4 Marks Questions - Maths STD 9 Questions - Vidyadip

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Question 14 Marks

Question 15: Two lines l and m intersect at the point 0 and P is a point on a line n passing through the point 0 such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m.

Answer

Giva two lines L and M intersect at the point O and P is a point on a line n passing through O such that P is equidistant from L and M, i.e., PQ = PR,

To prove N is the bisector of the angle formed by L and M i.e., n is the bisector of $\angle\text{QOR},$
Proof In $\triangle\text{OQP}\text{ and }\triangle\text{ORP} $,
$\angle\text{PQO}=\angle\text{PRO}=\text{90}^0$
[since,P in equidistant from L and M, so PQ and PR should be perpendicular to lines L and M respectively]
OP = OP [commo side]
P = PR [given]
$\therefore\triangle\text{OQP}\cong\triangle\text{ORP}$ [by RHS congruence rule]
$\Rightarrow\angle\text{POQ}=\angle\text{POR}$ [by CPCT]
Hence,n is the bisector of $\angle\text{QOR}.$Hence proved.

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Question 24 Marks

In D and E are points on side BC of a $\triangle\text{ABC}$ such that BD = CE and AD = AE. Show that $\triangle\text{ABC}\cong\triangle\text{ACE}.$

Answer

Given, $\triangle\text{ABC}$ in which BD = CE and AD = AE
$\triangle\text{ABC}\cong\triangle\text{ACE}$
In $\triangle\text{ADE},$ we have
$\text{AD}=\text{AE}$
$\angle\text{ADE}=\angle\text{AED}$
Now,
$\angle\text{ADE}+\angle\text{AEC}=180^{\circ}\ ...(\text{i})$
$\angle\text{AED}+\angle\text{AEC}=180^{\circ}\ ...(\text{ii})$
For eq.(i) and (ii),
$\angle\text{ADE}+\angle\text{ADB}=\angle\text{AED}+\angle\text{AEC}$
$\Rightarrow\angle\text{ADB}=\angle\text{AEC}$
In $\triangle\text{ABD},$ we have
$\text{BD}=\text{CE}$
So, by SAS criterion of congrurncr, we have
$\triangle\text{ABC}\cong\triangle\text{ACE}.$

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Question 34 Marks

ABCD is quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.

Answer

Given In quadrilateral ABCD, AB = AD and CB = CD.

constrion: To prove AC is the petpendicular bisector of BD.uction Join AC and BD.

Proof In $\triangle\text{ABC}\text{ and }\triangle\text{ADC},$

$\text{AB}=\text{AD}$ [given]

$\text{BC}=\text{CD}$ [given]

$\text{and}\ \text{AC}=\text{AC}$ [common side]

$\therefore \triangle\text{ABC}\cong\triangle\text{ADC}$ [by SSS congruence rule]

$\Rightarrow\ \angle1=\angle2$ [by CPCT]

now, in $\triangle\text{AOB}\text{ and }\triangle\text{AOD},\ \text{AB}=\text{AD}$ [given]

$\Rightarrow\ \angle1=\angle2$ [proved above]

$\text{and}\ \text{AO}=\text{AO} $ [common side]

$\therefore\ \triangle\text{AOB}\cong\triangle\text{AOD}$ [by SAS congruence rule]

$\Rightarrow\ \text{BO}=\text{DO}$ [by CPCT]

$\text{and}\ \angle3=\angle4$ [by CPCT]...(i)

But $\angle3+\angle4=180^\circ$[linear pair axiom]

$\angle3+\angle3=180^\circ$ [from Eq. (i)]

$\Rightarrow\ 2\angle3=180^\circ$

$\Rightarrow\ \angle3=\frac{180^\circ}{2}$

$\therefore\ \angle3=90^\circ$

i.e., AC is perpendicular bisector of BD.

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Question 44 Marks

In a triangle ABC, D is the mid-point of side AC such that $\text{BD}=\frac{1}{2}\text{AC}.$Show that $\angle\text{ABC}$ is a right angle.

