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aev [14]
2 years ago
9

Given: mAngleTRV = 60° mAngleTRS = (4x)° Prove: x = 30 3 lines are shown. A line with points T, R, W intersects with a line with

points V, R, S at point R. A line extends from point R to point Z between angle V R W. Angle V R T is 60 degrees and angle T, R, S is (4 x) degrees. What is the missing reason in step 3? A 2-column table with 6 rows is shown. Column 1 is labeled Statements with entries measure of angle T R V = 60 degrees and measure of angle T R X = (4 x) degrees, angle T R S and angle T R V are a linear pair, measure of angle T R S + measure of angle T R V = 180, 60 + 4 x + 180, 4 x =120, x = 30. Column 2 is labeled Reasons with entries given, definition of a linear pair, question mark, substitution property of equality, subtraction property of equality, division property of equality. substitution property of equality angle addition postulate subtraction property of equality addition property of equality
Mathematics
1 answer:
zmey [24]2 years ago
7 0

The reason behind the statement m∠TRS + m∠TRV = 180° is; Angle Addition Postulate

<h3>How to use angle addition postulate?</h3>

Angle addition postulate states that if D is the interior of ∠ABC, therefore, the sum of the smaller angles equals the sum of the larger angle, which from the attached image is;

m∠ABD + m∠DBC = m∠ABC.

From the attached image, we want to prove that x = 30°.

Now, T is the interior of straight angle ∠VRS.

m∠VRS = 180° (straight line angle)

Thus, from angle addition postulate, we can say that;

m∠TRS + m∠TRV = 180°.

Read more about two column proofs at;brainly.com/question/1788884

#SPJ1

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Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

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