3 Marks Question

Question 13 Marks

In following figure if AD is the bisector of $\angle\text{BAC},$ then prove that AB > BD.

Answer

Given ABC is a triangle such that AD is the bisector of $\angle\text{BAC},$
To prove AB > BD
Since, AD is the bisector of $\angle\text{BAC}.$
But $\angle\text{BAD}=\angle\text{CAD}\ ...(\text{i})$
$\therefore\angle\text{ADB} >\angle\text{CAD}$
A triangle is greater than of the opposite angle.
$\text{AB}>\text{BD}$

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Question 23 Marks

ABCD is a quadrilateral such that diagonal AC bisects the angles A and C. Prove that AB = AD and CB = CD.

Answer

Given in a quadrilateral ABCD, diagonal AC bisects the angles A and C.

To prove $\text{AB}=\text{CD}\ \text{ and }\text{CB}=\text{CD}$ Proof in $\triangle\ \text{ADC}\ \text{and}\ \triangle\ \text{ABC},$ $\angle\ \text{DAC}=\angle\ \text{BAC}$ $\big[\because$ AC is the bisector of $\angle\ \text{A}\ \text{and}\ \angle\ \text{C}\big]$ $\angle\ \text{DCA}=\angle\ \text{BCA}$ $\big[\because$ AC is the bisector of $\angle\ \text{A}\ \text{and}\ \angle\text{C}\big]$ $\text{and}\ \text{AC}=\text{AC}$ [common side] $\therefore\ \triangle\text{ADC}\cong\triangle\text{ABC}$ [byASA congruence rule] $\text{AD}=\text{AB}$ [by CPCT] $\text{and}\ \text{CD}=\text{CB}$ [by CPCT] Hence proved.

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Question 33 Marks

ABC is an isosceles triangle with AB = AC and D is a point on BC such that $\text{AD}\perp\text{BC}$ (see figure). To prove that $\angle\text{BAD} = \angle\text{CAD},$ a student proceeded as follows:

In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AB}=\text{AC}$
$\angle\text{B}=\angle\text{C}$
$\angle\text{ADM}=\angle\text{ADC}$
$\therefore\triangle\text{ABD}\cong\triangle\text{ADC}$
$\angle\text{BAD}=\angle\text{CAD}$
What is the defect in the above arguments?

Answer

In $\triangle\text{ABC,}$
$\text{AB}=\text{AC}$
$\Rightarrow \angle\text{ACB}=\angle\text{ABC}$

In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AB}=\text{AC}$
$\angle\text{ADB}=\angle\text{ACD}$
$\triangle\text{ABD}\cong\triangle\text{ACB}$
So, the defect in the given argument is that firstiy prove $\triangle\text{ABD}\cong\triangle\text{ACB}.$

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Question 43 Marks

In given $\text{l}\ ||\ \text{m}$ and M is the mid-point of a line segment AB. Show that M is also the mid-point of any line segment CD, having its end points on l and m, respectively.

Answer

Given in the $\text{l}\ ||\ \text{m}$ and M is the mid-point of a line segment AB. AM = BM. MC = MD $\text{l}\ ||\ \text{m}$ $\angle\text{BAC}=\angle\text{ABD}$ $\angle\text{AMC}=\angle\text{BMD}$ In $\triangle\text{AMC}$ and $\triangle\text{BMD,}$ $\angle\text{BAC}=\angle\text{ABD}$
$\text{AM}=\text{BM}$ $\angle\text{AMC}=\angle\text{BMD}$ $\text{MC}=\text{MD}$

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Question 53 Marks

In $\triangle\text{PQR},\angle\text{A}=\angle\text{Q}$ and $\angle\text{B}=\angle\text{R}.$ Which side of should be equal to side AB of so, that the two triangles are congruent? Give reason for your answer.

Answer

In triangle ABC and PQR, we have
$\angle\text{A}=\angle\text{Q}$
$\angle\text{B}=\angle\text{R}$
For the triangle to be congruent, we must AB = QR. They will be congruent by ASA rule.

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Question 63 Marks

AD is a median of the $\triangle\text{ABC}.$ Is it true that AB + BC + CA > 2AD? Give reason for your answer.

