RD Sharma Solutions Class 7 Congruence Exercise 16.4

RD Sharma Class 7 Solutions Chapter 16 Ex 16.4 PDF Free Download

RD Sharma Solutions Class 7 Chapter 16 Exercise 16.4

Exercise 16.4

Q1. Which of the following pairs of triangle are congruent by ASA condition?




We have,

Since \(\angle\) ABO = \(\angle\) CDO = 45° and both are alternate angles, AB // DC, \(\angle\) BAO = \(\angle\) DCO (alternate angle, AB // CD and AC is a transversal line)

\(\angle\) ABO = \(\angle\) CDO = 45° (given in the figure) Also, AB = DC (Given in the figure)

Therefore, by ASA \(\Delta AOB \cong \Delta DOC\)



In ABC ,

Now AB =AC (Given)

\(\angle\) ABD = \(\angle\) ACD = 40° (Angles opposite to equal sides)

\(\angle\) ABD + \(\angle\) ACD + \(\angle\) BAC = 180° (Angle sum property)

40° + 40° + \(\angle\) BAC=180°

\(\angle\)  BAC =180°- 80° =100°

\(\angle\) BAD + \(\angle\) DAC = \(\angle\)  BAC \(\angle\) BAD = \(\angle\) BAC – \(\angle\) DAC = 100° – 50° = 50°

\(\angle\) BAD = \(\angle\)  CAD = 50°

Therefore, by ASA, \(\Delta ABD \cong \Delta ADC\)



In \(\Delta\) ABC,

\(\angle\) A + \(\angle\) B + \(\angle\) C = 180°(Angle sum property)

\(\angle\) C = 180°- \(\angle\) A – \(\angle\) B \(\angle\) C = 180° – 30° – 90° = 60°


\(\angle\) P + \(\angle\) Q + \(\angle\) R = 180°(Angle sum property)

\(\angle\) P = 180° – \(\angle\) Q – \(\angle\) R \(\angle\) P = 180°- 60°- 90° = 30°

\(\angle\) BAC = \(\angle\) QPR = 30°

\(\angle\) BCA = \(\angle\) PRQ = 60°and AC = PR (Given)

Therefore, by ASA, \(\Delta ABC \cong \Delta PQR\)



We have only

BC = QR but none of the angles of \(\Delta\) ABC and \(\Delta\) PQR are equal.

Therefore, \(\Delta ABC  and cong \Delta PRQ\)

Q2. In figure, AD bisects A and AD and AD \(\perp\) BC.

(i) Is \(\Delta ADB \cong \Delta ADC\)?

(ii) State the three pairs of matching parts you have used in (i)

(iii) Is it true to say that BD = DC?



(i) Yes, \(\Delta ADB \cong \Delta ADC\), by ASA criterion of congruency.

(ii) We have used \(\angle\) BAD = \(\angle\) CAD \(\angle\) ADB = \(\angle\) ADC = 90°

Since, AD \(\perp\) BC and AD = DA

(iii) Yes, BD = DC since, \(\Delta ADB \cong \Delta ADC\)

Q3. Draw any triangle ABC. Use ASA condition to construct other triangle congruent to it.



We have drawn

\(\Delta\) ABC with \(\angle\) ABC = 65° and \(\angle\) ACB = 70°

We now construct \(\Delta PQR \cong \Delta ABC\) has \(\angle\) PQR = 65° and \(\angle\) PRQ = 70°

Also we construct \(\Delta\) PQR such that BC = QR

Therefore by ASA the two triangles are congruent

Q4. In \(\Delta\) ABC, it is known that \(\angle\)  B = C. Imagine you have another copy of  \(\Delta\) ABC 

(i) Is \(\Delta ABC \cong \Delta ACB\)

(ii) State the three pairs of matching parts you have used to answer (i).

(iii) Is it true to say that AB = AC?



(i) Yes \(\Delta ABC \cong \Delta ACB\)

(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.

Q5. In Figure, AX bisects \(\angle\) BAC as well as \(\angle\)  BDC. State the three facts needed to ensure that \(\Delta ACD \cong \Delta ABD\)



As per the given conditions, \(\angle\) CAD = \(\angle\) BAD and \(\angle\) CDA = \(\angle\) BDA (because AX bisects \(\angle\) BAC )

AD = DA (common)

Therefore, by ASA, \(\Delta ACD \cong \Delta ABD\)

Q6. In Figure, AO = OB and \(\angle\) A = \(\angle\) B. 

(i) Is \(\Delta AOC \cong \Delta BOD\)

(ii) State the matching pair you have used, which is not given in the question.

(iii) Is it true to say that \(\angle\) ACO = \(\angle\)BDO?



We have

\(\angle\) OAC = \(\angle\) OBD,


Also, \(\angle\) AOC = \(\angle\) BOD (Opposite angles on same vertex)

Therefore, by ASA \(\Delta AOC \cong \Delta BOD\)

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