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.
|Chapter Name||Properties of Triangles|
RD Sharma Solutions for Class 7 Chapter 16 Congruence Ex 16.4 Download PDF
Congruence Exercise 16.4
Which of the following pairs of triangle are congruent by ASA condition?
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
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
In Δ ABC,
∠A + ∠B + ∠C = 180°(Angle sum property)
∠C = 180°- ∠A – ∠B ∠C = 180° – 30° – 90° = 60°
∠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
We have only
BC = QR but none of the angles of ΔABC and ΔPQR are equal.
Therefore, ΔABC and Cong ΔPRQ
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?
(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
Draw any triangle ABC. Use ASA condition to construct other triangle congruent to it.
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
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?
(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.
In Figure, AX bisects ∠BAC as well as ∠BDC. State the three facts needed to ensure that ΔACD ≅ ΔABD
As per the given conditions, ∠CAD = ∠BAD and ∠CDA = ∠BDA (because AX bisects ∠BAC)
AD = DA (common)
Therefore, by ASA, ΔACD ≅ ΔABD
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?
∠OAC = ∠OBD,
AO = OB
Also, ∠AOC = ∠BOD (Opposite angles on same vertex)
Therefore, by ASA ΔAOC ≅ ΔBOD
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