Answer

We have to prove that $\angle\text{ABC}=90^\circ$

As is the mid-point of AC,

so, AD = DC

Also, $\text{BD}=\frac{\text{1}}{\text{2}}\text{AC}=\text{AD}$

$\big[\therefore$ D is the mid-point of AC$\big]$

$\therefore$ BD = AD = DC

In $\triangle\text{ABC},$ We have

BD = AD

$\therefore\angle1=\angle2\ ...(1)$

$\big[\because$ Angles opposite to equal sides are equal$\big]$

In $\triangle\text{BCD},$We have

BD = DC,

$\therefore\angle\text{3}=\angle4 ...(2)$

In $\triangle\text{ABC},$ We have

$\angle1+\angle\text{ABC}+\angle4=180^\circ$

$\Rightarrow\angle\text{1}+\angle\text{2}+\angle\text{3}\angle+\text{4}=180^\circ$

$\Rightarrow2(\angle\text{2}+\angle\text{3})=180^\circ)$

$\Rightarrow\angle2+\angle3=90^\circ$

$\Rightarrow\angle=90^\circ$

Hence proved.

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Question 54 Marks

S is any point on side QR of a $\triangle\text{PQR}.$ show that PQ + QR + RP > 2PS.

Answer

Given A point S on QR.

In $\triangle\text{PQS},$ we have
$\text{PQ}+\text{QS}>\text{PS}\ ...(\text{i})$
Now, in $\triangle\text{PSR},$ we have
$\text{RS}+\text{RP}>\text{PS}\ ...(\text{ii})$
From eq.(i) and (ii),
$\text{PQ}+\text{QS}+\text{RS}+\text{RP}>2\text{PS}$
$\Rightarrow\text{PQ}+\text{QR}+\text{RP}>2\text{PS}$

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Question 64 Marks

ABC is an isoscles triangle in which AC = BC. AD are respectively two altitudes to sides BC and BC. prive that AE = BD.

Answer

In $\triangle\text{ABC}$ and $\angle\text{DBC}$ we have

$\text{AB}=\text{AC}$

$\angle\text{ABD}=\angle\text{ACD}$

$\angle\text{BAD}=\angle\text{CAD}$

$\therefore\triangle\text{ADC}\cong\triangle\text{BCD}$

$\therefore\text{AB}=\text{AC}$

$\text{AC}-\text{CE}=\text{BC}-\text{CD}$

$\Rightarrow\text{AE}=\text{BD}$

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Question 74 Marks

Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Answer

Givan in $\triangle\text{ABC},$ AD is a median.

Construction produce AD to a point E such that AD = DE and join CE.

To prove AC + AB > 2AD

Proof in $\triangle\text{ABD}\text{ and } \triangle \text{ECD},$

$\text{AD}=\text{DE}$ [by construction]

$\text{BD}=\text{CD}$ [given AD is the media]

and $\angle\text{ADB}=\angle\text{CDE}$ [vertically opposite angle]

$\therefore\ \triangle\text{ABD}\cong\text{ECD}$ [by SAS congruence rule]

$\Rightarrow\ \text{AB}=\text{CE}$ [by CPCT]...(i)

Now, in $\triangle\text{AEC},$

$\text{AC}+\text{EC}>\text{AE}$

$\therefore\ \text{AC}+\text{AB}>2\text{AD} $ [Sum of two sides of a triangle is greater than the third side]

[from Eq. (i) and also taken that AD = DE]

Hence proved.

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Question 84 Marks

ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.

Answer

Given two $\triangle\text{ABC}$ and $\angle\text{DBC}$ are formed on the same base BC Such that A and D lie on the opposite side of AB = AC and DB = DC.
In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AB}=\text{AC}$
$\angle\text{ABD}=\angle\text{ACD}$
$\angle\text{BAD}=\angle\text{CAD}$

In $\triangle\text{ABO},$
$\text{AB}=\text{AC}$
$\angle\text{AOB}+\angle\text{AOB}=180^{\circ}$
$2\angle\text{AOB}=180^{\circ}$
$\angle\text{AOB}=\frac{180^{\circ}}{2}=180^{\circ}$
Hence, AD is the perpendicular of BC.

