RD Sharma solution class 7 chapter 16 Congruence Exercise 16.5

Exercise 16.5

Page-16.23

Question 1:

In each of the following pairs of right triangles, the measures of some parts are indicated along side. State by the application of RHS congruence condition which are congruent. State each result in symbolic form. (Fig. 46)

Answer 1:

i) $$\begin{gathered} \text{∠ADC}= \text{∠BCA} =90° \\ \text{AD} = \text{BC} \\ \text{and hyp AB} = \text{hyp AB} \\ \text{Therefore, by RHS, }△\text{ADB}\cong △\text{ACB.} \end{gathered}$$

ii)

$$\begin{gathered} \text{AD=AD} \text{(Common)} \\ \text{hyp AC} = \text{hyp AB} \text{(Given)} \\ \text{∠ADB}+\text{∠ADC} = 180°\text{(Linear pair)} \\ \angle \text{ADB}+90°=180° \\ \angle \text{ADB}=180°-90°=90° \\ \angle \text{ADB}=\angle \text{ADC} = 90° \\ \text{Therefore,} \text{by RHS,} \text{△ADB}=\text{△ADC} \end{gathered}$$

iii)
$$\begin{gathered} \text{hyp AO} = \text{hyp DO} \\ \text{BO} = \text{CO} \\ \angle \text{B} =\angle \text{C} =90° \\ \text{Therefore,} \text{by RHS,} \text{△AOB}\cong \text{△DOC} \end{gathered}$$

iv)
$$\begin{gathered} \text{Hyp AC} = \text{Hyp CA} \\ \text{BC}\hspace{0.17em}= \text{DC} \\ \text{∠ABC}=\text{∠ADC} = 90° \\ \text{Therefore,} \text{by RHS,} \text{△ABC}\cong \text{△ADC} \end{gathered}$$

v)
$$\begin{gathered} \text{BD} = \text{DB} \\ \text{Hyp AB} =\text{Hyp BC,} \text{as per the given figure.} \\ \text{∠BDA} +\text{∠BDC} = 180° \\ \text{∠BDA} +90° = 180° \\ \text{∠BDA}= 180°-90°= 90° \\ \text{∠BDA} =\text{∠BDC} = 90° \\ \text{Therefore,} \text{by RHS,} \text{△ABD}\cong \text{△CBD} \end{gathered}$$

Page-16.24

Question 2:

∆ ABC is isosceles with AB = AC. AD is the altitude from A on BC.
(i) Is ∆ ABD ≅ ACD?
(ii) State the pairs of matching parts you have used to answer (i).
(ii) Is it true to say that BD = DC?

Answer 2:

(i)Yes, $△ABD\cong △ACD$ by RHS congruence condition.

(ii) We have used Hyp AB = Hyp AC
AD = DA
and $\angle ADB = \angle ADC = 90°$ (AD$\perp$BC at point D)

(iii)Yes, it is true to say that BD = DC (c.p.c.t) since we have already proved that the two triangles are congruent.

Question 3:

∆ ABC is isoseles with AB = AC. Also, AD ⊥ BC meeting BC in D. Are the two triangles ABD and ACD congruent? State in symbolic form. Which congruence condtion do you use? Which side of ∆ ADC equls BD? Which angle of ∆ ADC equals ∠B?

Answer 3:

We have AB = AC  ......(1)
AD = DA (common)........(2)
and $\angle ADC=\angle ADB$ (AD$\perp$BC at point D)........(3)

Therefore from 1, 2 and 3, by RHS congruence condition,
 $$\begin{gathered} \text{△ABD}\cong \text{△ACD} \\ \text{Now,} \text{the triangles are congruent }. \text{Therefore,} \text{BD}= \text{CD.} \\ \text{And} \text{∠ABD}=\text{∠ACD} \text{(c.p.c.t).} \end{gathered}$$

Question 4:

Draw a right triangle ABC. Use RHS condition to construct another triangle congruent to it.

Answer 4:

Consider
$$\begin{gathered} \text{△ABC} \text{with} \text{∠B} \text{as right angle.} \\ \text{We now construct another right triangle on base BC,} \text{such that } \\ \text{∠C} \text{is a right angle and} \text{AB} = \text{DC} \\ \text{Also,} \text{BC} = \text{CB} \\ \text{Therefore,} \text{by RHS,} \text{△ABC}\cong \text{△DCB} \end{gathered}$$

Question 5:

In Fig. 47, BD and CE are altitudes of ∆ ABC and BD = CE.
(i) Is ∆ BCD ≅ ∆ CBE?
(ii) State the three pairs of matching parts you have used to answer (i).

Answer 5:

(i) Yes, $△BCD\cong △CBE$ by RHS congruence condition.
(ii) We have used hyp BC = hyp CB
BD = CE (given in question)
and $\angle BDC = \angle CEB =90°$.