3 Marks Question - Maths STD 9 Questions - Vidyadip

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17 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

In the adjoining figure, $ABCD$ is a quadrilateral. A line through $D,$ parallel to $AC,$ meets $BC$ produced in $P.$ Prove that $\text{ar}(\triangle\text{ABP})=\text{ar}(\text{quadrilateral ABCD}).$

Answer

Given: $ABCD$ is a quadrilateral in which through $D.$ A line drawn parallel to $AC$ which meets $BC$ produced in $P.$ To prove: $\text{ar}(\triangle\text{ABP})=\text{ar}(\text{quadrilateral ABCD})$

Proof: $\triangle\text{ACP}$ and $\triangle\text{ACD}$ have same base $AC$ and lie between parallel lines $AC$ and $DP.$
$\therefore\ \text{ar}(\triangle\text{ACP})=\text{ar}(\triangle\text{ACD})$
Adding $\text{ar}(\triangle\text{ABC})$ on both sides,
we get: $\text{ar}(\triangle\text{ACP})+\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})+\text{ar}(\triangle\text{ABC})$
$\Rightarrow\ \text{ar}(\triangle\text{ABP})=\text{ar}(\text{quadrilateral ABCD})$

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Question 23 Marks

In the adjoining figure, $CE || AD$ and $CF || BA.$ Prove that $\text{ar}(\triangle\text{CBG})=\text{ar}(\triangle\text{AFG}).$

Answer

$\triangle\text{BCF}$ and $\triangle\text{ACF}$ are on the same base $CF$ and between the same parallel lines $CF$ and $BA.$
$\therefore\ \text{ar}(\triangle\text{BCF})=\text{ar}(\triangle\text{ACF})$
$\Rightarrow\ \text{ar}(\triangle\text{BCF})-\text{ar}(\triangle\text{CGF})=\text{ar}(\triangle\text{ACF})-\text{ar}(\triangle\text{CGF})$
$\Rightarrow\ \text{ar}(\triangle\text{CBG})=\text{ar}(\triangle\text{AFG})$

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Question 33 Marks

In the adjoining figure, show that $ABCD$ is a parallelogram. Calculate the area of $||\ gm\ ABCD.$

Answer

$\text{Area of }\triangle\text{ABD}=\frac{1}{2}\times\text{base}\times\text{height}$
$=\Big(\frac{1}{2}\times5\times7\Big)\text{cm}^2=\frac{35}{2}\text{cm}^2$
$\text{Area of }\triangle\text{CBD}=\Big(\frac{1}{2}\times5\times7\Big)\text{cm}^2=\frac{35}{2}\text{cm}^2$
Since the diagonal $BD$ divides $ABCD$ into two triangles of equal area.
$\therefore ABCD$ is a parallellogram.
$\therefore$ Area of parallellogram $=\text{Area of }\triangle\text{ABD}+\text{Area of }\triangle\text{CBD}$
$=\Big(\frac{35}{2}+\frac{35}{2}\Big)\text{cm}^2=\frac{70}{2}\text{cm}^2$
$=35\text{cm}^2$
$\therefore$ Area of parallellogram = $35\ cm^2$.

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Question 43 Marks

In the adjoining figure, $BD || CA, E$ is the midpoint of $CA$ and $\text{BD}=\frac{1}{2}\text{CA}.$ Prove that $\text{ar}(\triangle\text{ABC})=2\text{ar}(\triangle\text{DBC}).$

Answer

$E$ is the midpoint of $CA.$
So, $AE = EC ...(1)$
Also,
$\text{BD}=\frac{1}{2}\text{CA}$ [Given]
So, $BD = AE ...(2)$
From $(1)$ and $(2)$ we have
$BD = EC$
$BD || CA$ and $BD = EC$
So, $BDEC$ is a parallelogram $BE$ acts as the median of $\triangle\text{ABC}.$
So,
$\text{ar}(\triangle\text{BCE})=\text{ar}(\triangle\text{ABE})=\frac{1}{2}\text{ar}(\triangle\text{ABC})\dots(1)$
$\text{ar}(\triangle\text{DBC})=\text{ar}(\triangle\text{BCE})\dots(2) [$Triangles on the same base and between the same parallels are equal in area$]$
From $(1)$ and $(2)$
$\text{ar}(\triangle\text{ABC})=2\text{ar}(\triangle\text{DBC})$

