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shutvik [7]
3 years ago
7

State whether the triangles could be proven congruent, if possible, by SSS or SAS.

Mathematics
1 answer:
Helen [10]3 years ago
3 0

Answer:

1. SSS

2. SAS

3. SAS

4. SSS

5. SAS

6. SAS

Step-by-step explanation:

1) Side FY ≅ Side CW  Given

Side FP ≅ Side CM     Given

Side YP ≅ Side MW    Given

∴ΔMCW ≅ ΔFPY by the Side-Side-Side (SSS) rule of congruency

2) ∠CBD ≅ ∠BCA given that both are alternate interior angles

Side EB ≅ Side EC and Side DB ≅ Side CA Given

ΔBED ≅ ΔAEC by Side-Angle-Side (SAS) rule of congruency

3. ∠SVU ≅ ∠SVT   Given

Side SV ≅ Side SV  by reflexive property

Side VT ≅ Side VU   Given

∴ ΔVSU ≅ ΔVST by Side-Angle-Side (SAS) rule of congruency

4. Side MN ≅ Side QP  Given

Side MQ ≅ Side NP     Given

Side NQ ≅ Side NQ    by reflexive property

∴ΔQNM ≅ ΔQNP by the Side-Side-Side (SSS) rule of congruency

5. Indirect proof

Side GL ≅ Side HL  Given

Side GJ ≅ Side HK     Given

∠JLG ≅ ∠HLK    vertically opposite angles

By sine rule

GJ/sin(∠JLG) = GL/n(∠GJL)

Similarly

HK/sin(∠HLK) = HL/n(∠HKL)

∴ ∠GJL ≅ ∠HKL

∴ ∠LGJ ≅ ∠LHK third angle of two triangles given the other two angles are congruent

∴ΔQNM ≅ ΔQNP by the Side-Angle-Side (SAS) rule of congruency

6. ∠XZY ≅ ∠XZW   Supplementary ∠s with ∠XZY = 90°

Side XZ ≅ Side XZ  by reflexive property

Side ZW ≅ Side ZY   Given

∴ ΔXYZ ≅ ΔXWZ by Side-Angle-Side (SAS) rule of congruency.

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                   \large\boxed{\large\boxed{161}}

Explanation:

Let's put the information in a table step-by step.

                                                (number of remaining students)

                                                        Juniors          Seniors

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  • Initially                                           J                     S
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  • Twice juniors as seniors         2(S - 15)
  • 3/4 of the juniors left              1/4×2(S - 15)
  • 1/3 of seniors left                                             2/3×(S - 15)

At the end, there were 8 more seniors than juniors:

  • 2/3×(S - 15) -  1/4×2(S - 15) = 8

Now you have obtained one equation, which you can solve to find S, the number of senior students, and then the number of junior students.

Solve the equation:

2/3\times (S - 15) -  1/4\times 2(S - 15) = 8

  • Mutilply all by 12:

8(S - 15)-6(S - 15)=96

  • Distribution property:

8S-120-6S-90=96

  • Addtion property of equalities:

8S-6S=96+120+90

  • Add like terms:

2S=306

  • Division property of equalities:

S=306/2=153

That is the number of senior students that came out to the information meeting, but the number of students remaining to perform in the school musical is (from the table above):

2/3\times (S-15)+1/4\times 2(S-15)

Just substitute S with 153 fo find the number of students that remained to perfom in the musical:

          2/3\times (153-15)+1/4\times 2(153-15)\\ \\ 2/3(138)+1/2(138)

          161

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