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mart [117]
3 years ago
7

Find The Area Of The Shape Shown Below

Mathematics
2 answers:
Vesnalui [34]3 years ago
8 0

Answer:

<u>32</u>

Step-by-step explanation:

I proceeded to divide the shape up into 2 triangle and 1 square. The added sum of the 2 divided triangles is 16 and the square is 16. Add those to to get <u>32</u>.

The following shape is a trapezoid

area formula:

A=

A=

A= (4)(12+4)

A= (4)(16)

A= (4)(16)

A= (64)

A=32

Also, is that Khan Academy?

svlad2 [7]3 years ago
6 0

Answer:

A=32

Step-by-step explanation:

The following shape is a trapezoid

area formula:

A=\frac{1}{2} h (b_{1} +b_{2} )

b_{1} =12\\b_{2}=4 \\h=4

A=\frac{1}{2} h (b_{1} +b_{2} )

A= \frac{1}{2}(4)(12+4)

A= \frac{1}{2}(4)(16)

A= \frac{1}{2}(4)(16)

A= \frac{1}{2}(64)

A=32

<u><em>I HOPE THIS HELPS YOU OUT:)</em></u>

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56,000.0

Step-by-step explanation:

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Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1
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\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}

From Left side:

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)

\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}

NOTE: sin²θ + cos²θ = 1

\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{2}{sin^2\theta-cos^2\theta}

\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}

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\dfrac{2sec^2\theta}{tan^2\theta-1}

Left side = Right side <em>so proof is complete</em>

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