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Question 94 Marks

Q is point on the side SR of a $\triangle\text{PSR}$ such that PQ = PR. prove that PS > PQ.

Answer

Given, PQ = PR

In $\triangle\text{PRQ},$ we have
$\text{PR} =\text{PQ}$
$\Rightarrow\angle\text{1}=\angle\text{R}$
But, $\angle\text{1}>\angle\text{S}$
$\Rightarrow\angle\text{R}>\angle\text{S}$
$\Rightarrow\text{PS}>\text{PR}$
$\Rightarrow\text{PS}=\text{PQ}$

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Question 104 Marks

ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.

Answer

Given in right angled $\triangle\text{ABC},\text{AB}= \text{AC}\text{ and }\text{CD}$ is the bisector of $\angle\text{C}.$ contruction draw $\text{DE}\ \bot\ \text{BC}.$

to prove $\text{AC}+\text{AD}= \text{BC}$

proof in right angled $\triangle\text{ABC},\text{AB}=\text{AC}\text{ and }\text{BC}$ is a hypotenuse [given]

$\angle1=\angle2$ [given,$\text{CD}$ isthe bisector of $\angle\text{C}$]

$\text{DC}=\text{DC}$ [common sides]

$\therefore\ \triangle\text{DAC}\cong\triangle\text{DEC}$ [by AAS congruence rule]

$\Rightarrow\ \text{DA}=\text{DE}.$ [by CPCT]...(i)

$\text{and}\ \text{AC}=\text{EC}$ ...(ii)

$\text{ln}\ \triangle\text{ABC},\ \text{AB}=\text{AC}$

$\angle\text{C}=\angle{B}$

[angle opposite to equal sides are equal]...(iii)

Again,in $\triangle\text{ABC},\ \angle\text{A}+\angle\text{B}+\angle\text{C}= 180^\circ$

[by angle sum property of a triangle]

$\Rightarrow\ 90^\circ+\angle\text{B}+\angle\text{B}=180^\circ$

$\Rightarrow\ 2\angle\text{B}=180^\circ-90^\circ$[from Eq.(iii)]

$\Rightarrow\ 2\angle\text{B}=90^\circ$

$\Rightarrow\ \angle\text{B}=45^\circ$

$\text{ln}\ \triangle\text{BED},\ \angle5=180^\circ-\big(\angle\text{B}+\angle{4}\big)$

[by angle sum property of a triangle]

$=180^\circ-(45^\circ+90^\circ)$

$=180^\circ-135^\circ=45^\circ$

$\therefore\ \angle\text{B}=\angle{5}$

$\Rightarrow\ \text{DE}=\text{BE}\ [\because$ sides opposite to equal angle are equal$]$ ...(iv)

from Eqs. (i) and (iv),

$\text{DA}=\text{DE}=\text{BE}$ ...(v)

$\because\text{BC}=\text{CE}+\text{EB}$

$=\text{CA}+\text{DA}$ [from Eqs. (ii) and (v)]

$\therefore\text{AD}+\text{AC}=\text{BC}$ Hence proved.

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Question 114 Marks

ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.

Answer

Given In quadrilateral ABCD, AB is the smallest and CD is largest side to find $\angle\text{B}>\angle\text{D}\text{ or }\angle\text{D}>\angle\text{B}.$

Construction Join BD.

Now, in $\triangle\text{ABD},\ \text{AD}>\text{AB}$ [since, AB is the smallest side in ABCD]

$\Rightarrow\ \angle{1}>\angle{3}$ [angle opposite to larger side is greater]...(i)

$\text{In }\triangle\text{ABCD},\ \text{CD}>\text{BC}$ [since, CD is the largest side in ABCD]

$\Rightarrow\ \angle{2}>\angle{4}$ [angle opposite to larger side is greater]...(ii)

On adding Eqs.(i) and (ii), we get

$\angle1+\angle2+\angle3+\angle4$

Hence, $\angle\text{B}>\angle\text{D}$

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Question 124 Marks

The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

Answer

Given, An object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in figure.
The image is as far behind the mirror as the object is in front of the mirror OB = OA.

