3 Marks Question

Question 1013 Marks

In the given figure, $\angle\text{ABC}=90^\circ$ and $\text{BD} \perp\text{AC.}$ If $AB = 5.7\ cm, BD = 3.8\ cm$ and $CD = 5.4\ cm,$ find $BC.$

Answer

We have, $\angle\text{ABC}=90^\circ$ and $\text{BD}\perp\text{AC}$
In $\triangle\text{ABC}$ and $\triangle\text{BDC}$
$\angle\text{ABC}=\angle\text{BDC} [$Each $90^\circ ]$
$\angle\text{C}=\angle\text{C} [$Commom$]$
Then, $\triangle\text{ABC}\sim\triangle\text{BDC} [$By $AA$ similarity$]$
$\therefore\frac{\text{AB}}{\text{BD}}=\frac{\text{BC}}{\text{DC}}[$ Corresponding parts of similar $\triangle$ are proportional$]$
$\Rightarrow\frac{5.7}{3.8}=\frac{\text{BC}}{5.4}$
$\Rightarrow\text{BC}=\frac{5.7}{3.8}\times8.1\text{cm}$

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Question 1023 Marks

In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides $AB$ and $AC$ respectively such that $DE || BC.$
If $\frac{\text{AD}}{\text{BD}}=\frac{4}{5}$ and $EC = 2.5\ cm,$ find $AE$.

Answer

Given: $\frac{\text{AD}}{\text{BD}}=\frac{4}{5}$ and $EC = 2.5\ cm$
by thales theorem,
$\frac{\text{AD}}{\text{BD}}=\frac{\text{AE}}{\text{CE}}$
$\Rightarrow\frac{4}{5}=\frac{\text{AE}}{2.5}\ \ (\because\text{CE}=\text{EC})$
$\Rightarrow\text{AE}=\frac{4\times2.5}{5}$
$\Rightarrow\text{AE}=2\text{cm}$

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Question 1033 Marks

In a $\triangle\text{ABC} D$ and $E$ are points on $AB$ and $AC$ respectively such that $DE || BC.$ If $AD = 2.4\ cm, AE = 3.2\ cm, DE = 2\ cm$ and $BC = 5\ cm,$ find $BD$ and $CE.$

Answer

In the $\triangle\text{ABC}, DE || BC$
$AD = 2.4\ cm, AE = 3.2\ cm, DE = 2\ cm$ and $BC = 5\ cm$

$\text{In}\ \triangle\text{ABC}$
$\because\text{DE}||\text{BC}$
$\therefore\triangle\text{ADE}\sim\triangle\text{ABC}$
$\Rightarrow\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}=\frac{\text{DE}}{\text{BC}}$
$\Rightarrow\frac{2.4}{\text{AB}}=\frac{3.2}{\text{AC}}=\frac{2}{5}$
Now, $\frac{2.4}{\text{AB}}=\frac{2}{5}\Rightarrow\text{AB}=\frac{2.4\times5}{2}=6\text{cm}$
$\text{DB}=\text{AB}-\text{AD}=6.0-2.4=3.6\text{cm}$
And $\because\frac{3.2}{\text{AC}}=\frac{2}{5}\Rightarrow\text{AC}=\frac{3.5\times5}{2}=8.75\text{cm}$
$\therefore\text{CE}=\text{AC}-\text{AE}=8.0-3.2=4.8\text{cm}=8\text{cm}$

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Question 1043 Marks

In the given figure, $DE || BC$
If $DE : BC = 3 : 5.$ Calculate the ratio of the areas of $\triangle\text{ADE}$ and the trapezium $BCED.$

Answer

$DE : BC = 3 : 5$ calculate the ratio of the areas of $\triangle\text{ADE}$ and the trapezium $BCED$
$\because\triangle\text{ADE}\sim\triangle\text{ABC}$
$\therefore\frac{\text{area}\triangle\text{ADE}}{\text{area}\triangle\text{ABC}}=\frac{\text{DE}^2}{\text{BC}^2}=\Big(\frac{3}{5}\Big)^2=\frac{9}{25}$
$\Rightarrow25\text{ area }\triangle\text{ADE}=9\text{ area }\triangle\text{ABC}$
$= 9($area $\triangle\text{ADE} +$ area trapezium $BCED)$
$= 9$ area $\triangle\text{ABC} + 9$ area trapezium $BCED$
$25$ area $\triangle\text{ADE} - 9$ area $\triangle\text{ADE}$
$= 9$ area trapezium $BCED$
$⇒ 16$ area $\triangle\text{ADE} = 9$ area trapezium $BCED$
$\Rightarrow\frac{\text{area }\triangle\text{ADE}}{\text{area trapezium BCED}}=\frac{9}{16}$
$\therefore$ Ratio $= 9 : 16.$

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Question 1053 Marks

In the given figure, we have $AB || CD || EF.$ If $AB = 6\ cm, CD = x\ cm, EF = 10\ cm, BD = 4\ cm$ and $DE = y\ cm,$ Calculate the values of $x$ and $y.$

