I need help with these 2 questions

9514 1404 393

Step-by-step explanation:

The conventional way to work these problems is to make use of the formulas for areas of a triangle, rectangle, and semicircle. If you've used these formulas for a while, you recognize that they give you certain relationships that may make these problems easy to do mentally.

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1. The area of a triangle is half the product of height and width, so is equivalent to the area of a rectangle either half as high, or half as wide. We note that the width of the sail is 5 cm, so half that is 2.5 cm--exactly the same as the height of the boat's hull. That means we can add the height of the sail to the length of the boat, and the total area can be considered to be the same as that of a rectangle that is

  2.5 cm high × (12 cm + 6 cm) long = (2.5)(18) cm² = 45 cm²

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2. The usual formula for the area of a circle is ...

  A = πr²

When expressed in terms of diameter, this becomes ...

  A = π(d/2)² = (π/4)d²

Then the area of a semicircle is half that, or (π/8)d². This is equivalent to the area of a rectangle that is "d" wide and "π/8·d" high. That is, the approximate area of the semicircle is that of a rectangle 4 cm high by (π/8·4 cm) = π/2 cm wide. In other words, the semicircle effectively adds π/2 cm to the left end of the 6 cm central rectangle of the figure.

As discussed above, the area of a triangle is equivalent to the area of a rectangle half as high. In this figure, the triangle is 10-6 = 4 cm wide, so can be considered to contribute 4/2 = 2 cm to the right end of the 6 cm central rectangle.

If we consider π ≈ 3, then the approximate area of the figure is ...

  (4 cm)(3/2 cm + 6 cm + 2 cm) = (4)(9.5) cm² = 38 cm²

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The exact value is 4(8+π/2) = 32+2π ≈ 38.283 cm².

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step-by-step explanation:

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