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3 years ago

Estimate the perimeter and area of the shaded figure to the nearest tenth.

Mathematics

1 answer:

Maru [420]3 years ago

6 0

Answer:

Perimeter = 18.7 units

Area = 13.5 units²

Step-by-step explanation:

Perimeter of ADEC = AD + DE + EC + AC

Length of AD = 3 units

By applying Pythagoras theorem in ΔDBE,

DE² = DB² + BE²

DE² = 3² + 3²

DE = √18

DE = 4.24 units

Length of EC = 3 units

By applying Pythagoras theorem in ΔABC,

AC² = AB² + BC²

AC² = 6² + 6²

AC = √72

AC = 8.49 units

Perimeter of ADEC = 3 + 4.24 + 3 + 8.49

= 18.73 units

≈ 18.7 units

Area of ADEC = Area of ΔABC - Area of ΔBDE

Area of ΔABC = $\frac{1}{2}(AB)(BC)$

= $\frac{1}{2}(6)(6)$

= 18 units²

Area of ΔBDE = $\frac{1}{2}(BD)(BE)$

= $\frac{1}{2}(3)(3)$

= 4.5 units²

Area of ADEC = 18 - 4.5

= 13.5 units²

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The length of a rectangle is 20 units more than its width. The area of the rectangle is x4−100.

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Step-by-step explanation:

Area of the rectangle $=(x^4-100)$

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3 years ago

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$\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = 1+1=2$

<u>Given the following data:</u>

$\frac{\sqrt{1 \;-cos^2x} }{sinx}$

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In Trigonometry, you should take note of the following mathematical expression:

$sin^2x + cos^2x = 1$

Therefore, we can obtain the following:

$sin^2x = 1 - cos^2x$ ...equation 1.

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Substituting the equations respectively, we have:

$\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = \frac{\sqrt{sin^2x} }{sinx} + \frac{\sqrt{cos^2x} }{cosx}\\\\$

Taking the square roots, we have:

$\frac{sinx}{sinx} + \frac{cosx}{cosx} = 1 +1\\\\1+1=2$

Therefore, for x such that 0 < x < $\frac{\pi}{2}$ , the expression is:

$\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = 1+1=2$

Read more on trigonometry here: brainly.com/question/4515552

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2 years ago

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