What is the area of a wall that is 28' wide by 10' high at the sides and has a triangular gable that measures 7' high?

Mathematics

1 answer:

Karolina [17]3 years ago

7 0

Answer:

378 ft²

Step-by-step explanation:

Given:

The Width of the wall = 28'

The Length of the wall = 10'

The Height of the gable = 7'

Now, the width of the gable will be equal to the width of the wall = 28'

therefore,

Area of the wall = Area of the rectangle section of wall + area of the gable

Total area of the wall = ( 28' × 10' ) + ( $\frac{1}{2}\times28'\times7'$ )

Total area of the wall = 280 + 98 = 378 ft²

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Find the equation of the line that is parallel to the given line and passes through the given point

Sati [7]

Answer:

$\huge\boxed{y=-\dfrac{1}{4}x-1\to x+4y=-4}$

Step-by-step explanation:

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$x=8,\ y=-3\\\\-3=-\dfrac{1}{4}(8)+b\\\\-3=-2+b\qquad\text{add 2 to both sides}\\\\-1=b\to b=-1$

$\text{Therefore the equation is:}\ y=-\dfrac{1}{4}x-1.\\\\\text{Convert to the standard form}\ Ax+By=C:\\\\y=-\dfrac{1}{4}x-1\qquad\text{multiply both sides by 4}\\\\4y=-x-4\qquad\text{add}\ x\ \text{to both sides}\\\\x+4y=-4$