3 Mark Question

Question 13 Marks

In the given figure, triangles ABC and DCB are right angled at A and D respectively and AC = DB Prove that $\triangle\text{ABC}\cong\triangle\text{DCB.}$

Answer

Given: In right triangles ABC and DCB right angled at A and D respectively and AC = DB.

To prove: $\triangle\text{ABC}\cong\triangle\text{DCB.}$
Proof: In right angled $\triangle\text{ABC}$ and $\triangle\text{DCB},$
hypotenuse BC = BC (common)
side AC = DB (given)
$\triangle\text{ABC}\cong\triangle\text{DCB.}$ (RHS condition)
Hence proved.

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Question 23 Marks

In fig. (i) $\text{PL}\bot\text{OA}$ and $\text{PM}\bot\text{OB}$ such that $\text{PL}=\text{PM}$. Is $\triangle\text{PLO}\cong\triangle\text{PMO}?$ Give reason in support of your answer.

Answer

In fig.$\text{PL}\bot\text{OA}$ and $\text{PM}\bot\text{OB}$ and $\text{PL}=\text{PM}$
Now in right $\triangle\text{PLO}$ and $\triangle\text{PMO}$,
Side $\text{PL}=\text{PM}$ (given)
Hypotenuse $\text{OP}=\text{OP}$ (common)
$\triangle\text{PLO}\cong\triangle\text{PMO}$ (RHS condition)
Yes $\triangle\text{PLO}\cong\triangle\text{PMO}$
Hence proved.

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Question 33 Marks

In the adjoining figure, $\triangle\text{ABC}$ is an isosceles triangle in which AB = AC. If E and F be the midpoints of AC and AB respectively, prove that BE = CF.

Answer

Given: $\triangle\text{ABC}$ is an isosceles triangle in which AB = AC.
E and F are the midpoints of AC and AB respectively.

To prove: BE = CF
Proof: IN $\triangle\text{BCF}$ and $\triangle\text{CBE},$
BC = BC (common)
BF = CE (Half of equal sides AB and AC)
$\angle\text{CBF}=\angle\text{BCF}$ (Angles opposite to equal sides)
$\triangle\text{BCF}\cong\triangle\text{CBE}$ (SAS condition)
CF = BE (c.p.c.t.)
or BE = CF
Hence proved.

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Question 43 Marks

If two triangles have their corresponding angles equal, are they always congruent? If not, draw two triangles which are not congruent but which have their corresponding angles equal.

Answer

Two triangles whose corresponding angles are equal, it is not necessarily that they should be congruent. It is possible if at least one side must be equal. Below given a pair of triangles whose angles are equal but these are not congruent.

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Question 53 Marks

In fig. (i) $\text{AD}=\text{BC}$ and $\text{AD}||\text{BC}$. Is $\text{AB}=\text{DC}?$ Give reasons in support of your answer.

Answer

In the figure,

$\text{AD}=\text{BC}$ and $\text{AD}||\text{BC}$
In $\triangle\text{ABC}$ and $\triangle\text{ADC}$,
$\text{AC}=\text{AC}$ (common)
$\text{BC}=\text{AB}$ (given)
$\angle\text{ACB}=\angle\text{CAD}$ (Alternate angles)
$\triangle\text{ABC}\cong\triangle\text{ADC}$ (SAS condition)
$\text{AB}=\text{DC}$ (c.p.c.t.)
Hence proved.

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Question 63 Marks

In the adjoining figure, $\text{AB}=\text{AD}$ and $\text{CB}=\text{CD.}$ Prove that $\triangle\text{ABC}\cong\triangle\text{ADC.}$

Answer

In the figure,
$\text{AB}=\text{AD},\text{CB}=\text{CD}$
To prove: $\triangle\text{ABC}\cong\triangle\text{ADC}$
Proof: In $\triangle\text{ABC}$ and $\triangle\text{ADC.}$
$\text{AC}=\text{AC}$ (common)
$\text{AB}=\text{AD}$ (given)
$\text{CB}=\text{CD}$ (given)
$\triangle\text{ABC}\cong\triangle\text{ADC}$ (SSS condition)
Hence proved.

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Question 73 Marks

In the adjoining figure, P and Q are two points on equal sides AB and AC of an isosceles triangle ABC such that AP = AQ. Prove that BQ = CP.

Answer

Given : In isosceles $\triangle\text{ABC,}$ AB = AC. P and Q are the points on AB and AC respectively such that AP = AQ. To prove: BQ = CP.

Proof: In $\triangle\text{ABQ}$ and $\triangle\text{ACP,}$
AB = AC (given)
AQ = AP (given)
$\angle\text{A}=\angle\text{A}$ (common)
$\triangle\text{ABQ}\cong\triangle\text{ACP}$ (SAS condition)
BQ = CP (c.p.c.t.)
Hence proved.

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Question 83 Marks

Draw $\triangle\text{ABC}$ and $\triangle\text{PQR}$ such that they are equal in area but not congruent.

Answer

In $\triangle\text{ABC},$

$\text{Area}=\frac{1}{2}\times\text{BC}\times\text{AL}$ $=\frac{1}{2}\times5\times4=10\text{ cm}^2$
and in $\triangle\text{PQR}$
$\text{Area}=\frac{1}{2}\times\text{QR}\times\text{PR}$ $=\frac{1}{2}\times5\times4=10\text{ cm}^2$
In these triangles,
Areas of both triangles are equal but are no congruent to each other.

