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Maths 4 Marks Questions

Triangles · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 75 questions.

Questions · Page 2 of 2

4 Marks Questions

Question 514 Marks
Two triangles $ABC$ and $PQR$ are such that $AB = 3\ cm, AC = 6\ cm$, $\angle\text{A}=70^\circ,\text{PR}=9\text{cm},\angle\text{P}=70^\circ$ and $\text{PQ}=4.5\text{cm}.$ show that $\triangle\text{ABC}\sim\triangle\text{PQR}$ and state the similarity criterion.
Answer

In $\triangle\text{ABC}$ and $\triangle\text{PQR},$
$\angle\text{A}=\angle\text{P}=70^\circ\ ...\ \text{(Given)}$
$\frac{\text{AB}}{\text{PQ}}=\frac{3}{4.5}=\frac{2}{3}$
$\frac{\text{AC}}{\text{PR}}=\frac{6}{9}=\frac{2}{3}$
$\Rightarrow\frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}$
so, $\triangle\text{ABC}\sim\triangle\text{PQR}$ ...(SAS criterion for similarity)
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Question 524 Marks
Find the length of each side of a rhombus whose diagonals are $24\ cm$ and $10\ cm$ long.
Answer

Let $A B C D$ be the rhobbus with diagonals $A C$ and $B D$ intersecting each other at $O$.
We know that the diagonals of a rhombus bisect each other at right angles.
$\therefore$ if $AC = 24cm$ and $BD = 10cm, AO = 12cm$ and $BO = 5cm$
Applying Pythagoras theorem in right- angled triangle AOB, we get:
$A B^2=A O^2+B O^2=12^2+5^2=144+25=169$
$AB = 13cm$
Hence, the lenght of each side of the given rhombus is $13cm$
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Question 534 Marks
For the following statments state whether true $(T)$ or false$(F):$
If O is any point inside a rectangle ABCD then $OA^2 + OC^2 = OB^2 + OD^2$
Answer
True.
Solution:

Construction: Draw $EF || AB$ through $O.$
In $\triangle\text{OEA}$ and $\triangle\text{OFC},$ by pythagoras theorem,
$O A^2=O E^2+A E^2 \text { and } O C^2=O F^2+C F^2$
Adding the two equations, we get
$O A^2+O C^2=O E^2+A E^2+O F^2+C F^2 \ldots \ldots \text { (i) }$
$\triangle \mathrm{OFB}$ and $\triangle \mathrm{ODE}$, by pythagoras theorem,
$O B^2=O F^2+F B^2 \text { and } O D^2=O E^2+D E^2$
Adding the two equation, we get
$O B^2+O D^2=O F^2+F B^2+O E^2+D E^2....(ii)$
By Construction since $EF || CD,$
$\mathrm{DE}=\mathrm{CF} \text { and } \mathrm{AE}=\mathrm{FB}$
So, from$(i)$ and $(ii)$,, we have
$O A^2+O C^2=O B^2+O D^2$
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Question 544 Marks
In the given figure, $A B C$ is a triangle and $P Q$ is a straight line meeting $A B$ in $P$ and $A C$ in $Q$. If $A P=1 cm . P B=3 cm, A Q=1.5 cm, Q C=4.5 cm$, Prove that area of $\triangle A P Q$ is $\frac{1}{16}$ of the area of $\triangle A B C$.
Answer
We have:
$\frac{\text{AP}}{\text{AB}}=\frac{1}{1+3}=\frac{1}{4}$ and $\frac{\text{AQ}}{\text{AC}}=\frac{1.5}{1.5+4.5}=\frac{1.5}{6}=\frac{1}{4}$
$\Rightarrow\frac{\text{AP}}{\text{AB}}=\frac{\text{AQ}}{\text{AC}}$
Also,
$\angle\text{A}=\angle\text{A}$
By $SAS$ similarity, we can conclude that $\triangle\text{APQ}\sim\triangle\text{ABC}$
$\frac{\text{ar}(\triangle\text{APQ})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{AP}^2}{\text{AB}^2}=\frac{1^2}{4^2}=\frac{1}{16}$
$\Rightarrow\frac{\text{ar}(\triangle\text{APQ})}{\text{ar}(\triangle\text{ABC})}=\frac{1}{16}$
$\Rightarrow\text{ar}(\triangle\text{APQ})=\frac{1}{16}\times\text{ar}(\triangle\text{ABC})$
Hence proved.
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Question 554 Marks
Two vertical poles of height 9m and $14\ m$ stand on a plane ground. If the distance between their feet is $12\ m$, find the distance between their tops.
Answer
Let $AB$ and $CD$ be the given vertical poles.
Then,
$AB = 9\ m, CD = 14\ m$ and $AC = 12\ m$

