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Congruence Exercise 16.5

Question: 1

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.

Solution:

i)

Congruence Exercise 16.5

∠ADC = ∠BCA = 90°

AD = BC and hyp AB = hyp AB

Therefore, by RHS ΔADB ≅ ΔACB

ii)

Congruence Exercise 16.5

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

iii)

Congruence Exercise 16.5

hyp AO = hyp DOBO = CO ∠B = ∠C = 90°

Therefore, by RHS, ΔAOB≅ΔDOC

iv)

Congruence Exercise 16.5

Hyp A = Hyp CABC = DC ∠ABC = ∠ADC = 90°

Therefore, by RHS, ΔABC ≅ ΔADC

v)

Congruence Exercise 16.5

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

 

Question: 2

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?

Solution:

(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.

 

Question: 3

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?

Solution:

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)

 

Question: 4

Draw a right triangle ABC. Use RHS condition to construct another triangle congruent to it.

Solution:

Congruence Exercise 16.5

Consider

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

 

Question: 5

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)

Congruence Exercise 16.5

Solution:

(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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