An equilateral triangle with the measure of its side is 6 cm. The area of the triangle is \(\sqrt{m}\) \(cm^2\). Find m
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an equilateral triangle of side 12 has its conrner cut off to form a regular hexagon
a) what is the area of the corner cut off?
b) find the are of the
the area of an equilateral triangle is \(8\sqrt{12}cm^2\)
What is the perimeter of the 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?
3 tháng 10 2017 lúc 17:12
Give an equilateral triangle with each side 9cm. Find the area of this triangle.
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
The area of an isosceles right triangle is 9 cm2.
Find the length of its hypotenuse.
Answer: The length of its hypotenuse is .... cm.
The area of an isosceles right triangle is 9 cm2.
Find the length of its hypotenuse.
Answer: The length of its hypotenuse is .... cm.
The lengths in cm of three sides of a triangle are three consecutive odd numbers. Knowing that the perimeter of the triangle is 39cm, find the length of the longest side of the triangle. Show your work.
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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