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Troyanec [42]
4 years ago
15

What is the area figure below??

Mathematics
1 answer:
Ivenika [448]4 years ago
8 0

area of trapezoid is 1500 & represented with expression Area = \frac{75+25}{2}(30) . Correct answer is c) .

<u>Step-by-step explanation:</u>

Here we have , a Trapezoid in figure. A trapezoid is a 4-sided flat shape with straight sides that has a pair of opposite sides parallel . A trapezoid:

  • has a pair of parallel sides
  • is an isosceles trapezoid when it has equal angles from a parallel side
  • is called a "trapezium" in the UK

Area of Trapezoid = \frac{a+b}{2} (h) , a & b are the parallel sides and , h is height of trapezoid  . Let's find area of Trapezoid:

⇒ Area = \frac{a+b}{2}(h)

⇒ Area = \frac{75+25}{2}(30)

⇒ Area = \frac{100}{2}(30)

⇒ Area = (50)(30)

⇒ Area = 1500

So, area of trapezoid is 1500 & represented with expression Area = \frac{75+25}{2}(30) . Correct answer is c) .

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3 years ago
8x-2=-9+7x need help
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Hello :) 

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Let's plug the x value into the equation to verify:

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3 years ago
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Katena32 [7]

Answer:

10) 8              11) 10

Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
Rational number between 1.2 and 1.3 is 1.25 right?
kherson [118]

Answer:

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Step-by-step explanation:


3 0
3 years ago
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When a boy pulls his sled with a rope, the rope makes an angle of 70 with the horizontal. If a pull of 16 pounds on the rope is
alexandr1967 [171]

The horizontal component of force is 5.47 lb

<u>Step-by-step explanation:</u>

Given that a boy pulls his sled with a rope.

The rope makes an angle with horizontal = 70 degree

If a pull on the rope is needed to move the sled = 16 pounds

We need to find the horizontal component of force

In given scenario, the force used in this scenario is a vector addition of the horizontal and the vertical force component. The horizontal force component is calculated by multiplying the applied force by the cosine of the angle of the applied force from the horizontal plane in which it is found:                     horizontal force component = applied force x cos 70 degree

Horizontal component force = f \cos \alpha

Here, given applied force, f = 16 pounds

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Horizontal component of force = 16×0.342

Hence, horizontal component of force = 5.47 lb

3 0
3 years ago
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