In this chapter, we provide RD Sharma Class 7 ex 16.5 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 16.5 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 16.5 Download PDF
Congruence Exercise 16.5
In each of the following pairs of right triangles, the measures of some part are indicated alongside. State by the application of RHS congruence conditions which are congruent, and also state each result in symbolic form.
∠ADC = ∠BCA = 90°
AD = BC and hyp AB = hyp AB
Therefore, by RHS ΔADB ≅ ΔACB
AD = AD (Common)
hyp AC = hyp AB (Given)
∠ADB + ∠ADC = 180° (Linear pair)
∠ADB + 90° = 180°
∠ADB = 180° – 90° = 90°
∠ADB = ∠ADC = 90°
Therefore, by RHS Δ ADB = Δ ADC
hyp AO = hyp DOBO = CO ∠B = ∠C = 90°
Therefore, by RHS, ΔAOB≅ΔDOC
Hyp A = Hyp CABC = DC ∠ABC = ∠ADC = 90°
Therefore, by RHS, ΔABC ≅ ΔADC
BD = DB Hyp AB = Hyp BC, as per the given figure,
∠BDA + ∠BDC = 180°
∠BDA + 90° = 180°
∠BDA= 180°- 90° = 90°
∠BDA = ∠BDC = 90°
Therefore, by RHS, ΔABD ≅ ΔCBD
Δ ABC is isosceles with AB = AC. AD is the altitude from A on BC.
i) Is ΔABD ≅ ΔACD?
(ii) State the pairs of matching parts you have used to answer (i).
(iii) Is it true to say that BD = DC?
(i) Yes, ΔABD ≅ ΔACD by RHS congruence condition.
(ii) We have used Hyp AB = Hyp AC
AD = DA
∠ADB = ∠ADC = 90° (AD ⊥ 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.
ΔABC is isosceles with AB = AC. Also. AD ⊥ BC meeting BC in D. Are the two triangles ABD and ACD congruent? State in symbolic form. Which congruence condition do you use? Which side of ADC equals BD? Which angle of Δ ADC equals ∠B?
We have AB = AC …… (i)
AD = DA (common) ……(ii)
And, ∠ADC = ∠ADB (AD ⊥ BC at point D) ……(iii)
Therefore, from (i), (ii) and (iii), by RHS congruence condition, ΔABD ≅ ΔACD, the triangles are congruent.
Therefore, BD = CD.
And ∠ABD = ∠ACD (c.p.c.t)
Draw a right triangle ABC. Use RHS condition to construct another triangle congruent to it.
Δ ABC with ∠B as right angle.
We now construct another triangle on base BC, such that ∠C is a right angle and AB = DC
Also, BC = CB
Therefore, BC = CB
Therefore by RHS, ΔABC ≅ ΔDCB
In figure, BD and CE are altitudes of Δ ABC and BD = CE.
(i) Is ΔBCD ≅ ΔCBE?
(ii) State the three pairs or matching parts you have used to answer (i)
(i) Yes, ΔBCD ≅ ΔCBE by RHS congruence condition.
(ii) We have used hyp BC = hyp CB
BD = CE (Given in question)
And ∠BDC = ∠CBE = 90o
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