In this chapter, we provide RD Sharma Class 7 ex 16.4 Solutions Chapter 16 Congruence for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 16.4 Solutions Chapter 16 Congruence Maths pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 16 |

Chapter Name | Properties of Triangles |

Exercise | Ex 16.4 |

Table of Contents

**RD Sharma Solutions for Class 7 Chapter 16 ****Congruence** **Ex 16.4 Download PDF**

**Congruence**

**Congruence Exercise 16.4**

**Question: 1**

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

**Solution:**

i)

We have,

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

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

Therefore, by ASA ΔAOB ≅ ΔDOC

ii)

In ABC,

Now AB =AC (Given)

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

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

40° + 40° + ∠BAC=180°

∠BAC =180°- 80° =100°

∠BAD + ∠DAC = ∠BAC ∠BAD = ∠BAC – ∠DAC = 100° – 50° = 50°

∠BAD = ∠CAD = 50°

Therefore, by ASA, ΔABD ≅ ΔADC

iii)

In Δ ABC,

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

∠C = 180°- ∠A – ∠B ∠C = 180° – 30° – 90° = 60°

In PQR,

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

∠P = 180° – ∠Q – ∠R ∠P = 180°- 60°- 90° = 30°

∠BAC = ∠QPR = 30°

∠BCA = ∠PRQ = 60° and AC = PR (Given)

Therefore, by ASA, ΔABC ≅ ΔPQR

iv)

We have only

BC = QR but none of the angles of ΔABC and ΔPQR are equal.

Therefore, ΔABC and Cong ΔPRQ

**Question: 2**

In figure, AD bisects A and AD and AD ⊥ BC.

(i) Is ΔADB ≅ ΔADC?

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

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

**Solution:**

(i) Yes, ΔADB≅ΔADC, by ASA criterion of congruency.

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

Since, AD ⊥ BC and AD = DA

(iii) Yes, BD = DC since, ΔADB ≅ ΔADC

**Question: 3**

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

**Solution:**

We have drawn

Δ ABC with ∠ABC = 65° and ∠ACB = 70°

We now construct ΔPQR ≅ ΔABC has ∠PQR = 65° and ∠PRQ = 70°

Also we construct ΔPQR such that BC = QR

Therefore by ASA the two triangles are congruent

**Question: 4**

In Δ ABC, it is known that ∠B = C. Imagine you have another copy of Δ ABC

(i) Is ΔABC ≅ ΔACB

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

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

**Solution:**

(i) Yes ΔABC ≅ ΔACB

(ii) We have used ∠ABC = ∠ACB and ∠ACB = ∠ABC again.

Also BC = CB

(iii) Yes it is true to say that AB = AC since ∠ABC = ∠ACB.

**Question: 5**

In Figure, AX bisects ∠BAC as well as ∠BDC. State the three facts needed to ensure that ΔACD ≅ ΔABD

**Solution:**

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

AD = DA (common)

Therefore, by ASA, ΔACD ≅ ΔABD

**Question: 6**

In Figure, AO = OB and ∠A = ∠B.

(i) Is ΔAOC ≅ ΔBOD

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

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

**Solution:**

We have

∠OAC = ∠OBD,

AO = OB

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

Therefore, by ASA ΔAOC ≅ ΔBOD

**All Chapter RD Sharma Solutions For Class 7 Maths**

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