Const Draw, $BE || AC.$
Then,
$CE = AB = 9m$ and $BE = AC = 12m$
$\therefore\text{ DE} = \text{(CD} -\text{ CE)}$
$= (14 - 9)$
$= 5\text{m}$
In right $\triangle\text{BED},$ we have
$\text{BD}^2=\text{BE}^2+\text{DE}^2$
$=\Big[(12)^2+(5)^2\Big]\text{m}^2$
$=(144+25)\text{m}^2$
$=169\text{m}^2$
$\Rightarrow\text{BD}=\sqrt{169}=13\text{m}$
Hence, the distance between their tops is $13m.$
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Question 564 Marks
In the given figure, $DE || BC$ and $DE : BC = 3 : 5$. Calculate the ratio of the areas of $\triangle\text{ADE}$ and the trapezium $BCED.$
Answer
It is given that $DE || BC.$
$\therefore\angle\text{ADE}=\angle\text{ABC}$ (Corresponding angles)
$\angle\text{AED}=\angle\text{ACB}$ (Corresponding angles)
Applying AA similarity theorem, we can conclude that $\triangle\text{ADE}\sim\triangle\text{ABC}.$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{ADE})}=\frac{\text{BC}^2}{\text{DE}^2}$
Subtracting 1 from both sides, we get:
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{ADE})}-1=\frac{5^2}{3^2}-1$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})-\text{ar}(\triangle\text{ADE})}{\text{ar}(\triangle\text{ADE})}=\frac{25-9}{9}$
$\Rightarrow\frac{\text{ar}(\triangle\text{BCED})}{\text{ar}(\triangle\text{ADE})}=\frac{16}{9}$
or, $\frac{\text{ar}(\triangle\text{ADE})}{\text{ar}(\triangle\text{BCED})}=\frac{9}{16}$
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Question 574 Marks
For the following statments state whether true $(T)$ or false$(F):$
The length of the line segment joining the midpoint of any two sides of a triangle is equal to half the length of the third side.
Answer
True.
Solution:

Construction: Produce $D E$ to $F$ that $D E=E F$.
By the Basic proportinality theorem,
$D E \| B C$
That is, $D F \| B C....(i)$
In $\triangle\text{ADE}$ and $\triangle\text{CEF},$
$\angle\text{AED}=\angle\text{CEF}$ ....(Vertically opposite angles)
$\text{DE}=\text{EF}$ ....(Costruction)
$\text{AE}=\text{EC} ....(E$ is mid-point of $AC)$
$\Rightarrow\triangle\text{ADE}\cong\triangle\text{CEF}$....(SAS congruence criterion)
$\angle\text{ADE}=\angle\text{CFE}$ ....(cpct)
$\Rightarrow\text{AD }||\text{ CF}$
that is, $\text{AD }||\text{ CF}\Rightarrow\text{BD }||\text{ CF}\dots(\text{ii})$
So, from $(i)$ and $(ii)$,
$BDFC$ is a parallelogram.
$\Rightarrow DF = BC$
So, $\text{DE}=\frac{1}{2}\text{DF}=\frac{1}{2}\text{BC}$
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Question 584 Marks
Show that the line segment which joins the midpoint of the oblique sides of a trapezium is parallel to the parallel sides.
Answer
Let the trapezium be $A B C D$ with $E$ and $F$ as the points of $A D$ and $B C$, respectively. Produce $A D$ and $B C$ to meet at $P$.

