Find the area of the trapezoid 2 2 2 2 4

It is 10.

Step-by-step explanation:

If we just find the area for each section and add them all up it would equal 10. The first triangle, 2x2=4 then divide that by 2 will equal you 2. Then the square is just 2x2 =4. Then the last triangle, 2x4=8 then divide that by 2 = 4. 2+4+4=10.

Correct answer is (10)

Step-by-step explanation:

since 2 triangles come together to form a square\rectangle

length * width = Area

we can find both areas of the triangles:

left triangle: 2*2 = 4 --> 4 devided in half = 2

square area = 4

right side triangle: 2*4 = 8 ---> devided in half = 4

next add up all the measurements for each piece:

left triangle + middle square + right side triangle = area

(2+4+4 = 10)

Hope I helped you out :)

Have a great day \ and good luck on your studies :D

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841