the area of the square ABCD is 64 cm2 ,the points M and N are the midpoints of the side AD and BC. the area of the quadrilateral MBND is cm2
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the side of each small square in the figure is 4 cm. Calculate the area of the quadrilateral ABC.
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Given a square with the length of one side is 8 cm and a isosceles triangle with the length of its base is 12 cm. If the area of the square is equal to the area of the isosceles triangle then what is the length of the height of the isosceles triangle, in cm?
The sum of area of all squares in the following figure is 56250 dm2.The area of the smallest square is ....cm2.
The are of the smallest square is 375000 cm2
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A box contains 50 blue square cards whose the side length are 2 cm, 4 cm, 6 cm, ..., 100 cm, respectively and 50 red square cards with side lengths are 1 cm, 3 cm, 5 cm, ..., 99 cm, respectively. The total area of the blue cards is greater than the total area of the red care is .... cm2
\(2^2+4^2+...+100^2-\left(1^2+3^2+...+99^2\right)\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...\left(100^2-99^2\right)\)
\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(100-99\right)\left(100+99\right)\)
\(=1+2+3+...+100\)
\(=\frac{100.\left(100+1\right)}{2}=5050\left(cm^2\right)\)
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Point B,DB,D and JJ are midpoints of the sides of right triangle ACGACG . Points K,E,IK,E,I are midpoints of the sides of triangle JDGJDG, etc. If the dividing and shading process is done 100 times (the first three are shown) and AC=CG=6AC=CG=6, then the total area of the shaded triangles is nearest
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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.
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)
A square of side 8cm has the same perimeter as a rectangle. If the ratio of the width of the rectangle to its length is 1:3 then its area is ...cm2?
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