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2 years ago

What is the lateral area of this regular octagonal pyramid? 84.9 cm² 120 cm² 169.7 cm² 207.8 cm²

Mathematics

2 answers:

ahrayia [7]2 years ago

3 0

Like jdoe0001 answered with 8[1/2(5)(6 $8[ \frac{1}{2} (5)(6 \sqrt{2} )]$ if you solve the equation he/she gave you then you get 167.70562748. so your answer will be 167.7 cm^2

Lesechka [4]2 years ago

3 0

Check the picture below.

the lateral area, namely the area of the sides, in this case will be the area of all the triangular faces of the OCTAgonal pyramid, OCTA=8, so there are 8 of them.

now, each triangular face, has a base of 5 and a height of 6√2, so we simply get the area of each triangle and sum them up,

$\bf \stackrel{\textit{area of the 8 triangles}}{8\left[ \cfrac{1}{2}(5)(6\sqrt{2}) \right]}
$

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For circle O, and m∠ABC = 55°. In the figure, ∠_____ and ∠____ have measures equal to 35°.

luda_lava [24]

Answer:

In the figure ∠ABO and ∠BCO have measures equal to 35°.

Step-by-step explanation:

<u><em>The complete question is</em></u>

For circle O, m CD=125° and m∠ABC = 55°

In the figure<____, (AOB, ABO, BOA) and <_____ (BCO, OBC,BOC) have measures equal to 35°

The picture in the attached figure

step 1

Find the measure of angle COB

we know that

$m\angle COB=arc\ CD$ ----> by central angle

we have

$arc\ CD=125^o$

therefore

$m\angle COB=125^o$

step 2

we know that

AB is a tangent to the circle O at point A

ABC and ABO are right triangles

In the right triangle ABC

Find the measure of angle BCA

Remember that

$m\angle BCA+m\angle\ ABC=90^o$ ---> by complementary angles in a right triangle

we have

$m\angle ABC=55^o$

substitute

$m\angle BCA+55^o=90^o$

$m\angle BCA=90^o-55^o=35^o\\$

step 3

In the triangle BCO

Find the measure of angle CBO

we know that

$m\angle CBO+m\angle COB+m\angle BCO=180^o$ ---> the sum of the interior angles in any triangle must be equal to 180 degrees

we have

$m\angle COB=125^o$

$m\angle BCO=m\angle BCA=35^o$ -----> have measure equal to 35 degrees

substitute

$m\angle CBO+125^o+35^o=180^o$

$m\angle CBO=180^o-160^o=20^o$

step 4

Find the measure of angle ABO

In the right triangle ABO

we know that

$m\angle ABC=m\angle CBO+m\angle ABO$ ----> by angle addition postulate

we have

$m\angle ABC=55^o$

$m\angle CBO=20^o$

substitute

$55^o=20^o+m\angle ABO$

$m\angle ABO=55^o-20^o=35^o$ ----> have measure equal to 35 degrees

therefore

In the figure ∠ABO and ∠BCO have measures equal to 35°.

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2 years ago

Grades 3, 4, and 5 have their annual field day together. Each grade level is given 16 gallons of water.

sineoko [7]

Answer:

Their will be enough water for each student to consume 2 cups of water.

Step-by-step explanation:

Grade 3, 4 and 5 have their annual field day together . Each grade level is given 16 gallons of water. This means the three grades have 16 × 3 = 48 gallons of water.

The total students involved is 350 student. The total student is the sum of the student in all three grades.

Since the water will be distributed to the students in cups let us convert gallons to cups.

1 gallons = 16 cups

48 gallons = ? cups

cross multiply

48 × 16 = 768 cups.

768 cups of water is available to be shared among the 350 students.

1 student is entitled to 2 cups of water.

350 students will consume 350 × 2 = 700 cups of water.

Therefore, their will be enough water for each student to consume 2 cups of water.

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Step-by-step explanation:

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