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Question 53 Marks

In the adjoining figure, $D$ and $E$ are respectively the midpoints of sides $AB$ and $AC$ of $\triangle\text{ABC}.$ If $PQ || BC$ and $CDP$ and $BEQ$ are straight lines then prove that$\text{ar}(\triangle\text{ABQ})=\text{ar}(\triangle\text{ACP}).$

Answer

Since $D$ and $E$ are the mid-points of $AB$ and $AC$ respectively, $DE || BC || PQ$ In $\triangle\text{ACP},$ $AP || DE$ and $E$ is the mid-point of $AC.$
$\Rightarrow D$ is the mid-point of $PC [$Converse of mid-point theorem$]$
$\Rightarrow\ \text{DE}=\frac{1}{2}\text{AP}$
$\Rightarrow\ \text{AP}=2(\text{DE})\dots(\text{i})$ In
$\triangle\text{ABQ}, AQ || DE$ and $D$ is the mid-point of $AB.$
$ \Rightarrow E$ is the mid-point of $BQ [$Converse of mid-point theorem$]$
$\Rightarrow\ \text{DE}=\frac{1}{2}\text{AQ}$
$\Rightarrow\ \text{AQ}=2(\text{DE})\dots(\text{ii})$
From $(i)$ and $(ii), AP = AQ$
Now, $\triangle\text{ACP}$ and $\triangle\text{ABQ}$ are on the equal bases $AP$ and $AQ$ and between the same parallels $BC$ and $PQ.$
$\Rightarrow\ \text{ar}(\triangle\text{ACP})=\text{ar}(\triangle\text{ABQ})$

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Question 63 Marks

If $P$ and $Q$ are any two points lying respectively on the sides $DC$ and $AD$ of a parallelogram $ABCD$ then show that $\text{ar}(\triangle\text{APB})=\text{ar}(\triangle\text{BQC}).$

Answer

We know $\text{ar}(\triangle\text{APB})=\frac{1}{2}\text{ar}(\text{ABCD})\dots(1)$ [If a triangle and a parallelogram are on the same base and between the same parallels then the area of the triangle is equal to half the area of the parallelogram]
Similarly, $\text{ar}(\triangle\text{BQC})=\frac{1}{2}\text{ar}(\text{ABCD})\dots(2)$ From $(1)$ and $(2)$
$\text{ar}(\triangle\text{APB})=\text{ar}(\triangle\text{BQC})$ Hence proved.

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Question 73 Marks

In a parallelogram $ABCD$, any point $E$ is taken on the side $BC. AE$ and $DC$ when produced meet at a point $M.$ Prove that $\text{ar}(\triangle\text{ADM})=\text{ar}(\text{ABMC}).$

Answer

Construction: Join $AC$ and $BM$ Let $h$ be the distance between $AB$ and $CD.$

$\text{A}(\triangle\text{ACD})=\frac{1}{2}\times\text{CD}\times\text{h}$
$\text{A}(\triangle\text{ABM})=\frac{1}{2}\times\text{AB}\times\text{h}$
$=\frac{1}{2}\times\text{CD}\times\text{h} [AB = CD,$ opposite sides of $||^m ABCD]$
$\Rightarrow\ \text{A}(\triangle\text{ABM})=\text{A}(\triangle\text{ACD})$
$\Rightarrow\ \text{A}(\triangle\text{ABM})+\text{A}(\triangle\text{ACM})=\text{A}(\triangle\text{ACD})+\text{A}(\triangle\text{ACM})$
$\Rightarrow\ \text{A}(\text{ABMC})=\text{A}(\triangle\text{ADM})$

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Question 83 Marks

In a parallelogram $A B C D$, it is being given that $A B=10\ cm$ and the altitudes corresponding to the sides $A B$ and $A D$ are $D L=6\ cm$ and $B M=8\ cm$, respectively. Find $A D$.