Answer

It is given that $AB || CD || EF$
$AB = 6\ cm, CD = x\ cm$ and $EF = 10\ cm$

We have to calculate the values of $x$ and $y.$
In $\triangle\text{ADB}$ and $\triangle\text{DEF},$ we have
$\angle\text{ADB}=\angle\text{EDF} ($Vertically opposite angles$)$
$\Rightarrow\angle\text{ABD}=\angle\text{DEF} ($Alternate interior angles$)$
So $\triangle\text{ADB}\sim\triangle\text{DEF}$
$\frac{\text{EF}}{\text{AB}}=\frac{\text{OE}}{\text{OB}}$
$\frac{10\text{cm}}{6\text{cm}}=\frac{\text{y}}{4\text{cm}}$
$6\text{cm}\times\text{y}=40\text{cm}$
$\text{y}=\frac{40\text{cm}}{6\text{cm}}$
$\text{y}=6.67\text{cm}$
Similarly in $\triangle\text{ABE}$ we have
$\frac{\text{CD}}{\text{AB}}=\frac{\text{DE}}{\text{DB}}$
$\frac{4}{6.7}\text{cm}=\frac{\text{x}}{6}\text{cm}$
$6.7\text{cm}\times\text{x}=6\text{cm}\times4\text{cm}$
$\text{x}=\frac{24}{6.7}\text{cm}$
$\text{x}=3.78\text{cm}$
Hence $x = 3.78\ cm$ and $y = 6.67\ cm.$

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Question 1063 Marks

If $\triangle\text{ABC}$ and $\triangle\text{BDE}$ are equilateral triangles, where $D$ is the midpoint of $BC,$ find the ratio of areas of $\triangle\text{ABC}$ and $\triangle\text{BDE}.$

Answer

We have,
$\triangle\text{ABC}$ and $\triangle\text{BDE}$ are equilateral triangles then both triangles are equiangular
$\therefore\triangle\text{ABC}\sim\triangle\text{BDE} [$By $AAA$ similarity$]$
By area of similar triangle theorem
$\frac{\text{ar}(\triangle\text{ABC)}}{\text{ar}(\triangle\text{BDE})}=\frac{\text{BC}^2}{\text{BD}^2}$
$=\frac{(2\text{BD})^2}{\text{BD}^2} [D$ is the mid-point of $BC]$
$=\frac{4(\text{BD})^2}{\text{BD}^2}$
$=\frac{4}{1}$

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Question 1073 Marks

If $a \triangle\text{ABC}, AD$ is the bisector of $\angle\text{A},$ Meeting side $BC$ at $D.$
If $AB = 5.6\ cm, AC = 6\ cm$ and $DC = 3\ cm,$ find $BC.$

Answer

$AB = 5.6\ cm, AC = 6\ cm$ and $DC = 3\ cm$

$\because AD$ is the bisector of $\angle\text{A}$
$\therefore\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
$\Rightarrow\frac{5.6}{6}=\frac{\text{BD}}{3}$
$\Rightarrow\text{BD}=\frac{3\times5.6}{6}=2.8$
$\therefore BC = BD + DC = 2.8 + 3 = 5.8\ cm.$

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Question 1083 Marks

If $D$ and $E$ are points on sides $AB$ and $AC$ respectively of a $∆ABC$ such that $DE || BC$ and $BD = CE.$ Prove that $\triangle\text{ABC}$ is isosceles.

Answer

Given: In $\triangle\text{ABC}, D$ and $E$ are points on the sides $AB$ and $AC$ such that $BD = CD$

To prove: $\triangle\text{ABC}$ is an isosceles triangle
Proof: By thales theorem
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
But $BD = CE .....(i)$
$\therefore\text{AD}=\text{AE}\ ....(\text{ii})$
Adding we get,
$AD + BD = AE + CE$
$\Rightarrow\text{AB}=\text{AC}$
$\therefore\triangle\text{ABC}$ is an isosceles triangle.
Hence proved.

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Question 1093 Marks

In fig. state if $PQ || EF.$

Answer

We have,
$DP = 3.9\ cm, PE = 3\ cm, DQ = 3.6\ cm$ and $QF = 2.4\ cm.$
Now, $\frac{\text{DP}}{\text{PE}}=\frac{3.9}{3}=\frac{1.3}{1}=\frac{13}{10}$
And, $\frac{\text{DQ}}{\text{QF}}=\frac{3.6}{2.4}=\frac{36}{24}=\frac{3}{2}$
$\Rightarrow\frac{\text{DP}}{\text{PE}}\neq\frac{\text{DQ}}{\text{QF}}$
So, $PQ$ is not parallel to $EF.$

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Question 1103 Marks

In right-angled triangle $ABC$ is which $\angle\text{C}=90^\circ,$ if $D$ is the mid-point of $BC,$ prove that $AB^2 = 4AD^2 - 3AC^2.$

Answer

We have,
$\angle\text{C}=90^\circ$ and $D$ is the mid-point of $BC$
In $\triangle\text{ACB},$ by pythagoras theorem
$ A B^2=A C^2+B C^2 $
$ \Rightarrow A B^2=A C^2+(2 C D)^2[D \text { is the mid-point of } B C]$
$ A B^2=A C^2+4 C D^2$
$ \Rightarrow A B^2=A C^2+4\left(A D^2-A C^2\right)[\text { In } \triangle A C D, \text { by pythagoras theorem }]$
$ \Rightarrow A B^2=A C^2+4 A D^2-4 A C^2 $
$ \Rightarrow A B^2=4 A D^2-3 A C^2$

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Maths 3 Marks Question