In $\triangle\text{PAB},\text{ DC }|| \text{ AB}.$
Applying Thales' theorem, we get:
$\frac{\text{PD}}{\text{DA}}=\frac{\text{PC}}{\text{CB}}$
Now, E and F are the midpoint of $AD$ and $BC$, respectively.
$\Rightarrow\frac{\text{PD}}{2\text{DE}}=\frac{\text{PC}}{2\text{CF}}$
$\Rightarrow\frac{\text{PD}}{\text{DE}}=\frac{\text{PC}}{\text{CF}}$
Applying the converse of Thales' theorem in $\triangle PEF$, we get that $DC \| EF$.
Hence,$ EF || AB$.
Thus. $E F$ is parallel to both $A B$ and $D C$.
This completes the proof.
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Question 594 Marks
For the following statments state whether true $(T)$ or false$(F):$
If two triangles are similar then their corresponding angles are equal and their corresponding sides are equal.
Answer
False.
Solution:
Two triangles are said to be similar to each other if:
  1. Their corresponding angles are equal.
  2. Their corresponding sides are proportional.
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Question 604 Marks
The perimeters of two similar triangles $A B C$ and $P Q R$ are$ 32\ cm$ and $24\ cm$ respectively. If $PQ = 12\ cm$, find $AB.$
Answer
$\triangle\text{ABC}$ and $\triangle\text{PQR}$ are similar triangles, therefore corresponding sides of both the triangle are proportional.
So, $\frac{\text{Perimeter of }\triangle\text{ABC}}{\text{Perimeter of }\triangle\text{PQR}}=\frac{\text{AB}}{\text{PQ}}$
Let, $AB = x cm$
Then, $\frac{\text{x}}{12}=\frac{32}{24}$
$\text{x}=\frac{32\times12}{24}=16\text{cm}$
Hence, $AB = 16cm$
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Question 614 Marks
In the given figure, $\angle ABC =90^{\circ}$ and $BD \perp AC$. If $BD =8 cm, AD =4 cm$, find $CD .$
Answer

Given that$ BD = 8\ cm, AD = 4\ cm$
In $\triangle\text{DBA}$ and $\triangle\text{DCB},$ we have
$\angle\text{BDA}=\angle\text{CDB}=90^\circ$
$\angle\text{DBA}=\angle\text{DCB}$ $\big[\text{each}=90^\circ-\angle\text{A}\big]$
$\therefore\triangle\text{DBA}\sim\triangle\text{DCB}$ (by AAA similaritiy)
$\therefore\frac{\text{BD}}{\text{CD}}=\frac{\text{AD}}{\text{BD}}$
$\Rightarrow\text{CD}=\frac{\text{BD}^2}{\text{AD}}$
$\Rightarrow\text{CD}=\frac{(\text{8)}^2}{\text{4}}=\frac{64}{4}=16\text{cm}$
Hence, $CD= 16\ cm$
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Question 624 Marks
Each of the equal sides of an isosceles triangle is $25\ cm$. Find the length of its altitude if the $14\ cm.$
Answer