Answer

Since $ABCD$ is a parallelogram and $DL$ is perpendicular to $AB .$

So, its area $=A B \times D L=(10 \times 6)\ cm ^2=60\ cm^2$
Also, in parallelogram $A B C D, B M \perp A D$
$\therefore$ Area of parallelogram $A B C D=A D \times B M 60=A D \times 8\ cm$
$\therefore A D \times 8=60$
$\Rightarrow AD=\frac{60}{8}=7.5\ cm$
$ \therefore AD=7.5\ cm$

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Question 93 Marks

$P$ is any point on the diagonal $AC$ of a parallelogram $ABCD.$ Prove that $\text{ar}(\triangle\text{ADP})=\text{ar}(\triangle\text{ABP}).$

Answer

Construction: Join $BD$.
Let the diagonals $AC$ and $BD$ intersect at point $O.$

Diagonals of a parallelogram bisect each other.
Hence, $O$ is the mid-point of both $AC$ and $BD.$
We know that the median of a triangle divides it into two triangles of equal area.
In $\triangle\text{ABD}, OA$ is the median.
$\Rightarrow\ \text{A}(\triangle\text{AOD})=\text{A}(\triangle\text{AOB})\dots(\text{i})$
In $\triangle\text{BPD}, OP$ is the median.
$\Rightarrow\ \text{A}(\triangle\text{OPD})=\text{A}(\triangle\text{OPB})\dots(\text{ii})$
Adding $(i)$ and $(ii),$ we get: $\text{A}(\triangle\text{AOD})+\text{A}(\triangle\text{OPD})=\text{A}(\triangle\text{AOB})+\text{A}(\triangle\text{OPB})$
$\Rightarrow\ \text{A}(\triangle\text{ADP})=\text{A}(\triangle\text{ABP})$

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Question 103 Marks

$D$ is the midpoint of side $BC$ of $\triangle\text{ABC}$ and $E$ is the midpoint of $BD.$ If $O$ is the midpoint of $AE,$ prove that $\text{ar}(\triangle\text{BOE})=\frac{1}{8}\text{ar}(\triangle\text{ABC})$

Answer

$D$ is the midpoint of side $BC$ of $\triangle\text{ABC}.$
$\Rightarrow AD$ is the median of $\triangle\text{ABC}.$
$\Rightarrow\ \text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$
$E$ is the midpoint of side $BD$ of $\triangle\text{ABD},$
$\Rightarrow AE$ is the median of $\triangle\text{ABD}.$
$\Rightarrow\ \text{ar}(\triangle\text{ABE})=\text{ar}(\triangle\text{AED})$
$=\frac{1}{2}\text{ar}(\triangle\text{ABD})=\frac{1}{4}\text{ar}(\triangle\text{ABC})$
Also, $O$ is the midpoint of side $AE,$
$\Rightarrow BO$ is the median of $\triangle\text{ABE},$
$\Rightarrow\ \text{ar}(\triangle\text{ABO})=\text{ar}(\triangle\text{BOE})=\frac{1}{2}\text{ar}(\triangle\text{ABE})$
$=\frac{1}{4}\text{ar}(\triangle\text{ABD})=\frac{1}{8}\text{ar}(\triangle\text{ABC})$
Thus, $\text{ar}(\triangle\text{BOE})=\frac{1}{8}\text{ar}(\triangle\text{ABC})$

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Question 113 Marks

In a triangle $ABC,$ the medians $BE$ and $CF$ intersect at $G.$ Prove that $\text{ar}(\triangle\text{BCG})=\text{ar}(\text{AFGE}.)$

