The interior angles of a quadrilateral( 4 sided figure) add up to 360 degrees.
114 + 84 + 94 + x = 360
292 + x = 360
x = 360 - 292
x = 68 <== ur missing angle
Answer:
The largest possible area of the garden is
feet²
Step-by-step explanation:
Let the length of the rectangular vegetable garden = l
and width of the garden = w
Since length of the fence = 2w + l
[since there is a wall at one length of the fence so no fencing on that side]
2w + l = 3L
l = (3L - 2w)
Now area of the rectangular garden = Length × width
A = (3L - 2w)×w
A = 3Lw - 2w²
For the maximum area of the garden ![\frac{dA}{dw}=0](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdw%7D%3D0)
![\frac{dA}{dw}=\frac{d}{dw}(3Lw-2w^{2})](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdw%7D%3D%5Cfrac%7Bd%7D%7Bdw%7D%283Lw-2w%5E%7B2%7D%29)
3L - 4w = 0
4w = 3L
w = ![\frac{3L}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B3L%7D%7B4%7D)
Now we plug in the value of w to calculate the area of the garden.
A = ![3L(\frac{3L}{4})-2(\frac{3L}{4})^{2}](https://tex.z-dn.net/?f=3L%28%5Cfrac%7B3L%7D%7B4%7D%29-2%28%5Cfrac%7B3L%7D%7B4%7D%29%5E%7B2%7D)
= ![\frac{9L^{2}}{4}-\frac{18L^{2}}{16}](https://tex.z-dn.net/?f=%5Cfrac%7B9L%5E%7B2%7D%7D%7B4%7D-%5Cfrac%7B18L%5E%7B2%7D%7D%7B16%7D)
= ![\frac{36L^{2}-18L^{2}}{16}](https://tex.z-dn.net/?f=%5Cfrac%7B36L%5E%7B2%7D-18L%5E%7B2%7D%7D%7B16%7D)
= ![\frac{18L^{2} }{16}](https://tex.z-dn.net/?f=%5Cfrac%7B18L%5E%7B2%7D%20%7D%7B16%7D)
= ![\frac{9L^{2} }{8}](https://tex.z-dn.net/?f=%5Cfrac%7B9L%5E%7B2%7D%20%7D%7B8%7D)
Therefore, the largest possible area of the garden is
feet²
Answer:
45.18
Step-by-step explanation:
to find one tenth (1/10) of a number, divide that number by 10
451.8 divided by 10 is
45.18
hope this helps :)
Answer:
494 cm^2
Step-by-step explanation:
Find area od the rectangle:
14 x 26 = 364
Find area of the triangle:
10 x 26 = 260
260/2 = 130
Add the areas together:
364 + 130 = 494