Questions and Answers #4 Congruence

Questions and Answers Sheet 4 Congruence Question #1 To decide: Whether the given information is enough to prove that △ 𝐿𝑀𝑃 ≅△ 𝑁𝑃𝑀 using SAS Congruency Theorem. If so write a proof if not then explain the reason. Given: The given figure is as follows. Answer: Calculation: Recollect the SSS Congruence Theorem: If three sides of a triangle are congruent to three sides of a second triangle, then the two triangles are congruent. From the given figure, it can be observed that 𝐿𝑀 ≅ 𝑁𝑃 . By Reflexive Property of Congruence theorem, 𝑀𝑃 ≅ 𝑃𝑀 . So, only two pairs of sides are congruent. Now need one more condition 𝐿𝑃 ≅ 𝑁𝑀 to prove the given triangles are congruent. Therefore, the △ 𝐿𝑀𝑃, and △ 𝑁𝑃𝑀 are not congruent. Question #2 To find: the name of the corresponding parts in the congruent triangles. And complete the congruence statement. Given: The congruence statement is △ 𝐶𝐴𝐵 . Answer: Calculation: According to the question, From the congruent triangle. 𝐴𝐵 = 𝐷𝐸 𝐵𝐶 = 𝐷𝐶 𝐴𝐶 = 𝐸𝐶 ∴△ 𝐴𝐵𝐶 ≅△ 𝐸𝐷𝐶 (by (CPCT)) ⇒△ 𝐶𝐴𝐵 ≅△ 𝐶𝐸𝐷 Hence, the complete statement is △ 𝐶𝐴𝐵 ≅△ 𝐶𝐸𝐷 . Question #3 To determine: The smallest nonnegative integer x that satisfies the given system of congruences. x3 ( mod 5) x3 ( mod 7 ) Answer: x3 ( mod 5) x3 ( mod 7 ) We see that the solution x is unique modulo 7.5 = 35 Now, 5(10) − 7(7) = 1. Thus, 𝑥 = 4.5(10) − 3.7(7) 𝑥 = 200 − 147 𝑥 = 53 ( mod 35) x = 18 Therefore, 𝑥 = 18. Question #4 To determine: The smallest nonnegative integer x that satisfies the given system of congruences. x 1 ( mod 4 ) x 8 ( mod 9) Answer: x 1 ( mod 4 ) x 8 ( mod 9) We see that the solution x is unique modulo 9.4 = 36. Now, 9(1) − 4(2) = 1. Thus, 𝑥 = 1.9(1) − 8.4(2) 𝑥 = 9 − 64 𝑥 = −55 x = −19 x = 17 ( mod 36) ( mod 36) Therefore, 𝑥 = 17. Question #5 To determine: The smallest nonnegative integer x that satisfies the given system of congruences. x3 ( mod 5) x7 ( mod8) Answer: x3 ( mod 5) x7 ( mod8) We see that the solution x is unique modulo 5.8 = 40. Now, 8(2) − 5(3) = 1. Thus, 𝑥 = 3.8(2) − 7.5(3) 𝑥 = 48 − 105 𝑥 = −57 ( mod 40) x = −17 ( mod 40 ) x = 23 Therefore, 𝑥 = 23. Question #6 To determine: The smallest nonnegative integer x that satisfies the given system of congruences. ( mod 8) ( mod 25) x6 x  17 Answer: x6 x  17 ( mod 8) ( mod 25) We see that the solution x is unique modulo 25.8 = 200. Now, 25(1) − 8(3) = 1. Thus, 𝑥 = 6.25(1) − 17.8(3) 𝑥 = 150 − 408 𝑥 = −258 x = −58 x = 142 ( mod 200 ) ( mod 40 ) Therefore, 𝑥 = 142. Question #7 To determine: The smallest nonnegative integer x that satisfies the given system of congruences. ( mod1917 ) x  75 ( mod 385 ) x3 Answer: ( mod1917 ) x  75 ( mod 385 ) x3 We see that the solution x is unique modulo 385.1917 = 738045. Now, 1917(48) − 385(239) = 1. Thus, 𝑥 = 75.1917(48) − 3.385(239) 𝑥 = 6901200 − 276045 𝑥 = 6625155 x = 720795 ( mod 738045) Therefore, 𝑥 = 720795. Question #8 To determine: The smallest nonnegative integer x that satisfies the given system of congruences. x  