Answer

Construction: Join $EF$

Since the line segment joining the mid-points of two sides of a triangle is parallel to the third side, $FE || BC.$
Clearly, $\triangle\text{BEF}$ and $\triangle\text{CEF}$ are on the same base $EF$ and between the same parallel lines.
$\therefore\ \text{ar}(\triangle\text{BEF})=\text{ar}(\triangle\text{CEF})$
$\Rightarrow\ \text{ar}(\triangle\text{BEF})-\text{ar}(\triangle\text{GEF})=\text{ar}(\triangle\text{CEF})-\text{ar}(\triangle\text{GEF})$
$\Rightarrow\ \text{ar}(\triangle\text{BFG})=\text{ar}(\triangle\text{CEG})\dots(\text{i})$
We know that a median of a triangle divides it into two triangles of equal area.
$\therefore\ \text{ar}(\triangle\text{BEC})=\text{ar}(\triangle\text{ABE})$
$\Rightarrow\ \text{ar}(\triangle\text{BGC})+\text{ar}(\triangle\text{CEG})$
$=\text{ar}(\text{quadrilateral AFGE})+\text{ar}(\triangle\text{BFG})$
$\Rightarrow\ \text{ar}(\triangle\text{BGC})+\text{ar}(\triangle\text{BFG})$
$=\text{ar}(\text{quadrilateral AFGE})+\text{ar}(\triangle\text{BFG}) [$Using $(i)]$
$\Rightarrow\ \text{ar}(\triangle\text{BGC})=\text{ar(quadrilateral AFGE)}$

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Question 123 Marks

In the adjoining figure, $DE \| BC.$ Prove that:
$i. \text{ar}(\triangle\text{ACD})=\text{ar}(\triangle\text{ABE})$
$ii. \text{ar}(\triangle\text{OCE})=\text{ar}(\triangle\text{OBD}).$

Answer

$\triangle\text{DEC}$ and $\triangle\text{DEB}$ lies on the same base and between the same parallel lines.
So, $\text{ar}(\triangle\text{DEC})=\text{ar}(\triangle\text{DEB})\dots(1)$
$i.$ On adding $\text{ar}(\triangle\text{ADE})$ in both sides of equation $(1),$ we get:
$\text{ar}(\triangle\text{DEC})+\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{DEB})+\text{ar}(\triangle\text{ADE})$
$\Rightarrow\ \text{ar}(\triangle\text{ACD})=\text{ar}(\triangle\text{ABE})$
$ii.$ On adding $\text{ar}(\triangle\text{ODE})$ in both sides of equation $(1),$ we get:
$\text{ar}(\triangle\text{DEC})-\text{ar}(\triangle\text{ODE})=\text{ar}(\triangle\text{DEB})-\text{ar}(\triangle\text{ODE})$
$\Rightarrow\ \text{ar}(\triangle\text{OCE})=\text{ar}(\triangle\text{OBD}).$

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Question 133 Marks

In the adjoining figure, $\text{MNPQ}$ and $\text{ABPQ}$ are parallelograms and $T$ is any point on the side $BP.$ Prove that
$i. \text{ar}(\text{MNPQ})=\text{ar}(\text{ABPQ})$
$ii. \text{ar}(\triangle\text{ATQ})=\frac{1}{2}\text{ar}\text{(MNPQ}).$

Answer

$i.$ We know that parallelograms on the same base and between the same parallels are equal in area.
So,
$\text{ar}(\text{MNPQ})=\text{ar}(\text{ABPQ})[$Same base $PQ$ and $MB \| PQ] ...(1)$
$ii.$ If parallelogram and a triangle are on the same base and between the same paralles then the area of the triangle is equal to half the area of the parallelogram.
So, $\text{ar}(\triangle\text{ATQ})=\frac{1}{2}\text{ar}(\text{ABPQ})[$Same base $AQ$ and $AQ \| BP] ...(2)$
From $(1)$ and $(2)$
$\text{ar}(\triangle\text{ATQ})=\frac{1}{2}\text{ar}(\text{MNPQ})$