Answer

In $\triangle\text{ABD,}$ we have
$\text{AB}+\text{BD}>\text{AD}\ ....(\text{i})$
In $\triangle\text{ACD,}$ we have
$\text{AC}+\text{CD}>\text{AD}\ ...(\text{ii})$
On adding eq.(i) and (ii),
$\Rightarrow(\text{AB}+\text{BD}+\text{AC}+\text{CD})>2\text{AD}$
$\Rightarrow(\text{AB}+\text{BD}+\text{CD}+\text{AC})>2\text{AD}$
$\Rightarrow\text{AB}+\text{BC}+\text{AC}>2\text{AD}$

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Question 73 Marks

Find all the angles of an equilateral troiangle.

Answer

In $\triangle\text{ABC},$ we have
$\text{AB}=\text{AC}$
$\Rightarrow \angle\text{C}=\angle\text{B}\ ...(\text{i})$
$\text{BC}=\text{AC}$
$\angle\text{A}=\angle\text{B}\ ...(\text{ii})$
Now,
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$
$\Rightarrow \angle\text{A}+\angle\text{A}+\angle\text{A}=180^{\circ}$
$\Rightarrow 3\angle\text{A}=180^{\circ}$
$\Rightarrow \angle\text{A}=\frac{180^{\circ}}{3}=60^{\circ}$
$\therefore \angle\text{A}=\angle\text{B}=\angle\text{C}=60^{\circ}$

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Question 83 Marks

D is any point on side AC of a $\triangle\text{ABC}$ with AB = AC. Show that CD > BD.

Answer

In $\triangle\text{ABC},$ we have
$\text{AB}=\text{AC}$
$\therefore\angle\text{ABC}=\angle\text{ACB}$
Now,
$\angle\text{DBC}=\angle\text{ABC}$
$\therefore \angle\text{DBC}<\angle\text{DCB}$
Hence, $\text{CD}>\text{BD}$

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Question 93 Marks

ABCD is a quadrilaeter in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.

Answer

In $\triangle\text{ABC}$ and $\triangle\text{CBD},$ we have

$\text{AB}=\text{BC}$
$\text{AD}=\text{CD}$
$\text{BD}=\text{BD}$
$\therefore\text{ABC}\cong\triangle\text{CBD}$
$\Rightarrow\angle\text{1}=\angle\text{2}$
$\angle3=\angle4$
Hence, BD bises both the angle ABC and ADC.

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Question 103 Marks

Is it possible to construct a triangle with lengths of its sides as 4cm, 3cm and 7cm? Give reason for your answer.

Answer

No, it is not possible to construct a triangle with lengths of its sides as 4cm, 3cm and 7cm because here we see that sum of the lengths of two sides is equal to third side i.e.,
4 + 3 = 7
As we know that, the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

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Question 113 Marks

In the given figure $\triangle\text{CDE}$ is an equilateral triangle formed on a side CD of a square ABCD. Show that $\triangle\text{ADE}\cong\triangle\text{BCE}.$

Answer

Given, $\triangle\text{CDE}$ is an equilateral triangle formes on a side CD of a ABCD.
$\triangle\text{ADE}\cong\triangle\text{BCE}$
In $\triangle\text{ADE}$ and $\triangle\text{BCE},$
$\text{DE}=\text{CE}$
$\angle\text{ADE}=\angle\text{BCE}$
$\text{AD}=\text{BC}$
$\triangle\text{ADE}\cong\triangle\text{BCE}$

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Question 123 Marks

In $\text{BA}\perp\text{AC},\text{DE}\perp\text{DF}$ such that BA = DE and BF = EC. Show that $\triangle\text{ABC}\cong\triangle\text{DEF.}$

Answer

Given, $\text{BA}\perp\text{AC},\text{DE}\perp\text{DF}$ such that BA = DE and BF = EC.$\triangle\text{ABC}\cong\triangle\text{DEF}$
$\text{BF}=\text{EC}$
On adding CF both sides, we have
$\text{BF}+\text{CF}=\text{EC}+\text{CF}$
$\text{BC}=\text{EF}$
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$
$\angle\text{A}=\angle\text{D}=90^{\circ}$
$\text{BA}=\text{DE}$
$\triangle\text{ABC}\cong\triangle\text{DEF}$

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