What area is this shape ?

9514 1404 393

  40 square units

Step-by-step explanation:

The area can be considered as that of two trapezoids, a rectangle, and a triangle. The total area is the sum of the areas of these.

Draw two horizontal lines, one beginning at the left-most point, and another 2 units below that. Draw a vertical line segment extending the vertical line at the top down to the first horizontal line. Now the figure is divided as described above.

The upper left trapezoid has bases of 3 and 5, and a height of 4. Its area is ...

  A = 1/2(b1 +b2)h

  A = (1/2)(3 +5)(4) = 16

The trapezoid between the two horizontal lines has bases of 8 and 7, and a height of 2. Its area is ...

  A = (1/2)(8 +7)(2) = 15

The rectangle at upper right has a base of 2 and a height of 3. Its area is ...

  A = bh

  A = 2(3) = 6

The triangle at the bottom has a base of 3 and a height of 2. Its area is ...

  A = 1/2bh

  A = 1/2(3)(2) = 3

Then the total area of all of these pieces is ...

  Area = 16 +15 +6 +3 = 40 . . . . square units

the answer is 40 square units

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

f(x) = y

f(-3) = -5

step-by-step explanation: