4 Marks Questions

Question 14 Marks

In the adjoining figure, $ABCD$ is a trapezium in which $AB || DC; AB = 7\ cm; AD = BC = 5\ cm$ and the distance between $AB$ and $DC$ is $4\ cm$. Find the length of $DC$ and hence, find the area of trap. $ABCD$.

Answer

In right angled $\triangle\text{ALD},$
$\text{DL}^2=\text{AD}^2=5^2-4^2$ $=25-16=9$
$\Rightarrow\ \text{DL}=\sqrt{9}=3\text{cm}$
Again, in right angled $\triangle\text{BMC},$
$\text{MC}^2=\text{CB}^2-\text{MB}^2$
$=5^2-4^2=25-16=9\text{cm}$
$\Rightarrow\ \text{MC}=\sqrt{9}=3\text{cm}$
$\text{LM}=\text{AB}=7\text{cm}$
Now, $\text{CD}=\text{DL}+\text{LM}+\text{MC}$
$=3+7+3=13\text{cm}$
Area of trapezium $ABCD$ $=\frac{1}{2}\times\text{Sum of parallel sides}\times\text{Height}$
$=\frac{1}{2}\times(\text{CD}+\text{AB})\times\text{AL}$
$=\frac{1}{2}\times(13+7)\times4$
$=20\times2$
$=40\text{cm}^2$

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Question 24 Marks

The given figure shows a pentagon $ABCDE$. $EG$, drawn parallel to $DA$, meets $BA$ produced at $G$, and $CF$, drawn parallel to $DB$, meets $AB$ produced at $F$. Show that: $\text{ar}(\text{pentagon ABCDE})=\text{ar}(\triangle\text{DGF}).$

Answer

Given: $ABCDE$ is a pentagon $EG$, drawn parallel to $DA$, meets $BA$ produced at $G$ and $CF$, drawn parallel to $DB$, meets $AB$ produced at $F$$$.
To prove: $\text{ar}(\text{pentagon ABCDE})=\text{ar}(\triangle\text{DGF}).$
Proof: Triangles on the same base and between the same parallels are equal in area.
Since $\triangle\text{DGA}$ and $\triangle\text{AED}$ have same base $AD$ and lie between parallels lines $AD$ and $EG$.
$\therefore\ \text{ar}(\triangle\text{DGA})=\text{ar}(\triangle\text{AED})\dots(1)$
Similarly, $\triangle\text{DBC}$ and $\triangle\text{BFD}$ have same base $DB$ and lie between parallels lines $BD$ and $CF$.
$\therefore\ \text{ar}(\triangle\text{DBF})=\text{ar}(\triangle\text{DBC})\dots(2)$
Adding both the sides of the equations $(1)$ and $(2),$ we have
$\therefore\ \text{ar}(\triangle\text{DGA})+\text{ar}(\triangle\text{DBF})+\text{ar}(\triangle\text{ABD})\\=\text{ar}(\triangle\text{AED})+\text{ar}(\triangle\text{BCD})+\text{ar}(\triangle\text{ABD})$
$\therefore\ \text{ar}(\triangle\text{DGF})=\text{ar}(\text{pentagon ABCDE})$

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Question 34 Marks

In the adjoining figure, the point $D$ divides the side $BC$ of $\triangle\text{ABC}$ in the ratio $m : n$. Prove that $\text{ar}(\triangle\text{ABD}):\text{ar}(\triangle\text{ADC})=\text{m}:\text{n}$.

Answer

Given: $ABC$ is a triangle in which $D$ is a point on $BC$ such that; $BD : DC = m : n$

