First of all we can draw a parallel line to divide the figure into a triangle and a rectangle as shown in the figure. To find the area of our rectangle, remember that the area of a rectangle is length times width, so
![A_{r} =lw](https://tex.z-dn.net/?f=A_%7Br%7D%20%3Dlw)
. Since we know for our figure that the length and width of our rectangle are 13cm and 6cm respectively, lets replace those values in our formula to get its area:
![A _{r} =(13cm)(6cm)](https://tex.z-dn.net/?f=A%20_%7Br%7D%20%3D%2813cm%29%286cm%29)
![A=78cm^{2}](https://tex.z-dn.net/?f=A%3D78cm%5E%7B2%7D%20)
Similarly, the area of a triangle is one half times base times height, so
![At=( \frac{1}{2})bh](https://tex.z-dn.net/?f=At%3D%28%20%5Cfrac%7B1%7D%7B2%7D%29bh%20)
. Since we know that our base is 8cm and our height 6cm, lets replace those values in our equation to find the area of our triangle:
![A_{t} =( \frac{1}{2})(8cm)(6cm)](https://tex.z-dn.net/?f=A_%7Bt%7D%20%3D%28%20%5Cfrac%7B1%7D%7B2%7D%29%288cm%29%286cm%29%20)
![A_{t} =24cm^{2}](https://tex.z-dn.net/?f=A_%7Bt%7D%20%3D24cm%5E%7B2%7D%20)
Now the only thing left is add our areas:
![A_{total} =78cm^{2}+24cm^{2} =102cm^{2}](https://tex.z-dn.net/?f=A_%7Btotal%7D%20%3D78cm%5E%7B2%7D%2B24cm%5E%7B2%7D%20%20%3D102cm%5E%7B2%7D%20)
We can conclude that the correct answer is <span>
A. 102</span>
Answer:
8.2 units
Step-by-step explanation:
Given that the only information for our triangle are 2 sides and 1 angle, we must use the Law of Cosines to find side BC
<u />
<u>Recall the Law of Cosines</u>
<u />![a^2=b^2+c^2-2bc\cdot cos(A)](https://tex.z-dn.net/?f=a%5E2%3Db%5E2%2Bc%5E2-2bc%5Ccdot%20cos%28A%29)
<u>Identify angles and sides</u>
<u />![m\angle A=23^\circ\\a=BC=?\\b=AC=14\\c=AB=6](https://tex.z-dn.net/?f=m%5Cangle%20A%3D23%5E%5Ccirc%5C%5Ca%3DBC%3D%3F%5C%5Cb%3DAC%3D14%5C%5Cc%3DAB%3D6)
<u>Solve for side "a"</u>
<u />![a^2=b^2+c^2-2bc\cdot cos(A)\\\\a^2=14^2+6^2-2(14)(6)\cdot cos(23^\circ)\\\\a^2=196+36-168cos(23^\circ)\\\\a^2=222-168cos(23^\circ)\\\\a=\sqrt{222-168cos(23^\circ)}\\ \\a\approx8.2](https://tex.z-dn.net/?f=a%5E2%3Db%5E2%2Bc%5E2-2bc%5Ccdot%20cos%28A%29%5C%5C%5C%5Ca%5E2%3D14%5E2%2B6%5E2-2%2814%29%286%29%5Ccdot%20cos%2823%5E%5Ccirc%29%5C%5C%5C%5Ca%5E2%3D196%2B36-168cos%2823%5E%5Ccirc%29%5C%5C%5C%5Ca%5E2%3D222-168cos%2823%5E%5Ccirc%29%5C%5C%5C%5Ca%3D%5Csqrt%7B222-168cos%2823%5E%5Ccirc%29%7D%5C%5C%20%5C%5Ca%5Capprox8.2)
Therefore, the length of line segment BC is about 8.2 units
The angles of ∠EFG and ∠GFH are 71° and 109°
<h3>What are linear pair angles?</h3>
Linear pair of angles are formed when two lines intersect each other at a single point.
In other words, a linear pair of angles is a pair of adjacent angles formed when two lines intersect each other.
Linear pair angles are supplementary. This means the sum of a linear pair angles is 180 degrees.
Therefore,
∠EFG + ∠GFH = 180
Therefore,
∠EFG = 4n + 15
∠GFH = 5n + 39
hence,
4n + 15 + 5n + 39 = 180
9n + 54 = 180
9n = 180 - 54
9n = 126
n = 126 / 9
n = 14
Hence,
∠EFG = 4n + 15 = 4(14) + 15 = 71°
∠GFH = 5n + 39 = 5(14) + 39 = 109°
learn more on linear pair angles here: brainly.com/question/28264317
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Since 52% is the same as 52/100, we can multiply 875 by 52/100 in order to find the green buttons.
52/100(875) = 455
455 green marbles