In the non-convex quadrilateral ABCDABCD shown below, ∠BCD∠BCD is a right angle, AB=12,BC=4,CD=3AB=12,BC=4,CD=3 and AD=13AD=13.
What is the area of quadrilateral of ABCDABCD?
Said quadrangle ABCD for the measurement of the angle A; B; C; D proportional voi5; 8; 13va 10.A / Count measure the angles of a quadrilateral ABCDb / Extend sides AB and DC intersect at E, which lasted two sides AD and BC in F. Two-rays catnhau sense of angles AED and corner cutting AFB O. Separate each corner giaccua AFB cutting edge CD and AB at M and N. Prove cuadoan MN O is the midpoint.
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Said quadrangle ABCD for the measurement of the angle A; B; C; D proportional voi5; 8; 13va 10.A / Count measure the angles of a quadrilateral ABCDb / Extend sides AB and DC intersect at E, which lasted two sides AD and BC in F. Two-rays catnhau sense of angles AED and corner cutting AFB O. Separate each corner giaccua AFB cutting edge CD and AB at M and N. Prove cuadoan MN O is the midpoint.
Said quadrangle ABCD for the measurement of the angle A; B; C; D proportional voi5; 8; 13va 10.A / Count measure the angles of a quadrilateral ABCDb / Extend sides AB and DC intersect at E, which lasted two sides AD and BC in F. Two-rays catnhau sense of angles AED and corner cutting AFB O. Separate each corner giaccua AFB cutting edge CD and AB at M and N. Prove cuadoan MN O is the midpoint.
Nói ABCD Tứ giác cho việc đo các góc A; B; C; D tỉ lệ voi5; số 8; 13va 10.A / Đếm đo các góc của một tứ giác ABCDb / Mở rộng bên AB và DC cắt nhau ở E, kéo dài hai bên AD và BC ở F. Hai tia catnhau cảm giác của góc AED và góc cắt AFB O. Tách mỗi góc giaccua AFB cắt cạnh CD và AB tại M và N. Chứng minh O là trung điểm của MN
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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Given acute triangle ABC(AB<AC). O is the midpoint of BC, BM and CN are the altitudes of triangle ABC. The bisectors of angle \(\widehat{BAC}\)and \(\widehat{MON}\)meet each other at D. AD intesects BC at E. Prove that quadrilateral BNDE is inscribed in a circle.s
( HELP ME )
1.Given the quadrilateral ABCD with two diagonals perpendicular and AB = 8cm, BC = 7cm, AD = 4cm. Evaluate CD.
2.Given three consecutive even natural numbers, which have the product of last two numbers is 80 greater than the product of first two numbers.
Find the largest number.
Answer: The largest number is
2) đặt 3 số có dạng a; a+2, a+4 rồi khai triển ra
số lớn nhất là 22
the side of each small square in the figure is 4 cm. Calculate the area of the quadrilateral ABC.
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)