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What is the area of this figure?

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in²

A parallelogram with a right triangle created inside it with a short leg length of 16 inches and a long leg length of 20 inches. A triangle attached to the top of the parallelogram has a short leg measurement of 8 inches.

What is the area of this figure Enter your answer in the box in A parallelogram with a right triangle created inside it with a short leg length of 16 inches and class=

Respuesta :

Check the picture below.

so the area of that figure is simply the area of that triangle and parallelogram added up.

[tex]\bf \stackrel{\textit{triangle's area}}{\cfrac{1}{2}(20)(8)}~~ + ~~\stackrel{\textit{parallelogram's area}}{(20)(16)}\implies 80+320\implies 400[/tex]

Ver imagen jdoe0001
micahdisu

Answer: The area of the figure is 400 square inches

Step-by-step explanation: Please refer to the diagram attached.

The diagram in the question shows a parallelogram ABCD attached to a right angled triangle EBC.

In order to compute the total area of the entire figure, the area of both detachments would have to be calculated separately and then summed up.

To calculate the area of a parallelogram ABCD, the formula is

Area = b x h

where b = base and h = perpendicular height.

The base is 20 inches while the height is 16 inches.

Therefore,

Area = 20 x 16

Area = 320 in^2

Next we calculate the area of triangle EBC. Area of a triangle is given as

Area = 1/2 base x height

Where the base = 20 (line AD is equal to line BC = 20) and height = 8. Therefore,

Area = 1/2 (20) x 8

Area = 10 x 8

Area = 80 in^2

Having calculated the area of the two parts of the entire figure, the area of the entire figure is given as

Area of parallelogram ABCD + Area of triangle EBC

We now have 320 + 80 = 400

Hence, area of the figure = 400 square inches

Ver imagen micahdisu