To prove: $\text{ar}(\triangle\text{ABD}):\text{ar}(\triangle\text{ADC})=\text{m}:\text{n}$
Proof: $\text{ar}(\triangle\text{ABD})=\frac{1}{2}\times\text{BD}\times\text{AL}$ And, $\text{ar}(\triangle\text{ADC})=\frac{1}{2}\times\text{DC}\times\text{AL}$
Now, $\text{BD}:\text{DC}=\text{m}:\text{n}$ $\therefore\ \text{BD}=\text{DC}\times\frac{\text{m}}{\text{n}}$
$\therefore\ \text{ar}(\triangle\text{ABD})=\frac{1}{2}\times\text{BD}\times\text{AL}$
$=\frac{1}{2}\times\Big(\text{DC}\times\frac{\text{m}}{\text{n}}\Big)\times\text{AL}$
$=\frac{\text{m}}{\text{n}}\times\Big(\frac{1}{2}\times\text{DC}\times\text{AL}\Big)$
$=\frac{\text{m}}{\text{n}}\times\text{ar}(\triangle\text{ADC})$
$\Rightarrow\ \frac{\text{ar}(\triangle\text{ABD})}{\text{ar}(\triangle\text{ADC})}=\frac{\text{m}}{\text{n}}$
$\Rightarrow\ \text{ar}(\triangle\text{ABD}):\text{ar}(\triangle\text{ADC})=\text{m}:\text{n}$

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Question 44 Marks

The base $BC$ of $\triangle\text{ABC}$ is divided at $D$ such that $\text{BD}=\frac{1}{2}\text{DC}.$ Prove that $\text{ar}(\triangle\text{ABD})=\frac{1}{3}\times\text{ar}(\triangle\text{ABC}).$

Answer

Given: $D$ is a point on $BC$ of $\triangle\text{ABC},$ such that $\text{BD}=\frac{1}{2}\text{DC}$
To prove: $\text{ar}(\triangle\text{ABD})=\frac{1}{3}\times\text{ar}(\triangle\text{ABC}).$
Construction: Draw $\text{AL}\perp\text{BC}.$
Proof: In $\triangle\text{ABC},$
we have: $BC = BD + DC \Rightarrow BD + 2BD = 3 \times BD$
Now, we have: $\text{ar}(\triangle\text{ABD})=\frac{1}{2}\times\text{BD}\times\text{AL}$
$\text{ar}(\triangle\text{ABC})=\frac{1}{2}\times\text{BC}\times\text{AL}$
$\Rightarrow\ \text{ar}(\triangle\text{ABC})=\frac{1}{2}\times3\text{BD}\times\text{AL}$
$\Rightarrow\ 3\times\Big(\frac{1}{2}\times\text{BD}\times\text{AL}\Big)$
$\Rightarrow\ \text{ar}(\triangle\text{ABC})=3\times\text{ar}(\triangle\text{ABD})$
$\therefore\ \text{ar}(\triangle\text{ABD})=\frac{1}{3}\text{ar}(\triangle\text{ABC})$

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Question 54 Marks

Calculate the area of trapezium $PQRS$, given in figure $(ii).$

Answer

$\text{RT}\perp\text{PQ}$ In right angled $\triangle\text{RTQ}$
$RT^2 = RQ^2 - TQ^2 = 17^2 - 8^2 = 289 - 64 = 225cm^2$
$\therefore\ \text{RT}=\sqrt{225}=15\text{cm}$
$\therefore$ Area of trapezium
$=\frac{1}{2}(\text{Sum of parallel sides})\times\text{Distance between them}$
$=\frac{1}{2}\times(\text{PQ}+\text{SR})\times\text{RT}$
$=\frac{1}{2}\times(16+8)\times15$
$=\Big(\frac{1}{2}\times24\times15\Big)\text{cm}^2$
$=180\text{cm}^2$
$\therefore$ Area of trapezium $= 180cm^2.$

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Question 64 Marks

$BD$ is one of the diagonals of a quadrilaterl $ABCD$.
If $\text{AL}\perp\text{BD}$ and $\text{CM}\perp\text{BD},$show that
ar$($quadrilaterl $ABCD)$ $=\frac{1}{2}\times\text{BD}\times(\text{AL}+\text{CM}).$

Answer

ar(quadrilaterl ABCD) $=\text{ar}(\triangle\text{ABD})+\text{ar}(\triangle\text{DBC})$
$\text{ar}(\triangle\text{ABD})=\frac{1}{2}\times\text{base}\times\text{height}$
$=\frac{1}{2}\times\text{BD}\times\text{AL}\dots(1)$
$\text{ar}(\triangle\text{DBC})=\frac{1}{2}\times\text{BD}\times\text{CL}\dots(2)$
From $(1)$ and $(2)$, we get:
ar$($quadrilaterl $ABCD)$ $=\frac{1}{2}\times\text{BD}\times\text{AL}+\frac{1}{2}\times\text{BD}\times\text{CL}$
$\Rightarrow $ ar$($quadrilaterl$ ABCD)$ $=\frac{1}{2}\times\text{BD}\times(\text{AL}+\text{CL})$
Hence, proved.

