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Solving SBT Lesson 2: Equal triangles (C8 SBT Math 7 Horizons)
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Solution 1 page 45 SBT Math 7 Creative horizon episode 2 – CTST

In Figure 12, find triangle equal to triangle ABH

Detailed instructions for solving Lesson 1

Solution method

Check the conditions of the two triangles ABH and KBH.

Detailed explanation

Consider triangle ABH and triangle KBH with:

General insurance

\(\widehat {AHB} = \widehat {KHB} = {90^o}\)

AB = KB

Derive: \(\Delta ABH = \Delta KBH(c – g – c)\)

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Solution 2 page 45 SBT Math 7 Creative horizon episode 2 – CTST

Are the two triangles in figure 13a and 13b congruent? Why?

Detailed instructions for solving Lesson 2

Solution method

Check for congruence cases of two congruent triangles.

Detailed explanation

* Figure 13a: Consider \(\Delta ACB\) and \(\Delta ECB\) with:

AC = EC

\(\widehat {ACB} = \widehat {EC{\rm{D}}}\)

BC = DC

Derive: \(\Delta ACB = \Delta EC{\rm{D}}(c – g – c)\)

* Figure 14a: Two triangles ABC and DBC are not congruent because two triangles ABC and DBC have no corresponding corresponding angle.

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Solution 3 page 45 SBT Math 7 Creative horizon episode 2 – CTST

State more conditions for the two triangles in figure 14a, 14b to be congruent in the case of side – angle – side.’

Detailed instructions for solving Lesson 3

Solution method

Consider the conditions on two sides and the included angle to give reasonable conditions

Detailed explanation

* Figure 14a: Consider \(\Delta AB{\rm{D}}\) and \(\Delta CB{\rm{D}}\) with:

AB = BC

\(\widehat {BA{\rm{D}}} = \widehat {BC{\rm{D}}}\)

To \(\Delta AB{\rm{D}}\) = \(\Delta CB{\rm{D}}\)(c – g – c) you need to add the condition AD = CD

* Figure 14b: Consider \(\Delta KNL\) and \(\Delta MNL\) with:

General capacity

\(\widehat {KNL} = \widehat {MNL}\) ( c – g – c) then need to add the condition that KN = MN

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Solution 4 page 45 SBT Math 7 Creative horizon episode 2 – CTST

Observe figure 15 and then change the sign ? with the appropriate triangle name

a) \(\Delta MNI = \Delta ?\)

b) \(\Delta INM = \Delta ?\)

c) \(\Delta ? = \Delta QIP\)

Detailed instructions for solving Lesson 4

Solution method

Consider the conditions of congruence of two triangles

Detailed explanation

a) \(\Delta MNI = \Delta PQI\)( because there is MI = PI, MN = PQ, NI = QI)

b) \(\Delta INM = \Delta IQP\) (because IN = IQ; NM = QP, IM = IP)

c) \(\Delta NIM = \Delta QIP\) (because NI = QI, IM = IP, NM = QP)

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Solve problems 5 pages 46 SBT Math 7 Creative horizon episode 2 – CTST

Given \(\Delta ABC = \Delta D{\rm{EF}}\) and \(\widehat {{A^{}}} = {44^o}\), EF = 7 cm, ED = 15 cm . Calculate the measure \(\widehat D\) and the lengths BC, BA.

Detailed instructions for solving Lesson 5

Solution method

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From two congruent triangles, the corresponding sides and angles are congruent.

Detailed explanation

We have \(\Delta ABC = \Delta D{\rm{EF}}\) so that:

\(\widehat D = \widehat {{A^{}}} = {44^o}{,^{}}BC = {\rm{EF = 7(cm)}}{{\rm{,}} ^{}}BA = E{\rm{D}} = 15(cm)\)

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Solve lesson 6 page 46 SBT Math 7 Creative horizon episode 2 – CTST

Are the pairs of triangles in Figure 16 congruent? If so, under what circumstances are they equal?

Detailed instructions for solving Lesson 6

Solution method

Check the conditions according to the congruence cases of two triangles

Detailed explanation

a) \(\Delta A{\rm{E}}B = \Delta CF{\rm{D}}\)according to the case of alignment – ​​angle – side

b) \(\Delta ABE = \Delta C{\rm{D}}F\) in the corner – side – angle case

c) \(\Delta ABE = \Delta C{\rm{D}}F\)in the edge-to-edge case

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Solution 7 page 46 SBT Math 7 Creative horizon episode 2 – CTST

Let \(\Delta ABC = \Delta D{\rm{EF}}\) and AB = 9 cm, AC = 7 cm, EF = 10 cm. Calculate the perimeter of triangle ABC.

Detailed instructions for solving Lesson 7

Solution method

From two congruent triangles, the corresponding sides and angles are congruent.

Detailed explanation

Since \(\Delta ABC = \Delta D{\rm{EF}}\) BC = EF = 10 (cm)

Therefore, the perimeter of triangle ABC is: 9 + 7 + 10 = 26 (cm)

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Solve problem 8 page 46 SBT Math 7 Creative horizon episode 2 – CTST

Given triangle ABC with AB = AC, take a point M on side BC such that BM = CM. Prove that triangles ABM and ACM are congruent.

Detailed instructions for solving Lesson 8

Solution method

Consider the angle conditions to see in which case two triangles are congruent?

Detailed explanation

Consider two triangles ABM and ACM with:

AB = AC

BM = CM (assumption)

AM is the common edge

Derive: \(\Delta ABM = \Delta ACM(c – c – c)\)

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Solve problem 9 page 46 SBT Math 7 Creative horizon episode 2 – CTST

Given the angle xOy. Take two points A and B on the ray Ox such that OA < OB. Take two points C and D on the ray Oy such that OA = OC, OB = OD. Let M be the intersection of AD and CB. Prove that:

a) AD = CB

b) \(\Delta MAB = \Delta MC{\rm{D}}\)

Detailed instructions for solving Lesson 9

Solution method

Prove that two triangles have two corresponding sides congruent.

– Considering the conditions of two triangles MAB and MCD, in what cases are those two triangles congruent?

Detailed explanation

a) Consider triangle AOD and triangle COB with:

OA = OC

\(\widehat O\) common

OB = OD

Derive: \(\Delta AO{\rm{D}} = \Delta COB(c – g – c)\)

b) We have: \(\Delta AO{\rm{D}} = \Delta COB(c – g – c)\)(prove above)

Consider triangle MAB and triangle MCD with:

Thus: \(\widehat {MBA} = \widehat {M{\rm{D}}C};\widehat {MAB} = \widehat {MC{\rm{D}}}\) equal)

AB = CD (because OA = OC, OB = OD)

Derive: \(\Delta MAB = \Delta MC{\rm{D(g – c – g)}}\)

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