$\therefore\text{CN}\perp\text{LM}$ and $\text{AB}\perp\text{LM}$
$\text{AB}\ ||\ \text{CN}$
$\angle\text{A}=\angle\text{i}\ ...(\text{i})$
$\angle\text{B}=\angle\text{r}\ ...(\text{ii})$
$\angle\text{i}=\angle\text{r}\ ...(\text{iii})$
From eq.(i), (ii) and (iii)
$\angle\text{A}=\angle\text{B}$
In $\triangle\text{COB},$
$\angle\text{B}=\angle\text{A}$
$\angle\text{1}=\angle\text{2}$
$\text{OB}=\text{OA}$
In $\triangle\text{OBC},$
$\angle\text{1}=\angle\text{2}$
$\angle\text{i}=\angle\text{r}$
On mulitiplying both sides of then adding both sides,
$90^{\circ}-\angle\text{i}=90^{\circ}-\angle\text{r}$
$\Rightarrow \angle\text{ACO}=\angle\text{BCO}$
$\Rightarrow \text{OC}=\text{OC}$
$\therefore\triangle\text{OBC}=\triangle\text{OAC}$
$\Rightarrow \text{OB}=\text{OA}$
Hence, the image is as far behind the mirror as the object is in front of the mirror.

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Question 134 Marks

Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. BO is produced to a point M. Prove that $\angle\text{MOC}=\angle\text{ABC}.$

Answer

Given Lines, OB and OC are angle bisectore of $\angle\text{B}$ and $\angle\text{C}$ an isosceles $\triangle\text{ABC}$ such that AB = AC which intersrct each other at O and BO is produced to M.
$\angle\text{MOC}=\angle\text{ABC}$

In $\triangle\text{ABC},$
$\text{AB}=\text{AC}$
$\frac{1}{2}\angle\text{ACB}=\frac{1}{2}\angle\text{ABC}$
$\angle\text{OCB}=\angle\text{OBC}$
Now,
$\angle\text{MOC}=\angle\text{OBC}+\angle\text{OCB}$
$\Rightarrow\angle\text{MOC}=\angle\text{OBC}+\angle\text{OBC}$
$\Rightarrow \angle\text{MOC}=2\angle\text{OBC}$
$\Rightarrow \angle\text{MOC}=\angle\text{ABC}$

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Question 144 Marks

Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD

Answer

Given ABCD is a quadilateral,

Construction join diagonais AC and BD.

To show AB + BC + CD + DA > AC + BD

In $\triangle\text{ABC},$ AB + BC > AC [Sum of two sides of a triangle is greater than the third side] ...(i)

In $\triangle\text{BCD},$ BC + CD > BD [Sum of two sides of a triangle is greater than the third side] ...(ii)

In $\triangle\text{CDA},$ CD + DA > AC [Sum of two sides of a triangle is greater than the third side] ...(iii)

In $\triangle\text{DAB},$ DA + AB + > BD [Sum of two sides of a triangle is greater than the third side] ...(iv)

On adding Eqs. (i), (ii), (iii) (iv), We get

2(AB + BC + CD + DA) > 2 (AC + BD)

⇒ AB + BC + CD + DA > AC + BD

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Question 154 Marks

In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

Answer

We have to prove that $\angle\text{ABC}=90^\circ$

As is the mid-point of AC,

so, AD = DC

Also, $\text{BD}=\frac{\text{1}}{\text{2}}\text{AC}=\text{AD}$

$\big[\therefore$ D is the mid-point of AC$\big]$

$\therefore$ BD = AD = DC

In $\triangle\text{ABC},$ We have

BD = AD

$\therefore\angle1=\angle2\ ...(1)$

$\big[\because$ Angles opposite to equal sides are equal$\big]$

In $\triangle\text{BCD},$We have

BD = DC,

$\therefore\angle\text{3}=\angle4 ...(2)$

In $\triangle\text{ABC},$ We have

$\angle1+\angle\text{ABC}+\angle4=180^\circ$

$\Rightarrow\angle\text{1}+\angle\text{2}+\angle\text{3}\angle+\text{4}=180^\circ$

$\Rightarrow2(\angle\text{2}+\angle\text{3})=180^\circ)$

$\Rightarrow\angle2+\angle3=90^\circ$

$\Rightarrow\angle=90^\circ$

Hence proved.