Let $\triangle\text{ABC}$ be the isosceles triangle and $AD$ be the altitude.
The height of an isosceles triangles is the same as its median.
So, $BD = DC = 7cm$
$\triangle\text{ADB}$ is a right.angled triangle.
By Pythagoras theorem,
$ A B^2=A D^2+B D^2 $
$ \Rightarrow A D^2=A B^2-B D^2 $
$ \Rightarrow A D^2=25^2-7^2 $
$ \Rightarrow A D^2=625-49 $
$ \Rightarrow A D^2=576 $
$ \Rightarrow A D=24 \mathrm{~cm}$
Hence, the length of the altitude is $24\ cm.$
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Question 634 Marks
In the given figure, $XY || AC$ and $XY$ divides $\triangle\text{ABC}$ into two regions, equal in area. Show that $\frac{\text{AX}}{\text{AB}}=\frac{(2-\sqrt{2})}{2}.$
Answer
In $\triangle\text{ABC}$ and $\triangle\text{BXY},$ we have:
$\angle\text{B}=\angle\text{B}$
$\angle\text{BXY}=\angle\text{BAC}$ (Corresponding angles)
Thus, $\triangle\text{ABC}\sim\triangle\text{BXY}$ (AA criterion)
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BXY})}=\frac{\text{AB}^2}{\text{BX}^2}=\frac{\text{AB}^2}{\text{(AB}-\text{AX})^2}\dots(\text{i})$
Also, $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BXY})}=\frac{2}{1}$ $\{\therefore(\triangle\text{BXY})=\text{ar}(\text{trapezium AXYC})\}\dots(\text{ii})$
From $(i)$ and $(ii)$,, we have:
$\Rightarrow\frac{\text{AB}^2}{(\text{AB}-\text{AX})^2}=\sqrt{2}$
$\Rightarrow\frac{(\text{AB}-\text{AX})}{\text{AB}}=\frac{1}{\sqrt{2}}$
$\Rightarrow1-\frac{\text{AX}}{\text{AB}}=\frac{1}{\sqrt{2}}$
$\Rightarrow\frac{\text{AX}}{\text{AB}}=1-\frac{1}{\sqrt{2}}$
$=\frac{\sqrt{2}-1}{\sqrt{2}}=\frac{(2-\sqrt{2})}{\sqrt{2}}$
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Question 644 Marks

The sides of certain triangles are given below. Determine them are right triangles:

$(\text{a} - 1)\text{cm},2\sqrt{\text{a}}\text{ cm},(\text{a} + 1)\text{cm}.$

Answer

For a given triangle to be a right angled, the sum of the squares of the two sides must be equal to the square of the largest side.

Let $\text{p}=(\text{a} - 1)\text{cm},\text{q}=2\sqrt{\text{a}}\text{ cm}$ and $\text{r}=(\text{a}+1)\text{cm}^2$

$\text{p}^2+\text{q}^2=\Big[(\text{a}-1)^2+\big(2\sqrt{\text{a}}\big)^2\Big]\text{cm}^2$

$=\big(\text{a}^2+1-2\text{a}+4\text{a}\big)\text{cm}^2$

$=\big(\text{a}^2+1+2\text{a}\big)\text{cm}^2=(\text{a}+1)^2\text{cm}^2$

$\text{r}^2=(\text{a}+1)^2\text{cm}^2$

$\therefore\text{p}^2+\text{q}^2=\text{r}^2$

Hence, the given triangle is a right triangle.

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Question 654 Marks
If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $2 A B=D E$ and $B C=6 cm$, find $E F$.
Answer
$\therefore\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}$
$\Rightarrow\frac{1}{2}=\frac{6}{\text{EF}}$
$\Rightarrow\text{EF}=12\text{cm}$
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Question 664 Marks
In the given figure, $\triangle\text{ABC}$ and $\triangle\text{DBC}$ have the same base $BC$. If $AD$ and $BC$ intersect at $O$, prove that $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DBC})}=\frac{\text{AO}}{\text{DO}}.$
Answer

