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Maru [420]
3 years ago
5

Cai tried to prove that \triangle FGH\cong \triangle HIJ△FGH≅△HIJtriangle, F, G, H, \cong, triangle, H, I, J. Statement Reason 1

FG=HI=6FG=HI=6F, G, equals, H, I, equals, 6 Given 2 FH=HJ=4FH=HJ=4F, H, equals, H, J, equals, 4 Given 3 \overline{FG} \parallel \overline{HI} FG ∥ HI start overline, F, G, end overline, \parallel, start overline, H, I, end overline Given 4 \angle HFG\cong\angle JHI∠HFG≅∠JHIangle, H, F, G, \cong, angle, J, H, I When a transversal crosses parallel lines, alternate interior angles are congruent. 5 \triangle FGH\cong \triangle HIJ△FGH≅△HIJtriangle, F, G, H, \cong, triangle, H, I, J Side-angle-side congruence What is the first error Cai made in his proof? Choose 1 answer: Choose 1 answer: (Choice A) A Cai used an invalid reason to justify the congruence of a pair of sides or angles. (Choice B) B Cai only established some of the necessary conditions for a congruence criterion. (Choice C) C Cai established all necessary conditions, but then used an inappropriate congruence criterion. (Choice D) D Cai used a criterion that does not guarantee congruence.
Mathematics
1 answer:
alexira [117]3 years ago
5 0

The first error made by Cai in his proof was: the use of an invalid reason to <em>justify </em>the congruence of a pair of sides or angles. (<em>Option A</em>)

<em><u>Recall:</u></em>

  • If two triangles have two pairs of congruent sides and a pair of included side that are congruent, both triangles are considered congruent by the Side-Angle-Side Congruence Theorem (SAS).

<em>△FGH and △HIJ are shown in the diagram attached below. Also, the diagram shows the </em><em>proof </em><em>of Cai.</em>

<em />

The first error that can be noted is that Cai the reason given for stating that \angle HFG \cong \angle JHI is false and invalid.

<HFG and <JHI are corresponding angles which makes them congruent. They are not alternate interior angles.

  • Therefore, the first error made by Cai in his proof was: the use of an invalid reason to <em>justify </em>the congruence of a pair of sides or angles. (<em>Option A</em>)

Learn more here:

brainly.com/question/13408604

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Answer:

1- The solution of I2x + 5I = 9 is {-7 , 2}

2- The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- The solution of I5 - xI = 6 is {-1 , 11}

4- The solution of I6x - 8I + 7 = 5 is ∅

5- The solution of Ix + 3I = 12 is {-15 , 9}

6- The solution of Ix - 3I = -12 is ∅

Step-by-step explanation:

* At first lets explain the meaning of IxI = a

- If IxI = a ⇒ then x = a or x = -a

- IxI never give a negative answer, because IxI means the

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Ex: I-2I is 2

* Now lets find the solution of each equation

1- ∵ I2x + 5I = 9

∴ 2x + 5 = 9 ⇒ subtract 5 from both sides

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2- ∵ I2x + 7I + 2 = 11 ⇒ Subtract 2 from both sides

∴ I2x + 7I = 9

∴ 2x + 7 = 9 ⇒ subtract 7 from both sides

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∴ 2x + 7 = -9 ⇒ subtract 7 from both sides

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* The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- ∵ I5 - xI = 6

∴ 5 -x = 6 ⇒ subtract 5 from both sides

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OR

∴ 5 -x = -6 ⇒ subtract 5 from both sides

∴ -x = -11 ⇒ divide both sides by -1

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* The solution of I5 - xI = 6 is {-1 , 11}

4- ∵ I6x - 8I + 7 = 5 ⇒ Subtract 7 from both sides

∴ I6x - 8I = -2

- I  I never give negative answer

* The solution of I6x - 8I + 7 = 5 is ∅

5- ∵ Ix + 3I = 12

∴ x + 3 = 12 ⇒ subtract 3 from both sides

∴ x = 9

OR

∴ x + 3 = -12 ⇒ subtract 3 from both sides

∴ x = -15

* The solution of Ix + 3I = 12 is {-15 , 9}

6- ∵ Ix - 3I = -12

- I  I never give negative answer

* The solution of Ix - 3I = -12 is ∅

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