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Question 164 Marks

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.

Answer

Given O is a point in the interior of a squaren ABCD such that an euailateral triangle.

Since, AOB is an triangle.
$\angle\text{OAB}=\angle\text{OBA}=60^{\circ}\ ...(\text{i})$
$\angle\text{DAB}=\angle\text{CBA}=90^{\circ}\ ...(\text{ii})$
From eq. (i) and (ii)
$\angle\text{DAB}-\angle\text{OAB}=\angle\text{CBA}=\angle\text{OBA}$
$\angle\text{DAO}=\angle\text{CBO}=90^{\circ}$
$\Rightarrow\text{AO}=\text{BO}$
$\Rightarrow \text{AD}=\text{BC}$
$\Rightarrow \text{OC}=\text{OD}$
Hence, an isosceles triangle.

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Question 174 Marks

Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)

Answer

Given, A quadrilateral ABCD.

To prove: AB + BC + CD + DA < 2(BD + AC)

Proof: In $\triangle\text{AOD}$ We have

$\therefore\text{OA}+\text{OB}>\text{AB }...(\text{i}) $

$\big[\therefore$ Sum of the lengths of any sides of a triangle must be greater than rhird side$\big]$

In $\triangle\text{BOC}$ We have

OB + OC + > BC ...(2) [Same reason]

In $\triangle\text{COD},$ We have

OC + OD + > CD ...(3) [Same reason]

In $\triangle\text{DOA},$ We have

OD + OA > DA ...(4) [Same reason]

Adding (1), (2), (3) and (4), We get

OA + B + OB + OC + OC + OD + OA + > AB + BC + CD +DA

⇒ 2(OA + OB + OC + OD) > AB + BC + CD + DA

⇒ 2{(OA + OC) + (OB + OD)} > AB + BC + CD + DA

⇒ 2(AC + BD) > AB + BC + CD + DA

⇒ AB + BC + CD + DA < 2(BD + AC)

Hence, proved.

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Question 184 Marks

P is point on the bisector of $\angle\text{ABC}.$ if the line throught P, parallel to BA meet BC at Q, prove thatBPQ is an isosceles triangle.

Answer

We have to prove thet BPQ is an isosceles triangle.

$\angle\text{1}=\angle\text{2}\ ...(\text{i})$
Now, PQ is paraller to BA and BP cuts them
$\therefore \angle\text{1}=\angle\text{3}\ ...(\text{ii})$
from (i) and (ii),
In $\triangle\text{BPQ},$ we have
$\angle\text{2}=\angle\text{3}$
$\therefore\text{PQ}=\text{BQ}$
Hence, BPQ is an isosceles triangle.

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Question 194 Marks

ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE.

Answer

Given $\triangle\text{ABC}$ is an isosceles triangle in which AB = AC and BD, CE are its two medians.
To show BD = CE.

In $\triangle\text{ABD}$ and $\triangle\text{ACE},$
$\text{AB}=\text{AC}$
$\angle\text{A}=\angle\text{A}$
$\text{AD}=\text{AC}$
$\frac{1}{2}\text{AB}=\frac{1}{2}\text{AC}$
$\text{AE}=\text{AD}$
AS D is the mid-point of AC and E is the mid-point of AB.
$\triangle\text{ABC}\cong\triangle\text{ACE}$

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Question 204 Marks

Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than $\frac{2}{3}$ of a right angle.

Answer

Consider $\triangle\text{ABC}$ in which BC is the longest side.

