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In the following pairs of triangle (Figures), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic.
1) In Δ ABC and Δ DEF
AB = DE = 4.5 cm (Side)
BC = EF = 6 cm (Side) and
AC = DF = 4 cm (Side)
Therefore, by SSS criterion of congruence, ΔABC ≅ ΔDEF
2)
In Δ ACB and Δ ADB
AC = AD (Side)
BC = BD (Side) and
AB = AB (Side)
Therefore, by SSS criterion of congruence, ΔACB ≅ ΔADB
3) In Δ ABD and Δ FEC,
AB = FE (Side)
AD = FC (Side)
BD = CE (Side)
Therefore, by SSS criterion of congruence, ΔABD ≅ ΔFEC
4) In Δ ABO and Δ DOC,
AB = DC (Side)
AO = OC (Side)
BO = OD (Side)
Therefore, by SSS criterion of congruence, ΔABO ≅ ΔODC
In figure, AD = DC and AB = BC
(i) Is ΔABD ≅ ΔCBD?
(ii) State the three parts of matching pairs you have used to answer (i).
Yes ΔABD = ΔCBD by the SSS criterion. We have used the three conditions in the SSS criterion as follows:
AD = DC
AB = BC and
DB = BD
In Figure, AB = DC and BC = AD.
(i) Is ΔABC ≅ ΔCDA?
(ii) What congruence condition have you used?
(iii) You have used some fact, not given in the question, what is that?
We have AB = DC
BC = AD
and AC = AC
Therefore by SSS ΔABC ≅ ΔCDA
We have used Side congruence condition with one side common in both the triangles.
Yes, have used the fact that AC = CA.
In ΔPQR ≅ ΔEFD,
(i) Which side of ΔPQR equals ED?
(ii) Which angle of ΔPQR equals angle E?
ΔPQR ≅ ΔEFD
(i) Therefore PR = ED since the corresponding sides of congruent triangles are equal.
(ii) ∠QPR = ∠FED since the corresponding angles of congruent triangles are equal.
Triangles ABC and PQR are both isosceles with AB = AC and PO = PR respectively. If also, AB = PQ and BC = QR, are the two triangles congruent? Which condition do you use?
It ∠B = 50°, what is the measure of ∠R?
We have AB = AC in isosceles ΔABC
And PQ = PR in isosceles ΔPQR.
Also, we are given that AB = PQ and QR = BC.
Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)
Hence, ΔABC ≅ ΔPQR
Now
∠ABC = ∠PQR (Since triangles are congruent)However, ΔPQR is isosceles.
Therefore, ∠PRQ = ∠PQR = ∠ABC = 50°
ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If ∠BAC = 40° and ∠BDC = 100°, then find ∠ADB.
∠BAD = ∠CAD (c.p.c.t)
∠BAD + ∠CAD = 40°/ 2 ∠BAD = 40°
∠BAD = 40°/2 =20°
∠ABC + ∠BCA + ∠BAC = 180° (Angle sum property)
Since ΔABC is an isosceles triangle,
∠ABC = ∠BCA ∠ABC +∠ABC + 40°= 180°
2 ∠ABC = 180° – 40° = 140° ∠ABC = 140°/2 = 70°
∠DBC + ∠ BCD + ∠ BDC = 180° (Angle sum property)
Since ΔABC is an isosceles triangle, ∠ DBC = ∠BCD ∠DBC + ∠DBC + 100o = 180°
2 ∠DBC = 180° – 100o = 80°
∠DBC = 80°/2 = 40°
In Δ BAD,
∠ABD + ∠BAD + ∠ADB = 180°(Angle sum property)
30° + 20° + ∠ADB = 180° (∠ADB = ∠ABC – ∠DBC), ∠ADB = 180°- 20° – 30°
∠ADB = 130°
∠ADB =130°
Δ ABC and ΔABD are on a common base AB, and AC = BD and BC = AD as shown in Figure. Which of the following statements is true?
(i) ΔABC ≅ ΔABD
(ii) ΔABC ≅ ΔADB
(iii) ΔABC ≅ ΔBAD
In ΔABC and ΔBAD we have,
AC = BD (given)
BC = AD (given)
and AB = BA (common)
Therefore by SSS criterion of congruency, ΔABC ≅ ΔBAD
There option (iii) is true.
In Figure, ΔABC is isosceles with AB = AC, D is the mid-point of base BC.
(i) Is ΔADB ≅ ΔADC?
(ii) State the three pairs of matching parts you use to arrive at your answer.
We have AB = AC.
Also since D is the midpoint of BC, BD = DC
Also, AD = DA
Therefore by SSS condition,
ΔADB ≅ ΔADC
We have used AB, AC : BD, DC AND AD, DA
In figure, ΔABC is isosceles with AB = AC. State if ΔABC ≅ ΔACB. If yes, state three relations that you use to arrive at your answer.
Yes, ΔABC ≅ ΔACB by SSS condition.
Since, ABC is an isosceles triangle, AB = BC, BC = CB and AC = AB
Triangles ABC and DBC have side BC common, AB = BD and AC = CD. Are the two triangles congruent? State in symbolic form, which congruence do you use? Does ∠ABD equal ∠ACD? Why or why not?
Yes,
Given,
Δ ABC and Δ DBC have side BC common, AB = BD and AC = CD
By SSS criterion of congruency, ΔABC ≅ ΔDBC
No, ∠ABD and ∠ACD are not equal because AB ≠ AC
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