1003 x  2974 ( mod17,369 ) ( mod 5472 ) Answer: x  1003 x  2974 ( mod17,369 ) ( mod 5472 ) We see that the solution x is unique modulo 17369.5472 = 95043168. Now, 17369(−2647) − 5472(8402) = 1. Thus, 𝑥 = 2974.17369(−2647) − 1003.5472(8402) 𝑥 = 46113671232 − 136731859682 𝑥 = −90618188450 x = −52992822 ( mod 950431168) Therefore, 𝑥 = −52992822. Question #9 To determine: The smallest nonnegative integer x that satisfies the given system of congruences. ( mod 4 ) x  8 ( mod 9 ) x  10 ( mod 25 ) x 1 Answer: ( mod 4 ) x  8 ( mod 9 ) x  10 ( mod 25 ) x 1 First let us solve the first two congruences. x 1 x 8 ( mod 4 ) ( mod 9 ) We see that the solution x is unique modulo 4.9 = 36. Now, 9 − 4(2) = 1. Thus, 𝑥 = 1.9 − 8.4(2) 𝑥 = 9 − 64 𝑥 = −55 x = −19 x = 17 Therefore, ( mod 36 ) ( mod 36 ) x = 17 ( mod 36) . Now, let us solve the below equations. x  17 x  10 ( mod 36 ) ( mod 25) We see that the solution x is unique modulo 36.25 = 900. Now, 25(13) − 36(9) = 1. Thus, 𝑥 = 17.25(13) − 10.36(9) 𝑥 = 5525 − 3240 𝑥 = 2285 x = 485 ( mod 900) Therefore, 𝑥 = 485. Question #10 Complete all statements and reasons for the following proof problem. Given: ∠𝑅 and ∠𝑉 are right angles; ∠1 ≅ ∠2 Prove: △ 𝑅𝑆𝑇 ≅△ 𝑉𝑆𝑇 Answer: Step 1 Given: ∠𝑅 = ∠𝑉 = 90∘ , ∠1 ≅ ∠2 For proving given thing, we take △ 𝑅𝑆𝑇 and △ 𝑉𝑆𝑇 Step 2 So, △ 𝑅𝑆𝑇 and △ 𝑉𝑆𝑇 ∠𝑅 = ∠𝑉 (given) ∠1 = ∠2 (given) 𝑆𝑇 = 𝑆𝑇 (common) So, by ASA congruency rule \triangle 𝑅𝑆𝑇 ≅△ 𝑉𝑆𝑇 Hence, △ 𝑅𝑆𝑇 and △ 𝑉𝑆𝑇 are congruence. Question #11 Complete the following statement about congruent for the given triangles. △ 𝐷𝐶𝐸 ≅? Answer: Step 1 To check the congruence of triangles We write in the same sequence in both the congruent triangles Using the congruence test on the triangles shown We collect the equal sides and angles Step 2 Here Side 𝐷𝐶 = 𝐵𝐴 Side 𝐷𝐸 = 𝐵𝐸 Side 𝐶𝐸 = 𝐴𝐸 and ∠𝐷 = ∠𝐵 and ∠𝐶 = ∠𝐴 Hence △ 𝐷𝐶𝐸 ≅△ 𝐵𝐴𝐸 Question #12 To find: congruent criteria used to prove that the triangles are congruent. Given information: 𝐴𝐵 ≅ 𝐷𝐶 and 𝐴𝐶 ≅ 𝐷𝐵 Answer: Consider △ 𝐴𝐵𝐶,△ 𝐷𝐶𝐵 𝐴𝐵 ≅ 𝐷𝐶 (Given) 𝐴𝐶 ≅ 𝐷𝐵 (Given) 𝐵𝐶 ≅ 𝐶𝐵 (common side of △ 𝐴𝐵𝐶,△ 𝐷𝐶𝐵 ) Therefore, △ 𝐴𝐵𝐶 ≅△ 𝐷𝐶𝐵 (SSS congruence criteria) Also. Congruent triangles basically overlap each other. So, △ 𝐴𝐵𝐶,△ 𝐷𝐶𝐵 is a pair of overlapping triangles. Question #13 To show that two sides are congruent. Given information: ∠𝐻𝐺𝐽 ≅ ∠𝐾𝐽𝐺 ∠𝐾𝐺𝐽 ≅ ∠𝐻𝐽𝐺 Answer: Formula used: The below property is used: Two triangles are congruent by ASA congruence rule. Proof: From figure, we get It is given that, ∠𝐻𝐺𝐽 ≅ ∠𝐾𝐽𝐺 ∠𝐾𝐺𝐽 ≅ ∠𝐻𝐽𝐺 By reflexive property, we get 𝐺𝐽 ≅ 𝐺𝐽 By ASA congruence rule, we get △ 𝐺𝐻𝐽 ≅△ 𝐽𝐾𝐺 As corresponding parts of congruent triangles are congruent, we get 𝐻𝐺 ≅ 𝐾𝐽 Question #14 Write a paragraph proof. Given: 𝐴𝐷 and 𝐸𝐶 intersect at B. Prove: △ 𝐴𝐵𝐸 ∼△ 𝐷𝐵𝐶 Answer: Step 1:To Prove Given: 𝐴𝐷 & 𝐸𝐶 intersect at B To determine: △ 𝐴𝐵𝐸 ∼△ 𝐷𝐵𝐶 Step 2:Proof In △ 𝐴𝐵𝐸 & △ 𝐷𝐵𝐶 𝐴𝐵 𝐷𝐵 𝐵𝐸 𝐵𝐶 𝐴𝐸 𝐷𝐶 = = = 12 4 15 5 10 3 1 3 = 3...(1) = 3 ...(2) = 10 10 3 = 3 ...