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lubasha [3.4K]
3 years ago
10

PLZ HELP I WILL MARK AS BRAINLIEST

Mathematics
1 answer:
-Dominant- [34]3 years ago
4 0

Answer:

  7 square units

Step-by-step explanation:

As with many geometry problems, there are several ways you can work this.

Label the lower left  and lower right vertices of the rectangle points W and E, respectively. You can subtract the areas of triangles WSR and EQR from the area of trapezoid WSQE to find the area of triangle QRS.

The applicable formulas are ...

  area of a trapezoid: A = (1/2)(b1 +b2)h

  area of a triangle: A = (1/2)bh

So, our areas are ...

  AQRS = AWSQE - AWSR - AEQR

  = (1/2)(WS +EQ)WE -(1/2)(WS)(WR) -(1/2)(EQ)(ER)

Factoring out 1/2, we have ...

  = (1/2)((2+5)·4 -2·2 -5·2)

  = (1/2)(28 -4 -10) = 7 . . . . square units

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Answer:

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Step-by-step explanation:

Solution:-

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- The value of θ = 128° was reported by Tamara suggests that the answer lies in the second quadrant of a cartesian plane where ( sin (θ) ) have positive values for "a" and (cos (θ) , tan (θ) have negative values for "a". So for cos (128):

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- The value of θ = 308° was reported by Uma suggests that the answer lies in the fourth quadrant of a cartesian plane where ( cos (θ) ) have positive values for "a" and (sin (θ) , tan (θ) have negative values for "a". So for cos (308):

Hence,                           0 < a < 1 , θ = 308°

- The angle θ reported by Uma and Salina are similar solution because of property law of complementary angles:

                                     cos (θ) = cos ( 360 - θ )

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