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ryzh [129]
3 years ago
7

What is the area of the figure? Enter your answer in the box. 400 in²

Mathematics
2 answers:
lesya692 [45]3 years ago
7 0

Answer:

should be 112 cm^2

Step-by-step explanation:

10cm*10cm+(1/2(4*6)

dem82 [27]3 years ago
5 0

Answer:

112

Step-by-step explanation:

To make this easier we can pull apart the two know shapes which is the square and triangle

the formula to find the area square is length x width

so 10 x 10=100 this is the area of the triangle

the picture doesn't show the measurements of the triangle but we can still find it

16 in goes from the top of the triangle to the bottom of the square and we know the right side of the square is 10 in so to find the length of the left side of the triangle we can subtract 10 from 16 which is 6.

now we need to find the bottom part of the triangle we can see that it measures 6 in to the right of the triangle and we know that the side of the square is 10 in so all we do is subtract 6 from 10 and you get 4

the formula to find the area of a triangle is height x base divided by 2

so 6 x 4 divided by 2 is 12

now we know the area of the triangle and square is 12 and 100 so we add these together to get 112

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About 2% of the population has a particular genetic mutation. 400 people are randomly selected. Find the mean for the number of
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The State Board of Education has $2,183 to buy new calculators. If each calculator costs $37, how many calculators can the board
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5 0
2 years ago
The coordinates of rhombus ABCD are A(–4, –2), B(–2, 6), C(6, 8), and D(4, 0). What is the area of the rhombus? Round to the nea
a_sh-v [17]
Check the picture below.

so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad 
C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2}
\\\\\\
AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\
-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad 
D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2}
\\\\\\
BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2}
\\\\\\
BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}

that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.

namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,

\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right]
\\\\\\
4[15]\implies 60

7 0
3 years ago
Read 2 more answers
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