Yes it would be as long as two sides of a triangle are congruent then it’s fine

- Given - <u>a </u><u>rectangle </u><u>with </u><u>length</u><u> </u><u>2</u><u>5</u><u> </u><u>feet </u><u>and </u><u>perimeter </u><u>8</u><u>0</u><u> </u><u>feet</u>
- To calculate - <u>width </u><u>of </u><u>the </u><u>rectangle</u>
We know that ,

where <u>b </u><u>=</u><u> </u><u>width </u><u>/</u><u> </u><u>breadth</u> of rectangle
<u>substituting</u><u> </u><u>the </u><u>values </u><u>in </u><u>the </u><u>formula </u><u>stated </u><u>above </u><u>,</u>

hope helpful ~
The limit (as 'x' approaches zero) of x²/(2x+1) is
0 / (0+1) = 0 .
You get this by substituting the limit in place of 'x'.
Sometimes that doesn't work. Sometimes it gives you
monstrosities like 0/0, or 1/0, or ∞/∞ .
But it works for this one. There's nothing wrong with 0/1 = zero .
14/1 x 3/5 = 42/5 which then equals 8 1/20