The DAV Maths Class 7 Solutions and DAV Class 7 Maths Chapter 9 Worksheet 2 Solutions of Construction of Triangles offer comprehensive answers to textbook questions.
DAV Class 7 Maths Ch 9 WS 2 Solutions
Question 1.
In ∆ ABC and ∆ DEF, AB = DE and BC = EF (See fig.). What additional information is required to make the two triangles congruent by SAS congruence condition ?
Answer:
In ∆ ABC and ∆ DEF,
AB = DE and BC = EF
included ∠ABC = ∠DEF
Hence, additional condition for congruency required is ∠ABC must be equal to
Question 2.
∆ ABC is isosceles with AB = AC (See fig.).
Line segment AD bisects ∠A and meets the base BC at D. Find the third pair of corresponding parts which make ∆ ADB ≅ ∆ ADC by SAS congruence condition. Is it true to say that BD = CD? Why?
Answer:
In ∆ ABD and ∆ ACD,
AB = AC [given]
AD = AD
And ∠B AD = ∠CAD
∆ABD = ∆ ACD
Hence the third condition required is ∠BAD = ∠CAD.
Question 3.
In the given figure, AB ∥ DC and AB = DC.
(i) Is ∠BAC = ∠DCA? Why?
Answer:
AB ∥ DC and AC is transversal (given)
∴ ∠BAC = ∠ACD (alternate angles)
AB = DC (given)
AC = AC (Common)
(ii) Is ∆ABC ≅ ∆CDA by SAS congruence condition?
Answer:
∴ ∆ABC ≅ ∆CDA (By SAS)
(iii) State the three facts that you have used to answer (ii).
Answer:
(a) AB = DC (given)
(b) AC = AC (Common)
(c) ∠BAC = ∠ACD (Alternate angles)
Question 4.
In the given figure, which pairs of triangles are congruent by SAS congruence condition?
Answer:
In ∆ABC and ∆ DEF,
AB = DE = 3 cm (given)
BC = DF = 4 cm (given)
∠ABC = ∠EDF = 70° (given)
∆ABC ≅ ∆DEF (By SAS)
Answer:
In ∆PQS and ∆RQS,
PS = QR = 4 cm (given)
QS = QS = 5.8 cm (given)
∠P = ∠R = 90° (given)
∆PQS ≅ ∆RQS (By RHS)
Answer:
In ∆PQR and ∆XYZ,
PR = XY = 4 cm (given)
QR = YZ = 2 cm (given)
∠PRQ = ∠X YZ = 55° (given)
∆PQR ≅ ∆XYZ (By SSS)
Answer:
In ∆ABD and ∆ABC,
AD = BC (given)
DB = AC (given)
AB = AB (Common)
∆ABD ≅ ∆ABC (By SAS)
Answer:
In ∆PQS and ∆RQS,
PS = QR = 4 cm (given)
QS = QS = 5.8 cm (given)
∠P = ∠R = 90° (given)
∆PQS ≅ ∆RQS (By RHS)
Answer:
In ∆ABD and ∆ABC,
AD = BC (given)
DB = AC (given)
AB = AB (Common)
∆ABD ≅ ∆ABC (Common)
Question 5.
In the given figure, AB = AD and ∠BAC = ∠DAC
(i) State in symbolic form, the congruence of two triangles ABC and ADC is true. Also, state the congruence condition used.
Answer:
In ∆ABC and ∆ADC,
AB = AD (given)
∠BAC = ∠DAC (given)
AC = AC (Common)
∆ABC ≅ ∆ADC (Common)
(ii) Complete each of the following, so as to make it true.
(a) ∠ABC = ________________
(b) ∠ACD = ________________
(c) Line segment AC bisects ________________ and ________________
Answer:
(a) ∠ABC = ∠ADC
(b) ∠ACD = ∠ACB
(c) ∠BAD and ∠BCD