Selina Concise Class 7 Math Chapter 19 Congruency: Congruent Triangles Exercise 19 Solutions
EXERCISE 19
(1) (i) In these triangles, corresponding sides are not equal. Hence, these are not congruent triangles.
(ii) In the first triangle,
Third angle = 180° – (40° + 30)
= 180° – 70° = 110°
Since, two sides of one triangle are equal to the two sides of the other triangle and the included angles are equal.
∴ The triangle are congruent by S.A.S.
(iii) In these triangles, corresponding sides are not equal. Hence, these are not congruent triangles.
(iv) Since, three sides of one triangle are equal to the three sides of the other triangle.
∴ The triangle are congruent by S.S.S.
(v) In these right triangles, one side and diagonal of the one are equal to the corresponding and diagonal are equal. Hence, these are congruent triangles by R.H.S.
(vi) In these triangles two sides and one angle of the one are equal to the corresponding sides and one angle of the other are equal. Hence, these are congruent triangles by S.S.A.
(vii)
(2) Proof:
In ∆ABD and ∆ACD,
AD = AD (common)
AB = BC (given)
BD = DC (given)
∴ ∆ABD ≅ ∆ACD (by S.S.S) (Proved)
(3) Given: In figure,
AB = AD, CB = CD
To prove: ∆ABC ≅ ∆ADC
∠B = ∠D
AC bisects angle DCB
Proof: In ∆ABC and ∆ADC,
AC = AC (common)
AB = AD (given)
CB = CD (given)
(i) ∴ ∆ABC ≅ ∆ADC (By SSS)
∴ ∠B = ∠D
∠BCA = ∠DCA
∴ AC bisects ∠DCB.
(4) Given: In figure,
AD = AC
BD = CD
To prove:
(i) ∆ ABD ≅ ∆ACD
(ii) ∠B = ∠C
(iii) ∠ADB = ∠ADC
(iv) ∠ADB = 90°
(i) ∆ ABD ≅ ∆ACD (By SSS)
(ii) ∠B = ∠C (c.p.c.t)
(iii) ∠ADB = ∠ADC
But ∠ADB + ∠ADC = 180° (linear pair)
∴ ∠ADB = ∠ADC = 180°/2 = 90°
(5) (i) IN ∆ACB and ∆ECD
AC = CE (given)
∠ACB = ∠DCE (vertically opposite angles)
BC = CD (given)
∴ ∆ACB ≅ ∆ECD (by SAS)
(ii) Hence, AB = ED (c.p.c.t.) (Proved)
(11) Proof:
In ∆ABC and ∆BCD,
CB = CB (common)
∠ABC = ∠BCD (each 90o)
AB = CD (given)
(i) ∆ABC ≅ ∆DCB (By S.A.S.)
(ii) AC = DB (c.p.c.t) (Proved)
(12) Proof:
In ∆AOD and ∆BOC,
OA = OB (given)
∠AOD = ∠BOC (vertically opposite)
OD = OC (given)
(i) ∆AOD ≅ ∆BOC (by S.A.S)
(ii) AD = BC (c.p.c.t)
(iii) ∠ADB = ∠ACB (c.p.c.t)
(iv) ∆ADB ≅ ∆BCA
∆ADB = ∆BCA (given)
AB = AB (common)
∴ ∆ADB ≅ ∆BCA (Proved)
(13) Consider ∆ADB and ∆CEB:
AB = BC (sides of same equilateral triangle)
∠ADB = ∠BEC (each 90o)
∠ABD = ∠BCE (each 60o)
∴∆ADB ≅ ∆BEC (By A.A.S)
Hence, (i) AD = BE (c.p.c.t)
(ii) BD = CE (c.p.c.t) (Proved)
(14) In figure AB = BC, AD = DC
∠ABD = 50o, ∠ADB = y – 7o
∠CBD = x + 5o, ∠CDB = 38o
To find: The value of x and y
In ∆ABD and ∆CBD
BD = BD (common)
AB = BC (given)
AD = CD (given)
∴ ∆ABD ≅ ∆CBD (By S.S.S)
∴ ∠ABD = ∠CBD
⇒ 50° = x + 5°
⇒ x + 5° = 50°
⇒ x = 50° – 5° = 45°
And ∠ADB = ∠CDB
⇒ y – 7° = 38°
⇒ y = 38° + 7°
⇒ y = 45°
(15) In ∆ABC
AD ⊥ BC, BD = CD (given)
In ∆ABD and ∆ACD
AD = AD (common)
∠ADB = ∠ADC (each 90°)
BD = CD (given)
(i) ∴ ∆ABD ≅ ∆CAD (by S.A.S)
(ii) ABV = AC (c.p.c.t.)
(iii) ∠B = ∠C (proved)
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