Construction: Draw $\text{AX}\perp\text{CO}$ and $\text{DY}\perp\text{BO}.$
As,
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DBC})}=\frac{\frac{1}{2}\times\text{AX}\times\text{BC}}{\frac{1}{2}\times\text{DY}\times\text{BC}}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DBC})}=\frac{\text{AX}}{\text{DY}}\dots(\text{i})$
In $\triangle\text{ABC}$ and $\triangle\text{DBC},\angle\text{AXY}=\angle\text{DYO}=90^\circ(\text{By construction})$
$\angle\text{AOX}=\angle\text{DOY}(\text{Vertically opposite angles})$$\therefore\triangle\text{AXY}\sim\triangle\text{DYO }(\text{By AA criterion})$
This completes the proof.
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Question 674 Marks
In a $\triangle\text{ABC},\text{AD}$ is the bisector of $\angle\text{A}.$
If $A B=10 cm, A C=14 cm$ and $B C=6 cm$, find $B D$ and $D C$.
Answer
It is given that $AD$ bisects $\angle\text{A}.$
Applying angle-bisector theorem in $\triangle\text{ABC},$ we get:
$\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$
Let BD be $x cm.$
Therefore, $DC = (6 - x)cm$
$\Rightarrow\frac{\text{x}}{6-\text{x}}=\frac{10}{14}$
$\Rightarrow14\text{x}=60-10\text{x}$
$\Rightarrow24\text{x}=60$
$\Rightarrow\text{x}=2.5\text{cm}$
Thus, $BD = 2.5cm$
$DC = 6 - 2.5 = 3.5cm$
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Question 684 Marks
In the adjoining figure, $A B C$ is a triangle in which $A B=A C$. If $D$ and $E$ are points on $A B$ and $A C$ respectively such that $A D=A E$, show that the points $B, C, E $and $D$ are concyclic.
Answer
Given:
$AD = AE ...(i)$
$AB = AC ...(ii)$
Subtracting $AD$ from both sides, we get:
$\Rightarrow AB - AD = AC - AD$
$\Rightarrow AB - AD = AC - AE$ (Since, $AD = AE)$
$\Rightarrow BD = EC ...(iii)$
Dividing equation $(i)$ by equation $(iii)$, we get:
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
Dividing the converse pf Thales' theorem, $DE || BC$
$\Rightarrow\angle\text{DEC}+\angle\text{ECB}=180^\circ$ (Sum of interior angles on the same side of a transversal line is $180^\circ )$
$\Rightarrow\angle\text{DEC}+\angle\text{CBD}=180^\circ$
$($Since, $AB = AC \Rightarrow \angle\text{B}=\angle\text{C})$
Hence, quadrilateral $BDED$ is cyclic.
Therefore, $B, C, E$ and $D$ are concyclic points.
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Question 694 Marks
In the given figure, $MN || BC$ and $AM : MB = 1 : 2$
Find $\frac{\text{area}(\triangle\text{AMN})}{\text{area}(\triangle\text{ABC})}.$
Answer
In $\triangle\text{ABC}$ and $\triangle\text{AMN},$
So, $\angle\text{M}=\angle\text{B}$ and $\angle\text{N}=\angle\text{C}$ ....(corresponding angles)
$\Rightarrow\triangle\text{ABC}\sim\triangle\text{AMN}$ ....(AA criterion for similarity)
$\Rightarrow\frac{\text{area}(\triangle\text{AMN})}{\text{area}(\triangle\text{ABC})}=\frac{\text{AM}^2}{\text{AB}^2}$
Now, $\frac{\text{AM}}{\text{MB}}=\frac{1}{2}$
$\Rightarrow\frac{\text{MB}}{\text{AM}}=2$
By Componendo,
$\Rightarrow\frac{\text{MB}+\text{AM}}{\text{AM}}=2+1$
$\Rightarrow\frac{\text{AB}}{\text{AM}}=3$
$\Rightarrow\frac{\text{AM}}{\text{AB}}=\frac{1}{3}$
So, $\Rightarrow\frac{\text{area}(\triangle\text{AMN})}{\text{area}(\triangle\text{ABC})}=\bigg(\frac{\text{AM}^2}{\text{AB}^2}\bigg)^2=\bigg(\frac{1}{3}\bigg)=\frac{1}{9}.$
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Question 704 Marks
Find the length of each side of a rhombus whose diagonals are $24\ cm$ and $10\ cm$ long.
Answer

