Give triangle of ABC have area is 540 cm2. M and N is 2 points in BC's side and each side is CM, MN, NB is equal. From M write paralle lines with AC. From N write parallel with AB, they cut at I. Connect IA, IB, IC. Calculate area of flower garden.
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with triangle ABC, d is the line passing through B, E of AC. Via E draw straight lines parallel to AB and BC cut d at M, N. D is the intersection of ME and BC. NE lines cut AB and MC at F and K. CMR AFN triangles are in the same form as the MDC triangle
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?
Given a square with the length of one side is 8cm and an isosceles triangle with the length of its base is 12 cm . If the area of the square equal of the area of the isosceles triangle then is the length of height of the isosceles triangle ?
M.n ơi kb vs mk nha ! Mk là thành viên ms nên chưa có bn !
Girl 2k5 -FA
In a triangle of area 100cm2 , the ratio between the length of one side and the corresponding height is 1:2.
What is the length of the height, in m?
Answer: The height is...m.
(write your answer by decimal in simplest form)
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
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từ li ke chuyển sang thì quá khứ đơn là gì các bạn
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An equilateral triangle with the measure of its side is 6 cm. The area of the triangle is \(\sqrt{m}\) \(cm^2\). Find m
30 tháng 8 2020 lúc 15:29
The area of triangle ABC is 300 . In triangle ABC, Q is the midpoint of BC, P is a point on AC between C and A such that CP = 3PA . R is a point on side AB such that the area of \(\Delta\)PQR is twice the area of \(\Delta\)RBQ . Find the area of \(\Delta\)PQR
Dịch thôi chứ ko bt làm:Diện tích tam giác ABC là 300. Trong tam giác ABC, Q là trung điểm BC, P là một điểm trên AC nằm giữa C và A sao cho CP = 3PA. R là một điểm trên cạnh AB sao cho diện tích của \(\Delta\)PQR gấp đôi diện tích của \(\Delta\)RBQ. Tìm diện tích của\(\Delta\) PQR
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)