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Question 74 Marks

$M$ is the mid-point of the side $AB$ of a parallelogram $ABCD$. If ar $(AMCD) = 24cm^2$, find $\text{ar}(\triangle\text{ABC}).$

Answer

Join $AC. AC$ divides parallelogram $ABCD$ into two congruent triangles of equal areas.
$\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar}(\text{ABCD})$
$M$ is the mid-point of $AB$. So, $CM$ is the median.
$CM$ divides $\triangle\text{ABC}$ in two triangles with equal area. $\text{ar}(\triangle\text{AMC})=\text{ar}(\triangle\text{BMC})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$
$\text{ar}(\text{AMCD})=\text{ar}(\triangle\text{ACD})+\text{ar}(\triangle\text{AMC})$
$=\text{ar}(\triangle\text{ABC})+\text{ar}(\triangle\text{AMC})=\text{ar}(\triangle\text{ABC})+\frac{1}{2}\text{ar}(\triangle\text{ABC})$
$\Rightarrow\ 24=\frac{3}{2}\text{ar}(\triangle\text{ABC})$
$\Rightarrow\ \text{ar}(\triangle\text{ABC})=16\text{cm}^2$

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Question 84 Marks

The vertex $A$ of $\triangle\text{ABC}$ is joined to a point $D$ on the side $BC$.
The mid-point of $AD$ is $E$.
Prove that $\text{ar}(\triangle\text{BEC})=\frac{1}{2}\text{ar}(\triangle\text{ABC}).$

Answer

Given:
$A$ $\triangle\text{ABC}$ in which $E$ is the mid-point of line segment $AD$ where $D$ is a point on $BC$.

To prove:
$\text{ar}(\triangle\text{BEC})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$
Proof:
Since $BE$ is the median of $\triangle\text{ABD}$
So,
$\text{ar}(\triangle\text{BDE})=\text{ar}(\triangle\text{ABE})$
$\text{ar}(\triangle\text{BDE})=\frac{1}{2}\text{ar}(\triangle\text{ABD})\dots(\text{i})$
As, $CE$ is median of $\triangle\text{ADC}$
$\text{ar}(\triangle\text{CDE})=\frac{1}{2}\text{ar}(\triangle\text{ACD})\dots(\text{ii})$
Adding $(i)$ and $(ii)$, we get
$\text{ar}(\triangle\text{BDE})=\text{ar}(\triangle\text{CDE})\\=\frac{1}{2}\text{ar}(\triangle\text{ABD})+\frac{1}{2}\text{ar}(\triangle\text{ACD})$
$\text{ar}(\triangle\text{BEC})=\frac{1}{2}[\text{ar}(\triangle\text{ABD})+\text{ar}(\triangle\text{ACD})]$
$=\frac{1}{2}\text{ar}(\triangle\text{ABC}).$

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Question 94 Marks

Calculate the area of quadrilateral $ABCD$, given in Figure $(i).$

Answer

$ABCD$ is a quadrilateral.

Now in rightarrow angled $\triangle\text{DBC},$
$DB^2 = DC^2 - CB^2 = 17^2 - 8^2 = 289 - 64 = 225cm^2$
$\therefore\ \text{DB}=\sqrt{225}=15\text{cm}$
So, $\text{Area of }\triangle\text{DBC}=\Big(\frac{1}{2}\times15\times8\Big)\text{cm}^2$
$=60\text{cm}^2$
Again, in right angled $\triangle\text{DAB},$
$AB^2 = DB^2 - AD^2 = 15^2 - 9^2 = 225 - 81 = 144cm^2$
$\therefore\ \text{AB}=\sqrt{144}=12\text{cm}$
$\therefore\ \text{Area of }\triangle\text{DAB}=\Big(\frac{1}{2}\times12\times9\Big)\text{cm}^2$
$=54\text{cm}^2$
So, Area of quadrilateral $ABCD$ $=\text{Area of }\triangle\text{DBC}+\text{Area of }\triangle\text{DAB}$
$= (60 + 54)cm^2 = 114cm^2$
$\therefore$ Area of quadrilateral $ABCD = 114cm^2.$

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