)= 45 độ. Tính BM.DN =... a^2
Given that ABCD is a rectangle with AB = 12 cm, AD = 6 cm. M and N are respectively midpoint of segments BC and CD. Find the area of triangle AMN in square centimeters.
You have to draw the geometry yourself.
\(A_{ABCD}=AB.AD=12.6=72\left(cm^2\right)\)
M is the midpoint of segment BC so we have: \(BM=MC=\frac{BC}{2}=\frac{6}{2}=3\left(cm\right)\)
For the midpoint of CD is N, we also have: \(DN=NC=\frac{CD}{2}=\frac{12}{2}=6\left(cm\right)\)
We have:
\(A_{AMN}=A_{ABCD}-\left(A_{ABM}+A_{NCM}+A_{ADN}\right)\\ =72-\left(\frac{1}{2}.AB.BM+\frac{1}{2}.NC.MC+\frac{1}{2}AD.DN\right)\\ =72-\left(\frac{1}{2}.12.3+\frac{1}{2}.6.3+\frac{1}{2}.6.6\right)\\ =72-45\\ =27\left(cm^2\right)\)
Thusly, the area of triangle AMN in square centimeters is 27.
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Given that ABCD is a rectangle with AB = 12 cm, AD = 6 cm. M and N are respectively midpoint of segments BC and CD. Find the area of triangle AMN in square centimeters.
Dịch: Cho ABCD là HCN có AB = 12cm, AD = 6 cm. M và N lần lượt là trung điểm của các cạnh BC và CD. Tính diện tích tam giác AMN với đơn vị cm2.
SABCD = \(AB\cdot AD=12\cdot6=72\left(cm^2\right)\)
SADN = \(\frac{AD\cdot DN}{2}=\frac{AD\cdot\frac{1}{2}CD}{2}=\frac{AD\cdot\frac{1}{2}AB}{2}=\frac{6\cdot\frac{1}{2}12}{2}=18\left(cm^2\right)\)
SABM = \(\frac{AB\cdot BM}{2}=\frac{AB\cdot\frac{1}{2}BC}{2}=\frac{AB\cdot\frac{1}{2}AD}{2}=\frac{12\cdot\frac{1}{2}6}{2}=18\left(cm^2\right)\)
SMNC = \(\frac{MC\cdot NC}{2}=\frac{\frac{1}{2}BC\cdot\frac{1}{2}CD}{2}=\frac{\frac{1}{2}AD\cdot\frac{1}{2}AB}{2}=\frac{\frac{1}{2}6\cdot\frac{1}{2}12}{2}=9\left(cm^2\right)\)
SABCD = SADN + SABM + SMNC + SAMN
\(\Leftrightarrow\)SAMN = SABCD - SADN - SABM - SMNC
\(\Rightarrow\) SAMN = 72 - 18 - 18 - 9
= 27 (cm2)
In triangle ABC the points D and E are the intersections of the angular bisectors from C and B with the sides AB and AC respectively. Points F and G on the extentions of AB and AC beyond B and C respectively, satisfy BF= CG= BC. Prove that FG//DE.
I do not know how to answer this question. Stupid that staged shows English
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give isosceles trapezoid abcd ( ab // cd,ab<cd) from a draw ah perpendicular ab , ah intersects bd at h . from b draw bk perpendicular ab , bk intersects ac at k a) what figure is quadrilateral ahkb?why? b) Given that E,F are the midpoints of AB, DC; I and G are respectively the intersection points of AC with BD , CH with DK . Prove that four points E,I,G,F are collincar
câu 1: A rectangle has a length of 60cm and a width of 30cm. It is cut into 2 indentical squares, 2 identical rectangles and a shaded small square. Find the area of the shaded square.
Find the area of the shaded square.
câu 2.The number of ordered pairs (x; y) where x, y ∈ N* such that x2y2 - 2(x + y) is perfect square is ..........
câu 3.Let ABCD be the square with the side length 56cm. If E and F lie on CD, C respectively such that CF 14cm and EAF 45o then CE ........cm.
câu 4.
Given P...
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câu 1: A rectangle has a length of 60cm and a width of 30cm. It is cut into 2 indentical squares, 2 identical rectangles and a shaded small square. Find the area of the shaded square.
Find the area of the shaded square.
câu 2.The number of ordered pairs (x; y) where x, y ∈ N* such that x2y2 - 2(x + y) is perfect square is ..........
câu 3.Let ABCD be the square with the side length 56cm. If E and F lie on CD, C respectively such that CF = 14cm and EAF = 45o then CE = ........cm.
câu 4.
Given P(x) = (x2 - 1/2 x - 1/2)1008
If P(x) = a2016x2016 + a2015x2015 + ..... + a1x + a0
then the value of the sum a0 + a2 + a4 + .... + a2014 is ...........
Find the value of P = k(x1) x k(x2) x k(x3) x k(x4) x k(x5) câu 8.The smallest value of
is .............
câu 9.Let ABCD be a trapezoid with bases AB, CD and O be the intersection of AC and BD. If the areas of triangle OAB, triangle OCD are 16cm2, 40cm2respectively and M is the midpoint of BD, then the area of the triangle AMD is .........cm2.
câu 10.Bottle A contains 15% syrup. Bottle B contains 40% syrup. When these 2 bottles of syrup are mixed, the syrup content is 30% and the total volume is 600ml. How much syrup is in the bottle A at first?
câu 11.Let ABCD.A'B'C'D' be a cube with AC' = √3cm. Find the total surface area of this cube.
câu 12.Let ABC be a triangle with AB = 3cm, AC = 7cm. The internal bisector of the angle BAC intersects BC at D. The line passing through D and parallel to AC cuts AB at E. Find the measure of DE.
câu 12.Given two numbers x, y such that (4y2 - 12y + 25)(4x2 + 6x + 4) = 28The ratio of y to x is ........ câu 13.Mr.Joseph drives car from A to B at a constant speed. If the speed of the car is increased by 20%, it takes him one hour less than the usual time. If he drives at the constant speed for the first 100km before increasing the speed by 30%, it also takes him one hour less than the usual. The distance of AB is ..........km. câu 14.A triangle ABC has  = 120o and the bisector AD (D ∈ BC). If AB = 40cm, AD = 30cm, then AC = ..... cm. câu 15.As shown in the figure, the length of BE is .............
câu 16.
Let f(x) the polynomial given by f(x) = (1 + 2x + 3x2 + 4x3 + 5x4 + 84x5)
Suppose that f(x) = ao + a1x + a2x2 + ..... + ..... + a50x50.
The value of T = a1 + a2 + .... + a50 is .........
The area of the trapezoid ABCD is .........cm2. câu 18.Find the least possible value of A = 4x2 - 3x + 1/4x + 2015, where x varies in the set of positive real numbers. The least possible value of A is câu 19.
Let ABCD be the square with the side length 56cm. If E and F lie on CD, BC respectively such that CF = 14cm and EAF = 45o then CE = ........cm.
Let ABCD be the square with the side length 56cm. If E and F lie on CD, C respectively such that CF = 14cm and EAF = 45o then CE = ........cm.