To prove $\angle\text{A}=\frac{2}{3}$ right angle

Proof In $\triangle\text{ABC},\ \text{BC}>\text{AB}.$

[consider BC is the largest side]

$\Rightarrow\ \angle\text{A}>\angle\text{C}$ ...(i)

[angle opposite the lngest side is greatest]

$\text{and}\ \text{BC}>\text{AC}$

$\Rightarrow\ \angle\text{A}>\angle\text{B},$

[angle opposite the longest side is greatest]

On adding Eqs. (i) and (ii), we get

$2\angle\text{A}>\angle\text{B}+\angle\text{C}$

$\Rightarrow\ 2\angle\text{A}+\angle\text{A}>\angle\text{A}+\angle\text{C}$ [adding $\angle\text{A}$ both sides]

$\Rightarrow\ 3\angle\text{A}>\angle\text{A}+\angle\text{B}+\angle\text{C}$

$\Rightarrow\ 3\angle\text{A}>180^\circ$ [sum of the angles of a triangle is 180°]

$\Rightarrow\ \angle\text{A}>\frac{2}{3}\times90^\circ$

i.e., $\angle\text{A}>\frac{2}{3}$ of a right angle Hence proved.

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Question 214 Marks

Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC Prove that AD = BC.

Answer

Given in trapezium ABCD, points M and N are the mid-points of parallel sides AB and DC respectively and join MN, which is perpendicular to AB and DC.

To prove: proof since, M is the mid-point of AB

$\therefore \text{AM} = \text{MB}$

Now, in $\triangle \text{AMN} \text{ and } \triangle \text{BMN},\ \text{AM} = \text{MB}$ [proved above]

$\angle 3 = \angle 4$ [each 90°]

$\text{MN} = \text{MN}$ [common side]

$\therefore \triangle \text{AMN} \cong \text{BMN}$ [by SAS congruence rule]

$\therefore \angle 1 = \angle 2$ [by CPCT]

On multiplying both sides of above equation by - 1 and than adding 90° both sides, we get

$90^\circ - \angle 1 = 90^\circ - \angle 2$

$\Rightarrow \angle \text{AND} = \angle \text{BNC}$

Now, in $\triangle \text{ADN} \text{ and } \triangle \text{BNC}$

$\angle \text{AND} = \angle \text{BNC}$ [from Eq.(i)]

$\text{AN} = \text{BN}$ $\big[\because\ \triangle\text{AMN}\ \cong\ \triangle\text{BMN}\big]$

$\text{and}\ \text{DN}\ =\ \text{NC}$ $\big[\because $ N is the mid-point of CD (given)$\big]$

$\therefore\ \triangle\text{ADN}\ \cong\ \triangle\text{BCN}$ [by SAS congruence rule]

Hence,$\text{AD}\ =\ \text{BC}$ [by CPCT]

Hence proved.

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Question 224 Marks

ABC is a right triangle with AB = AC. If bisector of meets BC at D,then prove that BC = 2AD.

Answer

In $\triangle\text{ABC}$ is a right angle triangle with AB = AC, is the Bisector.

$\text{BC}=2\text{AD}$
$\text{AB}=\text{AC}$
$\angle\text{C}=\angle\text{B}$
Now, in right angle $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$
$\Rightarrow 90^{\circ}+\angle\text{B}+\angle\text{B}=180^{\circ}$
$\Rightarrow 2\angle\text{B}=90^{\circ}$
$\Rightarrow \angle\text{B}=45^{\circ}$
$\Rightarrow \angle\text{B}=\angle\text{C}=45^{\circ}$
$\Rightarrow \angle\text{3}=\angle\text{4}=45^{\circ}$
$\Rightarrow \angle\text{1}=\angle\text{2}=45^{\circ}$
$\text{BD}=\text{AD},\text{DC}=\text{AD}$
Hence, $\text{BC}=\text{BD}+\text{CD}$
$=\text{AD}+\text{AD}$
$=2\text{AD}$

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4 Marks Questions - Maths STD 9 Questions - Vidyadip