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Levart [38]
2 years ago
14

Find the area of quadrilateral ABCD. [Hint: the diagonal divides the quadrilateral into two triangles.]

Mathematics
1 answer:
Vadim26 [7]2 years ago
7 0

Answer:

B) 28.53 unit²

Step-by-step explanation:

The diagonal AD divides the quadrilateral in two triangles:

  1. Triangle ABD
  2. Triangle ACD

Area of Quadrilateral will be equal to the sum of Areas of both triangles.

i.e.

Area of ABCD = Area of ABD + Area of ACD

Area of Triangle ABD:

Area of a triangle is given as:

Area = \frac{1}{2} \times base \times height

Base = AB = 2.89

Height = AD = 8.6

Using these values, we get:

Area = \frac{1}{2} \times 2.89 \times 8.6 = 12.43

Thus, Area of Triangle ABD is 12.43 square units

Area of Triangle ACD:

Base = AC = 4.3

Height = CD = 7.58

Using the values in formula of area, we get:

Area = \frac{1}{2} \times 4.3 \times 7.58 = 16.30

Thus, Area of Triangle ACD is 16.30 square units

Area of Quadrilateral ABCD:

The Area of the quadrilateral will be = 12.43 + 16.30 = 28.73 units²

None of the option gives the exact answer, however, option B gives the closest most answer. So I'll go with option B) 28.53 unit²

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Answer:

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$12820.50

Step-by-step explanation:

To obtain a rounded value of $28 (to nearest dollar) the lowest and highest possible value before rounding will be : 27.50 and 28.49

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Lower bound = $27.50

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Therefore. Cost of making 450 chairs ;

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Which label on the cone below represents the diameter?
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c

Step-by-step explanation:

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Answer:

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85.36 steps North from centre

Step-by-step explanation:

<em>Refer to attached</em>

Musah start point and movement is captured in the picture.

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or 360°, so bearing of 315° is same as North-West 45°.

<em />

<em>Note. According to Pythagorean theorem, 45° right triangle with hypotenuse of a has legs equal to a/√2.</em>

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<u>How far North Is Musah's final point from the centre?</u>

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given
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324pi=9(basearea)
divide both sides by 9
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the area of te base is 36pi square cm (put 36 in the blank since th pi is alredy there)
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The Median should be 89.5

And the mean should be 69.75

I hope that helped

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