[4 marks sum]

Question 14 Marks

State, whether the pairs of triangles given in the following figures are congruent or not:
Δ ABC in which AB = 2 cm, BC = 3.5 cm and ∠C = 80° and Δ DEF in which DE = 2 cm, DF = 3.5 cm and ∠D = 80°.

Answer

In $\triangle A B C, A B=2 cm , B C=3.5 cm$ and $\angle C=80^{\circ}$ and in $\triangle D E F, D E=$ $2 cm , DF =3.5 cm$ and $\angle D =80^{\circ}$

From the figure, we see that two corresponding sides are equal but their included angles are not equal.
Hence, these are not congruent triangles

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Question 24 Marks

Use the information given in the following figure to prove triangles $A B D$ and $C B D$ are congruent. Also, find the values of $x$ and $y$.

Answer

Given: In the figure AB = BC, AD = DC
∠ABD = 50, ∠ADB = y − 7°
∠CBD = x + 5°, ∠CDB = 38°
To find: The value of x and y
In Δ ABD and Δ CBD
BD = BD ........(common)
AB = BC ........(given)
AD = CD ...........(given)
∴ Δ ABD ≅ Δ CBD .........(SSS axiom)
∴ ∠ABD = ∠CBD
⇒ 50 = x + 50°
⇒ x = 50° − 5° = 45°
and ∠ADB = ∠CDB
⇒ y − 7° = 38°
⇒ y = 38° + 7° = 45°
Hence x = 45°, y = 45°

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Question 34 Marks

$A B C$ is an equilateral triangle, $A D$ and $B E$ are perpendiculars to $B C$ and $A C$ respectively. Prove that:
(i) $AD = BE$
(ii) $BD = CE$

Answer

In Δ ABC,
AB = BC = CA,
AD ⊥ BC, BE ⊥ AC.
Proof: In Δ ADC and Δ BEC
∠ADC = ∠BEC ...........(each 90°)
∠ACD = ∠BCE ..............(common)
and AC = BC ............(sides of an equilateral triangle)
∴ Δ ADC ≅ Δ BEC ...........(A.A.S. Axiom)
Hence (i) AD = BE ...........(c.p.c.t.)
and (ii) BD = CE ...........(c.p.c.t.)
Hence proved.

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Question 44 Marks

In the given figure, prove that:
(i) $\triangle AOD \cong \triangle BOC$
(ii) $A D=B C$
(iii) $\angle ADB =\angle ACB$
(iv) $\triangle ADB \cong \triangle BCA$

Answer

Proof:
In Δ AOD and Δ BOC
OA = OB ........(given)
∠AOD = ∠BOC .............(vertically opposite angles)
OD = OC ............(given)
(i) ∴ Δ AOD ≅ Δ BOC ................(S.A.S. Axiom)
Hence (ii) AD = BC ..........(c.p.c.t.)
and (iii) ∠ADB = ∠ACB .....(c.p.c.t.)
(iv) Δ ADB ≅ Δ BCA
Δ ADB = Δ BCA ..............(Given)
AB = AB .................(Common)
∴ Δ AOB ≅ Δ BCA
Hence proved.

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Question 54 Marks

Prove that:
(i) $\triangle ABD \cong \triangle ACD$
(ii) $\angle B=\angle C$
(iii) $\angle ADB =\angle ADC$
(iv) $\angle A D B=90^{\circ}$

Answer

Given: In the figure,
$
\begin{aligned}
& A D=A C \\
& B D=C D
\end{aligned}
$

To prove:
(i) $\triangle ABD \cong \triangle ACD$
(ii) $\angle B=\angle C$
(iii) $\angle ADB =\angle ADC$
(iv) $\angle ADB =90^{\circ}$
Proof: In $\triangle A B D$ and $\triangle A C D$
$A D=A D$ .................(common)
$A D=A C$ .................(given)
$B D=C D$ ................(given)
(i) $\therefore \triangle ABD \cong \triangle ACD$ ..............(SSS axiom)
(ii) $\therefore \angle B=\angle C$ (c.p.c.t.)
(iii) $\angle ADB =\angle ADC$ (c.p.c.t.)
But $\angle ADB +\angle ADC =180^{\circ}$ ............(Linear pair)
$\begin{aligned} & \therefore \angle ADB =\angle ADC \\ & \text { (iv) } \angle ADB =\angle ADC \\ & =\frac{180^{\circ}}{2} \\ & =90^{\circ}\end{aligned}$

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Question 64 Marks

Prove that:
(i) $\triangle A B C \cong \triangle A D C$
(ii) $\angle B=\angle D$
(iii) $A C$ bisects angle $D C B$

Answer

Given: In the figure,
$
A B=A D, C B=C D
$
To prove: $\triangle A B C \cong \triangle A D C$
$
\angle B=\angle D
$
AC bisects angle DCB

Proof: In $\triangle A B C$ and $\triangle A D C$,
$A C=A C$ ..............(common)
$A B=A D$ ..............(given)
$C B=C D$ ..............(given)
(i) $\therefore \triangle ABC \cong \triangle ADC$ .................(SSS aciom)
(ii) $\therefore \angle B=D$ .................(c.p.c.t.)
$
\angle B C A=\angle D C A
$
(iii) $\therefore A C$ bisects $\angle D C B$

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MATHS [4 marks sum]

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