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3 years ago

A square pyramid and its net are shown below. What is the surface area of the pyramid?

Mathematics

2 answers:

Allushta [10]3 years ago

5 0

Answer:

736 Maybe

Step-by-step explanation:

Well what you gotta do is the triangles on the sides can be moved to form mini rectangles. When you do this you get 8 by 15 rectangles and since you have 4 of then you multiply those by 4. 8*15= 120 and 120*4 = 480. Then all you need to fing is the center square pretty easy to find that square right. Its just 16*16 = 256. And 480+256=736.

I hope this helps.

antoniya [11.8K]3 years ago

5 0

The answer is 736 squared centimeters.

Step by step:

Sides-

16x15=240

240/2=120

120x4=480

Base-

16x16=256

Add them-

256+480=736

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Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

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c = 120 m - 61 m

c = 59 m

↓ ↓

A → → → → → → → → → → → C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

$a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0$

Angle B is calculated as follows;

$Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0$

Angle C is calculated as follows;

$Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0$

Therefore, the largest angle of the field is 149⁰.

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