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kati45 [8]
3 years ago
7

A square is cut into two congruent triangles by drawing the diagonal between two corners. If the area of each triangle is 162 cm

,what is the length of the side of the square?

Mathematics
1 answer:
Pavel [41]3 years ago
5 0
Ok so area=162
since the triangles sides were cut out of a square, base=height
area=1/2bh=162
b=h
area=1/2b^2=162
times 2 both sides
b^2=324
sqrt both sides
b=18=h
the legnth of one side is 18cm
see diagram

another way, is total aera of square=areas of  both triangles=162+162=324
remember area of square=side^2 so
324=side^2
sqrt both sides
18=side

answer is 18cm

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Answer

You can multiply the first equation by 4 and the second equation by 3.

You can multiply the first equation by 4/3.

You can multiply the first equation by 3.

Explanation

When solving a system of equations by elimination, you want to add or subtract the equations to "get rid" of a variable.

To do that, one of the variables in both equations have to have the same coefficient.

The first answer possible gives x the coefficient of 12 for both equations. You would get 12x+4y=52 and 12x-9y=39. You could subtract those equations to get 13y=13.

The second way gives x the coefficient of 4. You would multiply the first equation by 4/3 to get 4x+4/3y=52/3. You can subtract to get one variable, and then solve from there. Although, multiplying for 4/3 is annoying, so it's not suggested.

You can also "get rid" the the y. Multiply the first equation by 3 to get 9x+3y=39. You can add these equations. When you add 9x+3y=39 and 4x-3y=13 you get 13x=52.

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3 years ago
Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

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now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

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take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

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take 2a^2t^2+2a+1=0

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The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

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The only solution, we have t=0.

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