RD Sharma solution class 7 chapter 16 Congruence Exercise 16.4

Exercise 16.4

Page-16.19

Question 1:

Which of the following pairs of triangles are congruent by ASA condition?

 

Answer 1:

1)  We have

$$\begin{gathered} \text{Since ∠ABO = ∠CDO} = 45° \text{and both are alternate angles}, \\ \text{AB}\parallel \text{DC} \\ \text{∠BAO} = \text{∠DCO} (\text{alternate angle }, \text{AB}\parallel \text{CD} \text{and AC is a transversal line)} \\ \text{∠ABO} = \text{∠CDO} = 45°\text{ (given in the figure)} \\ \text{Also, AB} = \text{DC} (\text{Given in the figure}) \\ \text{Therefore}, \text{by ASA} △\text{AOB} \cong △\text{DOC} \end{gathered}$$

2)  
$$\begin{gathered} \text{In △ABC ,} \\ \text{Now AB = AC (Given)} \\ \text{∠ABD = ∠ACD = 40° (Angles opposite to equal sides)} \\ \text{∠ABD +∠ACD+∠BAC=180° (Angle sum property)} \\ \text{40°+40°+∠BAC=180°} \\ \text{∠BAC=180°-80°=100°} \\ \text{∠BAD +∠DAC=∠BAC} \\ \text{∠BAD=∠BAC-∠DAC=100°-50°=50°} \\ \text{∠BAD =∠CAD = 50°} \\ \text{Therefore, by ASA, △ABD ≅△ADC} \end{gathered}$$
3)
   $$\begin{gathered} \text{In∆ABC,} \\ \text{∠A+∠B+∠C=180°(Angle sum property)} \\ \text{∠C=180°-∠A-∠B} \\ \text{∠C=180°-30°-90°=60°} \\ \text{In∆PQR,} \\ \text{∠P+∠Q+∠R=180°(Angle sum property)} \\ \text{∠P=180°-∠Q-∠R} \\ \text{∠P=180°-60°-90°=30°} \\ \text{∠BAC = ∠QPR = 30°} \\ \text{∠BCA=∠PRQ = 60°} \\ \text{and AC = PR (Given)} \\ \text{Therefore, by ASA, △ABC ≅△PQR} \end{gathered}$$

4)  
$$\begin{gathered} \text{We have only BC =QR but none of the angles of △ABC AND △PQR are equal.} \\ \text{Therefore, △ABC≇ △PRQ} \end{gathered}$$

Question 2:

In Fig. 37, AD bisects ∠A and AD ​⊥ BC.
(i) Is ∆ ADB ≅ ∆ ADC?
(ii) State the three pairs of matching parts you have used in (i).
(iii) Is it true to say that BD = DC?

Answer 2:

$$\begin{gathered} \text{(i) Yes, △ADB} \cong \text{△ADC, by ASA criterion of congruency} \\ \text{(ii) }\text{We have used }\text{∠BAD}\mathit{ }\mathit{=}\mathit{\text{∠}}\text{CAD} \\ \text{∠ADB=∠ADC = 90° since AD⊥BC} \\ \text{and AD = DA} \\ \text{(iii) Yes, BD\hspace{0.17em}= DC since, △ADB ≅△ADC} \end{gathered}$$

Page-16.20

Question 3:

Draw any triangle ABC. Use ASA condition to construct another triangle congruent to it.

Answer 3:

We have drawn
$$\begin{gathered} △\mathrm{ABC} \mathrm{with} \angle \mathrm{ABC} = 60°\mathrm{and} \angle \mathrm{ACB} = 70° \\ \mathrm{We} \mathrm{now} \mathrm{construct} △\mathrm{PQR}\cong △\mathrm{ABC} \\ △\mathrm{PQR} \mathrm{has} \angle \mathrm{PQR} =60° \mathrm{and} \angle \mathrm{PRQ} = 70° \\ \mathrm{Also} \mathrm{we} \mathrm{construct} △\mathrm{PQR} \mathrm{such} \mathrm{that} \mathrm{BC} =\mathrm{QR} \\ \mathrm{Therefore} \mathrm{by} \mathrm{ASA} \mathrm{the} \mathrm{two} \mathrm{triangles} \mathrm{are} \mathrm{congruent} \end{gathered}$$

Question 4:

In ∆ ABC, it is known that ∠B = ∠C. Imagine you have another copy of ∆ ABC
(i) Is ∆ ABC ≅ ∆ ACB?
(ii) State the three pairs of matching parts you have used to answer (i).
(iii) Is it true to say that AB = AC?

Answer 4:

(i) Yes $△ABC \cong △ACB$.
(ii) We have used $\angle ABC=\angle ACB and \angle ACB =\angle ABC again$.
Also BC = CB


(iii) Yes, it is true to say that AB = AC since $\angle ABC=\angle ACB$.

Question 5:

In Fig. 38, AX bisects ∠BAC as well as ∠BDC. State the three facts needed to ensure that ∆ ABD ≅ ∆ ACD.

Answer 5:

$$\begin{gathered} \text{As per the given conditions,∠CAD=∠BAD} \\ \text{and ∠CDA=∠BDA (because AX bisects ∠BAC )} \\ \text{AD=DA (common)} \\ \text{Therefore, by ASA, △ACD≅△ABD} \end{gathered}$$

Question 6:

In Fig. 39, AO = OB and ∠A = ∠B.


(i) Is ∆ AOC ≅ ∆ BOD?
(ii) State the matching pair you have used, which is not given in the question.
(iii) Is it true to say that ∠ACO = ∠BDO?

Answer 6:

We have
$$\begin{gathered} \text{∠OAC =∠OBD, AO = OB} \\ \text{Also, ∠AOC = ∠BOD (Opposite angles on same vertex) } \\ \text{Therefore, by ASA △AOC ≅△BOD} \end{gathered}$$