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kotykmax [81]
3 years ago
5

Find the area of the rhombus. 

Mathematics
2 answers:
Fudgin [204]3 years ago
6 0

Remember that the formula for the area of a rhombus is \dfrac{pq}{2}, where p and q are the diagonals of the rhombus.


We can already find that one diagonal has length 6 (based on the fact that the diagram shows that half of the diagonal is 3). However, we will need to find the length of the other diagonal in order to find the area.


Also note that the diagonals in a rhombus are perpendicular to each other, meaning that they form a right angle. Based on this fact, we can realize that a triangle forms with one side being the side of the rhombus (length 5), another side being one half of the unknown diagonal, and another side being half of the known diagonal (length 3). This right triangle has a hypotenuse of 5 and a smaller side of 3, which means that the other side must be 4. This can be determined by the Pythagorean Theorem (the length of the larger leg will be coined x):

3^2 + x^2 = 5^2

x^2 = 16

x = 4


Since we have determined one half of our unknown diagonal is 4, we can then realize that the length of the entire diagonal is 8. Using this and the fact that the length of the other diagonal is 6, we can finally find the area of our rhombus:

\dfrac{6 \cdot 8}{2} = \boxed{24}


The area of the rhombus is 24 units^2.

Inessa [10]3 years ago
3 0

A=24


Find the area by multiplying both diagonals and dividing them by 2. Like so: And you might be wondering how to do that. By using the pythagorean theorem and some simple addition, you could get the answer. One diagonal is 8 and the other is 6....so, 8x6=48....48/2=24


Please let me know if you have any other questions!


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4. To solve this problem, we divide the two expressions step by step:

\frac{x+2}{x-1}* \frac{x^{2}+4x-5 }{x+4}
Here we have inverted the second term since division is just multiplying the inverse of the term.

\frac{x+2}{x-1}* \frac{(x+5)(x-1)}{x+4}
In this step we factor out the quadratic equation.


\frac{x+2}{1}* \frac{(x+5)}{x+4}
Then, we cancel out the like term which is x-1.

We then solve for the final combined expression:
\frac{(x+2)(x+5)}{(x+4)}

For the restrictions, we just need to prevent the denominators of the two original terms to reach zero since this would make the expression undefined:

x-1\neq0
x+5\neq0
x+4\neq0

Therefore, x should not be equal to 1, -5, or -4.

Comparing these to the choices, we can tell the correct answer.

ANSWER: \frac{(x+2)(x+5)}{(x+4)}; x\neq1,-4,-5

5. To get the ratio of the volume of the candle to its surface area, we simply divide the two terms with the volume on the numerator and the surface area on the denominator:

\frac{ \frac{1}{3} \pi  r^{2}h }{ \pi  r^{2}+ \pi r \sqrt{ r^{2}  +h^{2} }  }

We can simplify this expression by factoring out the denominator and cancelling like terms.

\frac{ \frac{1}{3} \pi r^{2}h }{ \pi r(r+ \sqrt{ r^{2} +h^{2} } )}
\frac{ rh }{ 3(r+ \sqrt{ r^{2} +h^{2} } )}
\frac{ rh }{ 3r+ 3\sqrt{ r^{2} +h^{2} } }

We then rationalize the denominator:

\frac{rh}{3r+3 \sqrt{ r^{2} + h^{2} }}  * \frac{3r-3 \sqrt{ r^{2} + h^{2} }}{3r-3 \sqrt{ r^{2} + h^{2} }}
\frac{rh(3r-3 \sqrt{ r^{2} + h^{2} })}{(3r)^{2}-(3 \sqrt{ r^{2} + h^{2} })^{2}}}=\frac{3 r^{2}h -3rh \sqrt{ r^{2} + h^{2} }}{9r^{2} -9 (r^{2} + h^{2} )}=\frac{3rh(r -\sqrt{ r^{2} + h^{2} })}{9[r^{2} -(r^{2} + h^{2} )]}=\frac{rh(r -\sqrt{ r^{2} + h^{2} })}{3[r^{2} -(r^{2} + h^{2} )]}

Since the height is equal to the length of the radius, we can replace h with r and further simplify the expression:

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By examining the choices, we can see one option similar to the answer.

ANSWER: \frac{r(1 -\sqrt{2 })}{-3 }
8 0
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