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Question 93 Marks

In the adjoining figure, ABC is a triangle in which AD is the bisector of $\angle\text{AD}.$ If $\text{AD}\bot\text{BC},$ show that $\triangle\text{ABC}$ is an isosceles.

Answer

Given: In $\triangle\text{ABC},\text{AD}$ is the bisector of $\angle\text{A}$ i.e. $\angle\text{BAD}=\angle\text{CAD}$
$\text{AD}\bot\text{BC}.$
To prove: $\triangle\text{ABC}$ is an isosceles
Proof: In $\triangle\text{ADB}$ and $\triangle\text{ADC.}$
$\text{AD}=\text{AD}$ (common)
$\angle\text{BAD}=\angle\text{CAD}\text{ (AD}$ is the bisector of $\angle\text{A})$
$\angle\text{ADB}=\angle\text{ADC}($ each $90^\circ,\text{AD}\bot\text{BC})$
$\triangle\text{ADM}\cong\triangle\text{ADC}$ (ASA condition)
$\text{AB}=\text{AC}$ (c.p.c.t.)
Hence $\triangle\text{ABC}$ is an isosceles triangle.
Hence proved.

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Question 103 Marks

In the adjoining figure, $\triangle\text{ABC}$ is an isosceles triangle in which AB = AC. Also, D is a point such that BD = CD. Prove that AD bisects $\angle\text{A}$ and $\angle\text{D.}$

Answer

Given: In $\triangle\text{ABC,}$
AB = AC.
D is point such that BD = CD.
AD, BD and CD are joined.
To prove: Ad bisects $\angle\text{A}$ and $\angle\text{D}$
Proof: In $\triangle\text{ABD}$ and $\triangle\text{CAD,}$
AD = AD (Common)
AB = AC (given)
BD = CD (given)
$\triangle\text{ABD}\cong\triangle\text{CAD}$ (SSS condition)
$\angle\text{BAD}=\angle\text{CAD}$ (c.p.c.t.)
and $\angle\text{BDA}=\angle\text{CDA}$ (c.p.c.t.)
Hence, AD is the bisector of $\angle\text{A}$ and $\angle\text{D}.$
Hence proved.

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Question 113 Marks

Are two triangles congruent if two sides and an angle of one triangle are respectively equal to two sides and an angle of the other? If not then under what conditions will they be congruent?

Answer

In two triangles, if two sides and and included angle of the one equal to the corresponding two sides and included angle, then the two triangles are congruent.
If another angle except included angles are equal to each other and two sides are also equal these are not congruent. In the above figures, in $\triangle\text{ABC}$ and $\triangle\text{PQR}$, two corresponding sides and one angle are equal, but these are not congruent.

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Question 123 Marks

In the given figure, $\triangle\text{ABC}$ is an isosceles triangle in which AB = AC. If AB and AC are produced to D and E respectively such that BD = CE, prove that BE = CD.

Answer

Given: $\triangle\text{ABC}$ is an isosceles triangle in which AB = AC.
AB and AC are produced to D and E respectively such that BD = CE.
BE and CD are joined.
To prove: BE = CD.
Proof: Ab = Ac and BD = CE
Adding we get:
AB + BD = AC + CE
AD = AE
Now, in $\triangle\text{ACD}$ and $\triangle\text{ABE}$
AC = AB (given)
$\angle\text{A}=\angle\text{A}$ (common)
$\triangle\text{ACD}\cong\triangle\text{ABE}$ (SSA condition)
CD = BE (c.p.c.t)
Hence, BE = CD.

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Question 133 Marks

In the adjoining figure, $\text{AB}=\text{AC}$ and $\text{BD}=\text{DC}.$ Prove that $\triangle\text{ADB}\cong\triangle\text{ADC}$ and hence show that.

$\angle\text{ADB}=\angle\text{ADC}=90^\circ$
$\angle\text{BAD}=\angle\text{CAD}$

Answer

In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AD}=\text{AD}$ (common)
$\text{AB}=\text{AC}$ (given)
$\triangle\text{ABD}\cong{}\triangle\text{ADC}$ (SSS condition)
$\angle\text{BAD}=\angle\text{CAD}$ (c.p.c.t.)
and $\angle\text{ADB}=\angle\text{ADC}$ (c.p.c.t.)
But $\angle\text{ADB}+\angle\text{ADC}=180^\circ$ (Linear pair)
$\angle\text{ADB}=\angle\text{ADC}=90^\circ$
Hence proved.

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Question 143 Marks

In the given figure, $\text{PA}\bot\text{AB},\text{QB}\bot\text{AB}$ and PA = QB. Prove that $\triangle\text{OAP}\cong\triangle\text{OBQ.}$ Is OA = OB?

Answer

Given: In the figure,$\text{PA}\bot\text{AB},\text{QB}\bot\text{AB}$ and PA = QB.
To prove: $\triangle\text{OAP}\cong\triangle\text{OBQ,}$
Is OA = OB?
Proof: In $\triangle\text{OAP}$ and $\triangle\text{OBQ,}$
$\angle\text{A}=\angle\text{B}$ (each 90°)
AP = BQ (given)
$\angle\text{AOP}=\angle\text{BOQ}$ (vertically opposite angles)
$\triangle\text{OAP}\cong\triangle\text{OBQ}$ (AAS condition)
OA = OB (c.p.c.t.)
Hence proved.

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