Please answer this it's confusing

Approximately 1336.50301206033 square cm

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Explanation:

If we know the three side lengths of a triangle, then we can use Heron's formula to get the area

Heron's formula is

with S being the semiperimeter, or half the perimeter, so . We add up all three sides and divide by 2.

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The smaller triangle has side lengths of a = 19, b = 40, c = 41. So we get a semiperimeter of S = (a+b+c)/2 = (19+40+41)/2 = 50

Plug this along with the a,b,c values to get an area of...

which is approximate

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Do the same for the larger triangle as well

S = (a+b+c)/2 = (19+180+181)/2 = 190

Interestingly enough, we get an exact area value this time as opposed to some approximation.

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The difference between the two results is the area of the shaded region

1710 - 373.496987939661 = 1336.50301206033

This is approximate since the first result was approximate.

K = 1710 cm^2  - K = 373.497 cm^2  = 1337 cm^2

Step-by-step explanation:

Why don't you just look for the whole area and deduct the non-shaded area?

Total

Sides:

a = 180 cm

b = 181 cm

c = 19 cm

Angles:

A = 83.9744 °

B = 90 °

C = 6.02558 °

Other:

P = 380 cm

s = 190 cm

K = 1710 cm^2

r = 9 cm

R = 90.5 cm

Non shaded area:

Sides:

a = 40 cm

b = 41 cm

c = 19 cm

Angles:

A = 73.519 °

B = 79.3849 °

C = 27.0961 °

Other:

P = 100 cm

s = 50 cm

K = 373.497 cm^2

r = 7.46994 cm

R = 20.8569 cm

where:

A = angle A

B = angle B

C = angle C

a = side a

b = side b

c = side c

P = perimeter

s = semi-perimeter

K = area

r = radius of inscribed circle

R = radius of circumscribed circle

d. vertex

step-by-step explanation:

c.   x=2.4cm

step-by-step explanation:

similar triangles are triangles which have the same shape but not the same size. this means their angle measures are equal but their side lengths are not instead they are proportional. to solve, we will set up a proportion and solve.

a proportion is two equal ratios set equal to each other. we form the ratios by finding corresponding sides (sides which match to each other on both triangles by their position). out ratios will be one side of the small triangle over the corresponding side on the big triangle as shown below:

.

to solve the proportion, we'll cross multiply by multiplying numerator with denominator across the equal sign.