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Oksana_A [137]
3 years ago
6

Alex has a grid in the shape of a rectangle. He has right triangles. Each right triangle has an area of 25 square millimeters an

d each right triangle fits on the grid. If the length of the grid is 20 centimeters and the width is 12 centimeters, how many right triangles can Alex fit on the grid?
Mathematics
1 answer:
VladimirAG [237]3 years ago
5 0
First I would convert 20 and 12 to millimeters which would be 20= 200mm 12= 120mm. You would then find the area if the grid which would be 200•120=
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Step-by-step explanation:

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stellarik [79]

Answer:

c. 10 cm

Step-by-step explanation:

The parallel sides of trapizium are the bases so

base 1(B1) =9cm , base(B2)= 12cm

area(a) = 105 {cm}^{2}

the perpendicular distance means the height so lets find the height

area of trapizium= <u>B1 + B2</u> × h

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>2

{105cm}^{2}(the \: given \: area)  =  \frac{9cm(the \: given \: base) + 12cm(the \: given \: base)}{2}  \times h

{105cm}^{2}  =  \frac{21cm}{2}  \times h

now we crisscross

{105cm}^{2} \times 2 = 21cm \times h

{210cm}^{2}  = 21cm \times h

\frac{ {210cm}^{2} }{21cm}  =  \frac{21cm \: }{21cm}  \times h

h = 10 cm

hope u like it ‍❤️‍‍❤️‍‍❤️‍

For any question comment me

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Explain how to use the combine values strategy to find 223-119
stich3 [128]
223
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In stead of doing 3-9 and 2-1 do 23-19 then 2-1
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