If AB = 9 units, QS = 6 units, and AR = QR = RS, what is the area of the heart shape?

Step-by-step explanation:

AR = QR = RS  = (1/2) QS  = 3

So by the Pythagorean Theorem, AS = AQ  =

√ [3^2  + 3^2]   =  √[9 + 9 ]  = √18  

And the area of the two  semi-circles= the area of a circle with a radius of AS/2 =  

pi [AS/2]^2 =

pi [√18/2]^2  =  18pi/4  =  (9/2)pi  units^2

And the area of the rest of the figure is just that of a rhombus  =

The product of the diagonals / 2   =

AB*QS/2  = 9 * 6 / 2  = 27 units^2

Sothe total area  =

[(9/2) pi + 27]   units^2   =  about  =  41.1 units^2  [rounded]

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]