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Maths 5 Marks Questions

Triangles · CBSE STD 10 (CBSE - English Medium). 76 questions.

Questions · Page 2 of 2

5 Marks Questions

Question 515 Marks
In $\triangle\text{ABC},\text{D}$ and $\text{E}$ are the midpoint of AB and AC respectively. Find the ratio of the areas of $\triangle\text{ADE}$ and $\triangle\text{ABC}.$
Answer
It is given that D and E are midpoint of AB and AC.
Applying midpoint theorem, we can conclude that DE || BC.
Hence, by B.P.T we get:
$\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$
Also, $\angle\text{A}=\angle\text{A}$
Applying SAS similarity theorem, we can conclude that $\triangle\text{ADE}\sim\triangle\text{ABC}.$
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
$\therefore\frac{\text{ar}(\triangle\text{ADE})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{DE}^2}{\text{BC}^2}$
$=\frac{\big(\frac{1}{2}\text{BC}\big)^2}{\text{BC}^2}$
$=\frac{1}{4}$
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Question 525 Marks
Two triangles ABC and PQR are such that AB = 3cm, AC = 6cm, $\angle\text{A}=70^\circ,\text{PR}=9\text{cm},\angle\text{P}=70^\circ$ 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 535 Marks
Find the length of each side of a rhombus whose diagonals are 24cm and 10cm long.
Answer

Let ABCD be the rhobbus with diagonals AC and BD 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:
$AB^2 = AO^2 + BO^2 = 12^2 + 5^2 = 144 + 25 = 169$
AB = 13cm
Hence, the lenght of each side of the given rhombus is 13cm
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Question 545 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
Construction: Draw EF || AB through O.
In $\triangle OEA$ and $\triangle OFC$, 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
$O A^2+O C^2=O E^2+A E^2+O F^2+C F^2 \ldots \ldots \text { (i) }$
$\triangle OFB$ and $\triangle ODE$, 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
$O B^2+O D^2=O F^2+F B^2+O E^2+D E^2$
By Construction since EF || CD,
$D E=C F \text { and } A E=F B$
So, from (i) and (ii), we have
$O A^2+O C^2=O B^2+O D^2$
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Question 555 Marks
In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1cm. PB = 3cm, AQ = 1.5cm, QC = 4.5cm, Prove that area of $\triangle\text{APQ}$ is $\frac{1}{16}$ of the area of $\triangle\text{ABC}.$
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 565 Marks
Two vertical poles of height 9m and 14m stand on a plane ground. If the distance between their feet is 12m, find the distance between their tops.
Answer
Let AB and CD be the given vertical poles. Then, AB = 9m, CD = 14m and AC = 12m
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 575 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 585 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 DE to F that DE = EF. By the Basic proportinality theorem, DE || BC That is, DF || BC ....(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. ⇒ DF = BC So, $\text{DE}=\frac{1}{2}\text{DF}=\frac{1}{2}\text{BC}$
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Question 595 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 ABCD with E and F as the points of AD and BC, respectively. Produce AD and BC 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 $\triangle\text{PEF},$ we get that DC || EF. Hence, EF || AB. Thus. EF is parallel to both AB and DC. This completes the proof.
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Question 605 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 615 Marks
The perimeters of two similar triangles ABC and PQR are 32cm and 24cm respectively. If PQ = 12cm, 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 625 Marks
In the given figure, $\angle\text{ABC}=90^\circ$ and $\text{BD}\perp\text{AC}.$ If BD = 8cm, AD = 4cm, find CD.
Answer

Given that BD = 8cm, AD = 4cm
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= 16cm
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Question 635 Marks
Each of the equal sides of an isosceles triangle is 25cm. Find the length of its altitude if the 14cm.
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,
$AB^2 = AD^2 + BD^2$
$\Rightarrow AD^2 = AB^2- BD^2$
$\Rightarrow AD^2= 25^2 - 7^2$
$\Rightarrow AD^2 = 625 - 49$
$\Rightarrow AD^2 = 576$
$\Rightarrow AD = 24cm$
Hence, the length of the altitude is 24cm.
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Question 645 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 655 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 665 Marks
If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that 2AB = DE and BC = 6cm, find EF.
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 675 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 685 Marks
In a $\triangle\text{ABC},\text{AD}$ is the bisector of $\angle\text{A}.$
If AB = 10cm, AC = 14cm and BC = 6cm, find BD and DC.
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 695 Marks
In the adjoining figure, ABC is a triangle in which AB = AC. If D and E are points on AB and AC respectively such that AD = AE, 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:
⇒ AB - AD = AC - AD
⇒ AB - AD = AC - AE (Since, AD = AE)
⇒ 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 180º)
$\Rightarrow\angle\text{DEC}+\angle\text{CBD}=180^\circ$ $($Since, AB = AC ⇒ $\angle\text{B}=\angle\text{C})$
Hence, quadrilateral BDED is cyclic.
Therefore, B, C, E and D are concyclic points.
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Question 705 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 715 Marks
Find the length of each side of a rhombus whose diagonals are 24cm and 10cm 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 13cm.
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Question 725 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 735 Marks
If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that 2AB = DE and BC = 6cm, find EF.
Answer
Given $\triangle\text{ABC}\sim\triangle\text{DEF}$ and 2AB = DE and BC = 6cm
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}$
$\Rightarrow\frac{1}{2}=\frac{6}{\text{EF}}$
$\Rightarrow\text{EF}=3\text{cm}$
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Question 745 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 755 Marks
In the given figure, D is the midpoint of side BC and $\text{AE}\perp\text{BC}.$ If BC = a, AC = b, AB = c, ED = x, AD = p and AE = 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 765 Marks
In a trapezium ABCD, it is given that AB || CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that $\text{ar}(\triangle\text{AOB})=84\text{cm}^2.$ Find $\text{ar}(\triangle\text{COD}).$
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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5 Marks Questions - Page 2 - Maths STD 10 Questions - Vidyadip