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Question 143 Marks

In the adjoining figure, $ABCD$ is a quadrilateral in which diagonal $BD = 14 \ cm.$ If $\text{AL}\perp\text{BD}$ and $\text{CM}\perp\text{BD}$such that $AL = 8\ cm$ and $CM = 6\ cm,$ find the area of quadrilateral $ABCD.$

Answer

Area of $\triangle\text{BAD}=\frac{1}{2}\times\text{BD}\times\text{AL}$
$=\Big(\frac{1}{2}\times14\times8\Big)\text{cm}^2$
$=56\text{cm}^2$ Area of $\triangle\text{CBD}=\frac{1}{2}\times\text{BD}\times\text{CM}$
$=\Big(\frac{1}{2}\times14\times6\Big)\text{cm}^2$
$=42\text{cm}^2$

$\therefore$ Area of quadrilateral $ABCD=\text{Area of }\triangle\text{ABD}+\text{Area of }\triangle\text{CBD} = (56 + 42)cm^2 = 98cm^2$

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Question 153 Marks

Show that a diagonal divides a parallelogram into two triangles of equal area.

Answer

Let $ABCD$ be a parallelogram and $BD$ be its diagonal.
To prove: $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{CDB})$
Proof: In $\triangle\text{ABD}$ and $\triangle\text{CDB},$
we have : $AB = CD [$Opposite sides of a parallelogram$] AD = CB [$Opposite sides of a parallelogram$] BD = DB [$Common$]$
i. e., $\triangle\text{ABD}\cong\triangle\text{CDB} [SSS$ criteria$]$
$\therefore\ \text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{CDB})$

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Question 163 Marks

$ABCD$ is a parallelogram in which $BC$ is produced to $P$ such that $CP = BC$, as shown in the adjoining figure. $AP$ intersects $CD$ at $M.$ If ar$(DMB) = 7\ cm^2$, find the area of parallelogram $ABCD.$

Answer

In $\triangle\text{ADM}$ and $\triangle\text{PCM},$
$\angle\text{ADM}=\angle\text{PCM}$ [alternate angles]
$AD = CP [AD = BC = CP]$
$\angle\text{AMD}=\angle\text{PMC}$ [vertically opposite angles]
$\therefore\ \triangle\text{ADM}\cong\triangle\text{PCM}$
$\Rightarrow\ \text{A}(\triangle\text{ADM})=\text{A}(\triangle\text{PCM})$ And, $DM = CM [C.P.C.T.]$
$\Rightarrow BM$ is the median of $\triangle\text{BDC}.$
$\Rightarrow\ \text{A}(\triangle\text{DMB})=\text{A}(\triangle\text{CMB})$
$\Rightarrow\ \text{A}(\triangle\text{BDC})=2\times\text{A}(\triangle\text{DMB})$
$\Rightarrow\ 2\times7=14\text{cm}^2$
Now, A(parallelogram ABCD) $=2\times\text{A}(\triangle\text{BDC})$
$=2\times14=28\text{cm}^2$

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Question 173 Marks

In the adjoining figure, the diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect at $O$. If $BO = OD,$ prove that $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ADC}).$

Answer

Given: A quadrilateral $ABCD$ in which diagonal $AC$ and $BD$ intersect at $O$ and $BO = OD.$

To prove: $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ADC})$
Proof: Since $OB = OD [$Given$]$
So, $AO$ is the median of $\triangle\text{ABD}$
$\therefore\ \text{ar}(\triangle\text{AOD})=\text{ar}(\triangle\text{AOB})\dots(\text{i})$
As $OC$ is the median of $\triangle\text{CBD}$
$\therefore\ \text{ar}(\triangle\text{DOC})=\text{ar}(\triangle\text{BOC})\dots(\text{ii})$
Adding both sides of $(i)$ and $(ii),$
we get $\text{ar}(\triangle\text{AOD})+\text{ar}(\triangle\text{DOC})=\text{ar}(\triangle\text{AOB})+\text{ar}(\triangle\text{BOC})$
$\therefore\ \text{ar}(\triangle\text{ADC})=\text{ar}(\triangle\text{ABC})$

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