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NARA [144]
3 years ago
8

The area of the triangle above is 21. What is the value of x?

Mathematics
2 answers:
d1i1m1o1n [39]3 years ago
6 0
The answer is 6
I got it wrong on the test because I picked 7 automatically my mind thought of 7x3 but it's the wrong math



MakcuM [25]3 years ago
6 0

Answer:

x=6

Step-by-step explanation:

We have been given an image of a triangle. We have been given that area of the triangle is 21. We are asked to find the value of x.

We will use area of triangle formula to solve for x.

A=\frac{1}{2}\times {Base* Height}

Upon substituting our given values we will get,

21=\frac{1}{2}\times ({(x+1)*x})

21*2=\frac{1}{2}*2\times ({(x+1)*x})

42=x^2+x

x^2+7x-6x-42=0

x(x+7)-6(x+7)=0

(x+7)(x-6)=0

(x+7)=0\text{ or }(x-6)=0

x=-7\text{ or }x=6

Since length can not be negative, therefore, the value of x is 6.

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Answer:

17) x=8

18) x=\sqrt{189} or x= 3\sqrt{21}

Step-by-step explanation:

So the rule is a^{2} +b^{2} =c^{2}, "c" being the hypothenuse, or the long line that is opposite to the right angle.

17) We know that both values of x are equal to each other, which makes everything 10x easier!

                                               x^{2} +x^{2} =(8\sqrt{2})  ^{2}

(by the way we know the x values are our a and b values because they are legs! the way I like to remember the legs is that they are connected to the right angle box, and therefore support the hypothenuse)

<em>simplify</em> (╥︣﹏╥)    

                                                2x^2=8^2(\sqrt{2} )^{2}  

                                                  2x^2=64(2)

                                                   2x^2=128    

                                                  x^{2}= 128/2  

                                                     x^{2} =64

                                                        x=8      

18) Just pretend that the flipped triangle doesn't exist. It's parallel to the other triangle with values on it, and basically servers no purpose other than being parallel to the sister triangle :)

Anyways, since we know the hypothenuse (15) but we don't know one of our leg values (x), we're going to change our equation a bit!

                                                    c^{2} - b^{2} = a^{2}

It doesn't matter if you put the one leg value in a or b, just as long as you stick to that same equation you started with the entire time!

                                                   15^{2} -6^{2} =x^{2}

                                                   225-36=x^{2}  

                                                     x^{2} =189

                                               x=\sqrt{189}=3\sqrt{21}

        The more you do these, the easier they'll get, so don't worry!

                           

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<span>Let the age of the 4th person be F
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