(3) So, we have, 𝐴𝐵 𝐵𝐸 𝐴𝐸 = = 𝐷𝐵 𝐵𝐶 𝐷𝐶 So, by SSS congruence, △ 𝐴𝐵𝐸 ∼△ 𝐷𝐵𝐶 Hence, Proved. Question #15 A Counterexample to the statement, If with 𝑚 ≥ 2 , then ab ( mod m ) , where a,b,c, and m are integers ac  bc ( mod m ) . Answer: Here we need to just find one counter example that contradicts the validity of the given statement. Suppose we set values 𝑎 = 0, 𝑏 = 1, 𝑐 = 3, 𝑚 = 4 in ac  bc ( mod m ) We get, 0  1  1 3 0=3 ( mod 3) ( mod 3) 0=0 Thus, this congruence is satisfied. Next let us plug these values in a\equiv b\pmod m We get, 0 1 ( mod 3) , which is obviously incorrect. Thus a counterexample could be the set of value 𝑎 = 0, 𝑏 = 1, 𝑐 = 3, 𝑚 = 4 Question #16 A Counterexample to the statement, If ab ( mod m ) andc  d ( mod m ) , where a,b,c,d, and m are integers with c and d positive and 𝑚 ≥ 2 , then ac  bd ( mod m ) . Answer: Here we need to just find one counter example that contradicts the validity of the given statement. Suppose we set values 𝑎 = 2, 𝑏 = 2, 𝑐 = 4, 𝑑 = 9, and 𝑚 = 5 in ab ( mod m ) We get, 22 ( mod 5) , thus this congruence is satisfied. Next let us plug these values in cd ( mod m ) We get, 49 ( mod 5) , observe that 44 ( mod 5) , again this congruence is also satisfied. Lastly we plug these values into 9 = 4 + 1 × 5 , thus ac  bd ( mod m ) We get, 24  29 16  512 ( mod 5) ( mod 5) , observe that 16 = 1 + 3 × 5 and 512 = 2 + 102 × 5 , thus 1 ( mod 5)  2 ( mod 5) , which is incorrect Thus a counterexample could be the set of values 𝑎 = 2, 𝑏 = 2, 𝑐 = 4, 𝑑 = 9, and 𝑚 = 5 Question #17 State whether the △ 𝐴𝐵𝐶 is congruent to another triangle by SSS, SAS, or ASA postulates. Given information: We have given two triangles shown below. Answer: From the given figure we see that in both the triangles △ 𝐴𝐵𝐶 and △ 𝑃𝑁𝑌 𝐴𝐶 = 𝑁𝑌 ∠𝐴 = ∠𝑁 = 90∘ But neither 𝐴𝐵 = 𝑃𝑁 nor any other angles are equal. Therefore, none of the postulates SSS, SAS or ASA hold. So we can say that no congruence can be deduced. Question #18 To Proof: If 𝐷𝐹 ≅ 𝐸𝐺 then 𝑥 = 10 Given: Given: 𝐷𝐹 ≅ 𝐸𝐺 Answer: Since 𝐷𝐹 ≅ 𝐸𝐺 Therefore by using the defination of congruence. 𝐷𝐹 = 𝐸𝐺 Thus 11 = 2𝑥 − 9 11 + 9 = 2𝑥 2𝑥 = 20 𝑥 = 10 Hence proved. Question #19 To explain: How to construct a triangle that is congruent to △ 𝑂𝐵𝑇 . Answer: Here, △ 𝑂𝐵𝑇 is an obtuse triangle as ∠𝑂𝐵𝑇 > 90∘ . According to SSS congruence criteria, if all the sides of one triangle are congruent to the corresponding sides of other triangle then the two triangles are said to be congruent. Construct △ 𝑃𝑄𝑅 such that 𝑃𝑄 ≅ 𝑂𝐵, 𝑃𝑅 ≅ 𝑂𝑇, 𝑄𝑅 ≅ 𝐵𝑇 . So, △ 𝑂𝐵𝑇 ≅△ 𝑃𝑄𝑅 (by SSS congruence criteria) Question #20 To determine:The measure of angle 6. Given: ABDC is a rectangle. 𝑚∠1 = 38 Answer: Calculation: The diagonals of a rectangle are congruent and also bisect each other. So, triangle AEC is an isosceles triangle. According to Isosceles Triangle Theorem, ∠1 ≅ ∠𝐴𝐶𝐵 . Also, according to the definition of a rectangle, ¯{𝐴𝐶}||¯{𝐵𝐷} . Using Alternate Interior Angles Theorem; ∠6 ≅ ∠𝐴𝐶𝐵 . By Transitive Property of Congruence; ∠6 ≅ ∠1 . So, 𝑚∠6 = 𝑚∠1 𝑚∠6 = 38