[3 marks sum]

Question 13 Marks

In the given figure prove that:
(i) $P Q=R S$
(ii) $P S=Q R$

Answer

Proof:
In Δ PQR and Δ PSR,
PR = PR ..........(common)
∠PRQ = ∠RPS ............(given)
∠PQR = ∠PSR .......(given)
∴ Δ PQR ≅ Δ PSR .................(A.A.S. Axiom)
Hence
(i) PQ = RS .............(c.p.c.t.)
(ii) QR = PS ...............(c.p.c.t.)
or PS = QR
Hence proved.

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Question 23 Marks

In the given figure ;
$
\angle 1=\angle 2 \text { and } A B=A C \text {. }
$
Prove that:
(i) $\angle B=\angle C$
(ii) $BD = DC$
(iii) $A D$ is perpendicular to $B C$.

Answer

Proof:
In Δ ADB and Δ ADC,
AB = AC .........(given)
∠1 = ∠2 ............(given)
AD = AD .............(common)
∴ Δ ADB ≅ Δ ADC ................(S.A.S. Axiom)
(i) Hence ∠B = ∠C ..............(c.p.c.t.)
(ii) BD = DC ...............(c.p.c.t.)
(iii) and ∠ADB = ∠ADC ............(c.p.c.t.)
But ∠ADB + ∠ADC = 180° ....(Linear pair)
∴ ∠ADB = ∠ADC = 90°
Hence AD is perpendicular to BC.
Hence proved.

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Question 33 Marks

In the given figure, prove that: $\triangle A B D \cong \triangle A C D$

Answer

Proof:
In Δ ABD and Δ ACD,
AD = AD ..............(common)
AB = AC ...............(given)
BD = DC ...............(given)
∴ Δ ABD ≅ Δ ACD .............(S.S.S. Axiom)
Hence proved.

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Question 43 Marks

The given figure shows a triangle $A B C$ in which $A D$ is perpendicular to side $B C$ and $B D=C D$. Prove that:
(i) $\triangle A B D \cong \triangle A C D$
(ii) $A B=A C$
(iii) $\angle B=\angle C$

Answer

(i) In the given figure Δ ABC
AD ⊥ BC, BD = CD
In Δ ABD and Δ ACD
AD = AD ............(common)
∠ADB = ∠ADC ...............(each 90°)
BD = CD ...........(Given)
∴ Δ ABD ≅ Δ CAD .........(By SAS Rule)
(ii) Side AB = AC .........(c.p.c.t.)
(iii) ∠B = ∠C
Reason, since Δ ADB ≅ Δ ADC
∴ ∠B = ∠C
Hence proved.

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Question 53 Marks

State, whether the pairs of triangles given in the following figures are congruent or not:

Answer

In the first A, third angle
= 180° – (40°+ 30°)
= 180° – 70°
= 110°
Now in these two triangles the sides and included angle of the one are equal to the corresponding sides and included angle.
Hence these are congruent triangles
(S.A.S. axiom)

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Question 63 Marks

In the given figure, prove that:
(i) $\triangle ABC \cong \triangle DCB$
(ii) $AC = DB$

Answer

Proof:
In Δ ABC and Δ DCB,
CB = CB ...........(common)
∠ABC = ∠BCD ..........(each 90°)
and AB = CD .............(given)
(i) ∴ Δ ABC ≅ Δ DCB ..............(S.A.S. Axiom)
(ii) Hence AC = DB ............(c.p.c.t.)
Hence proved.

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Question 73 Marks

(i) $\triangle X Y Z \cong \triangle X P Z$
(ii) $Y Z=P Z$
(iii) $\angle Y X Z=\angle P X Z$

Answer

In right Δ XYZ and Δ XPZ,
Side XY = SIde XP .......(given)
Hypotenuse XZ = hypotenuse XZ .........(common)
(i) ∴ Δ XYZ ≅ Δ XPZ ...........(R.H.S. Axiom)
Hence (ii) YZ = PZ .............(c.p.c.t.)
and (iii) ∠YXZ = ∠PXZ ...........(c.p.c.t.)
Hence proved.

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MATHS [3 marks sum]

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