The area of trapezoid ABCD is 180 cm2 .The altitude is 8 cm,AD is 10 cm,and BC is 17 cm.
What is AB,in centimeters?
given isosceles trapezoid ABCD (AB//CD), AC is perpendicular to BD and the length of the height of the ABCD is 7 cm. What is the area of the isosceles trapezoid ABCD?
toan quoc te nha
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The area of trapezoid ABCD is 180 .The altitude is 8 cm,AD is 10 cm,and BC is 17 cm.
What is AB,in centimeters?
Answer: cm.
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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.
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 the figure, ABCD is a parallelogram, K is the midpoint of side AD, AB=2.5cm, BC=5cm, CH=4cm.
What is the area of the trapezoid BCDK?
đã biết học nhu rồi mà còn ra câu tiếng anh
dịch ra là thế này: Trong hình, ABCD là một hình bình hành, K là trung điểm của cạnh AD, AB = 2,5cm, BC = 5cm, CH = 4cm. diện tích của hình thang BCDK là gì?
rất tiếc em mới học lớp 7
không cần kết bạn Người ta thi toán tiếng anh đó đồ lạc hậu
a. The area of a trapezoid is 20m the larger base is 55dm and the smaller base is 45dm. Find the height of trapezoid.
b. Find the average of two bases of the trapezoid, known that: the area of trapezoid is 7 m2 and its height is 2m.
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
ai biết chuyển sang thì quá khớ đơn là gì ko
từ li ke chuyển sang thì quá khứ đơn là gì các bạn
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.