Check the picture below.
since in a rhombus the diagonals bisect each other, thus EC = EA.
now, the rhombus is simply 4 congruent triangles, we know the base and height of one of them, thus
![\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=8\\ h=15 \end{cases}\implies A=\cfrac{1}{2}(8)(15)\implies A=60 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of all 4 triangles}}{4(60)\implies 240}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20triangle%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dbh~~%20%5Cbegin%7Bcases%7D%20b%3D8%5C%5C%20h%3D15%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%288%29%2815%29%5Cimplies%20A%3D60%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20all%204%20triangles%7D%7D%7B4%2860%29%5Cimplies%20240%7D)
Answer:
Step-by-step explanation:
She has 290 cents.
Nickels: 6x+1
Quarters: x
Dimes: 4x
Now that we've converted it algebraically, we can add them
6x+1 + x + 4x = 11x+1 = 290
11x = 289
x = 289/11
let me know if incorrect, i will fix
Answer:
more ... The same shape and size, but we are allowed to flip, slide or turn. In this example the shapes are congruent (you only need to flip one over and move it a little). Angles are congruent when they are the same size (in degrees or radians). Sides are congruent when they are the same length.
Step-by-step explanation:
happy to help ya:)
Answer:
A
Step-by-step explanation:
The length of side walk is 500 feet
<em><u>Solution:</u></em>
Given that, A rectangle park measures 300 ft by 400 ft
Length = 300 feet
Width = 400 feet
A sidewalk runs diagonally from one comer to the opposite corner
We have to find the length of side walk
Which means, we have to find the length of diagonal of rectangle
<em><u>The diagonal of rectangle is given by formula:</u></em>

Where,
d is the length of diagonal
w is the width and l is the length of rectangle
<em><u>Substituting the values in formula, we get</u></em>

Thus length of side walk is 500 feet