In an rhombus, the diagonals are perpendicular bisectors of each other, and side are equal to eachother.
So, $\text{AO}=\frac{1}{2}\text{AC}=12\text{cm}$
$\text{OD}=\frac{1}{2}\text{BD}=5\text{cm}$
In right-angled $\triangle\text{AOD},$
$\text{AD}^2=\text{AO}^2+\text{OD}^2$
$\Rightarrow\text{AD}^2=\text{12}^2+\text{5}^2$
$\Rightarrow\text{AD}^2=\text{144}+\text{25}$
$\Rightarrow\text{AD}^2=\text{169}$
$\Rightarrow\text{AD}=\text{13}\text{cm}$
So, the length of the each side of the rhombus is $13\ cm.$
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Question 714 Marks
State the $SSS$-criterion for similarity of trianglrs.
Answer
If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.
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Question 724 Marks
If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $2AB = DE$ and $BC = 6c\ m$, find $EF.$
Answer
Given $\triangle\text{ABC}\sim\triangle\text{DEF}$ and $2AB = DE$ and $BC = 6\ cm$
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}$
$\Rightarrow\frac{1}{2}=\frac{6}{\text{EF}}$
$\Rightarrow\text{EF}=3\text{cm}$
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Question 734 Marks
For the following statments state whether true$(T)$ or false$(F):$
In a $\triangle\text{ABC},\text{AB}=6\text{cm},\angle\text{A}=45^\circ$ and $\text{AC}=8\text{cm}$ and in a $\triangle\text{DEF},\text{DF}=9\text{cm},\angle\text{D}=45^\circ$and $\text{DE}=12\text{cm},$ then $\triangle\text{ABC}\sim\triangle\text{DEF}.$
Answer
False.
Solution:
Given that,
$\angle\text{A}=45^\circ,\text{AB}=6\text{cm}$ and $\text{AC}=8\text{cm}$
$\angle\text{D}=45^\circ,\text{DF}=9\text{cm}$ and $\text{DE}=12\text{cm}$
Consider, $\triangle\text{ABC}$ and $\triangle\text{DFE},$
$\angle\text{A}=\angle\text{D}=45^\circ$
$\frac{\text{AB}}{\text{DE}}=\frac{6}{12}=\frac{1}{2}$
$\frac{\text{AC}}{\text{DF}}=\frac{8}{9}$
$\Rightarrow\frac{\text{AB}}{\text{DF}}\not=\frac{\text{AC}}{\text{DF}}$
Thus, the triangles are not similar.
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Question 744 Marks
In the given figure, $D$ is the midpoint of side $B C$ and $A E \perp B C$. If $B C=a, A C=b, A B=c, E D=x, A D=p$ and $A E=h$, prove that.

$\text{b}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}$
Answer
Given: D is the midpoint of side $\text{BC},\text{AE}\perp\text{BC},\text{BC}=\text{a},\text{AC}=\text{b},\text{AB}=\text{c},\text{ED}=\text{x},\text{AD}=\text{p}$ and $\text{AE}=\text{h}$
In $\triangle\text{AEC},\angle\text{AEC}=90^\circ$
$\text{AD}^2=\text{AE}^2+\text{ED}^2$ (by pythagoras theorem)
$\Rightarrow\text{p}^2=\text{h}^2+\text{x}^2$
In $\triangle\text{AEC},\angle\text{AEC}=90^\circ$
$\text{AC}^2=\text{AE}^2+\text{EC}^2$ (by pythagoras theorem)
$\text{b}^2=\text{h}^2+\Big(\text{x}+\frac{\text{a}}{\text{2}}\Big)^2=\big(\text{h}^2+\text{x}^2\big)+\text{ax}+\frac{\text{a}^2}{4}$
$\therefore\text{b}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}\dots(1)$
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Question 754 Marks
In a trapezium $A B C D$, it is given that $A B \| C D$ and $A B=2 C D$. Its diagonals $A C$ and $B D$ intersect at the point $O$ such that $\operatorname{ar}(\triangle A O B)=84 cm^2$. Find $\operatorname{ar}(\triangle COD )$.
Answer

The diagonals of a trapezium divide each other proportionally.
$\angle\text{CDO}=\angle\text{OBA}$ ...(alternate angles)
$\angle\text{COD}=\angle\text{AOB}$ ...(vertically opposite angles)
$\Rightarrow\triangle\text{COD}=\triangle\text{AOB}$ ...(AA criterion for similarity)
$\Rightarrow\frac{\text{ar}(\triangle\text{COD})}{\text{ar}(\triangle\text{AOB})}=\frac{\text{CD}^2}{\text{AB}^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{COD})}{84}=\frac{1^2}{2^2}$
$\Rightarrow\text{ar}(\triangle\text{COD})=21\text{cm}^2$
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4 Marks Questions - Page 2 - Maths STD 10 Questions - Vidyadip