# RD Sharma Solutions Class 7 Congruence Exercise 16.4

## RD Sharma Solutions Class 7 Chapter 16 Exercise 16.4

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

#### Exercise 16.4

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

i)

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 $\Delta AOB \cong \Delta DOC$

ii)

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, $\Delta ABD \cong \Delta ADC$

iii)

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°

In PQR,

$\angle$ 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, $\Delta ABC \cong \Delta PQR$

iv)

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 )

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,

AO = OB

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

Therefore, by ASA $\Delta AOC \cong \Delta BOD$

#### Practise This Question

If x = a cos θ,y=b sin θ,then d3ydx3 is equal to