Find the area of the shaded part?

Answer 1

#32 "cm"^2#

The area #A# of a triangle with base #b# and height #h# is given by #A=1/2bh#.
If we treat the side of length #8+4=12# as a base of the large triangle, then as the line with length #6# forms a right angle with that side, the triangle has a height of #6#. Thus the area of the large triangle is #1/2(12)(6) = 36#.
Similarly, if we treat the length #4# side of the white triangle as its base, then it has a height of #2#, meaning its area is #1/2(4)(2) = 4#.
As the area of the shaded section is the difference between the area of the large triangle and the area of the white triangle, we have our desired area as #36 - 4 = 32 "cm"^2#.
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Answer 2

See below.

Supposing no tricks and using #A = (b h)/2# we have
#A = A_1 - A_2 = ((4+8)xx6)/2-(4 xx 2)/2 = 36-4=32#

Now using Heron's formula with

#p = (8+7+12)/2# #A_1 = sqrt(p(p-8)(p-7)(p-12)) approx 26.91#
which is different from the former #36#. So the triangle's figure is a trick.
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Answer 3

#color(red)("Is this question correct in every detail?")#

Area of the shaded portion #ul("could be:")" " 32cm^2#

Using the general principle that the area of a triangle is:

#1/2xx" base" xx "height"#
The overall triangle area #->1/2xx (4+8)xx6 = 36color(white)(.)cm^2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The smaller triangle area #-> 1/2 xx 4 xx 2 = 4color(white)(.)cm^2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The area of the shaded portion #-> (36-4) cm^2 = 32 cm^2" "???? #
#color(red)("===================================")# #color(red)("Checking a few things")#

Using Pythagoras it should be the case that:

base of the whole #=8+4=12= sqrt(7^2-6^2) +sqrt(8^2-6^2)#
RHS #-> sqrt(13)+sqrt(28)#
#sqrt(13)+sqrt(28) ~~8.9 !=" length of the base" = 12#
#color(red)("Conclusion: There is contradicting information in the question")#
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Answer 4

To find the area of the shaded part, we need to subtract the area of the unshaded region from the area of the entire shape. If the shaded region is a simple shape like a rectangle or circle, we can calculate its area directly. However, if the shaded region is irregular, we may need to break it down into smaller, simpler shapes and calculate their areas separately before summing them up.

Once we have the areas of the shaded and unshaded regions, we subtract the area of the unshaded region from the area of the entire shape to find the area of the shaded part. This calculation gives us the desired result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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