2 Marks Questions

Question 12 Marks

In Figure $4,$ a triangle $A B C$ is drawn to circumscribe a circle of radius $3 \ cm$ , such that the segments $B D$ and $D C$ are respectively of lengths $6 \ cm$ and $9 \ cm$ . If the area of $\triangle A B C$ is $54$ , then find the lengths of sides $A B$ and $A C$.

Answer

Let us suppose that the circle touches the sides $A B$ and $A C$ of the triangle of the triangle at point $F$ and $E$, respectively.

Let the length of the line segment $A F$ be $x \ cm$.
According to the theorem, that the lengths of tangents drawn form an external point to circle are equal.
In $\triangle A B C$,
$C E=C D=9 \ cm$
$($Tangents of the circle from point $C)$
$B F=B D=6 \ cm$
$($Tangents of the circle from point $B )$
$A B=A F+F B=(x+6) \ cm$
$B C=B D+D C=6+9=15 \ cm$
$C A=C E+E A=(9+x) \ cm$
Area of $\triangle O B C=\frac{1}{2} \times B C \times O D$
$=\frac{1}{2} \times 15 \times 3=\frac{45}{2} \ cm^2$
Area of $\triangle O C A=\frac{1}{2} \times A C \times O E$
$=\frac{1}{2} \times(x+9) \times 3=\frac{3}{2}(x+9) \ cm^2$
Area of $\triangle O A B=\frac{1}{2} \times A B \times O F$
$=\frac{1}{2} \times(x+6) \times 3=\frac{3}{2}(x+6) \ cm^2$
Area of $\triangle A B C=$ Area of $\triangle O B C+$ Area of $\triangle O C A+$ Area of $\triangle O A B$
$\Rightarrow 54=\frac{45}{2}+\frac{3}{2}(x+9)+\frac{3}{2}(x+6)$
Multiply the above equation by $2 ,$
$\Rightarrow 108=45+3 x+27+3 x+18$
$\Rightarrow 18=6 x$
Divide above equation by $6 ,$
$\Rightarrow x=3$
So, $A B=x+6=3+6=9 \ cm$
$A C=9+x=9+3=12 \ cm$
Hence, the lengths of sides $A B=9 \ cm$ and $AC =12 \ cm$

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Question 22 Marks

State which of the two triangles given in the figure are similar. Also state the similarity criterion used.

Answer

In $\triangle A B C$ and $\triangle F E D$ we have,
$\angle A B C=\angle F E D$
$\frac{A B}{F E}=\frac{6}{4.5}=\frac{60}{45}=\frac{4}{3}$
And, $\frac{B C}{E D}=\frac{4}{3}$
$\Rightarrow \frac{A B}{F E}=\frac{B C}{E D}$
$\Rightarrow \triangle A B C \sim \triangle F E D$ by $\text{SAS}$ Rule
Since one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

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Question 32 Marks

In the figure, $\triangle O A D \sim \triangle O B C$, If $\angle A O C=90^{\circ}$ and $\angle O B C=30^{\circ}$, find $\angle O D A$ and $\angle C O B$

Answer

If $\angle AOC =90^{\circ}$,
then $\angle C O B=90^{\circ}$ since $A O B$ is a straight line and sum of angles on a straight lines are $180.$
In $\triangle O B C, \angle OBC =30^{\circ}, \angle COB =90^{\circ}$,
By using angle sum property of a triangle,
$\angle OCB=180^{\circ}-(\angle COB+\angle OBC)$
$=180^{\circ}-\left(90^{\circ}+30^{\circ}\right)$
$\angle O C B=180^{\circ}-120^{\circ}$
$\angle O C B=60^{\circ}$
Now since, $\triangle O A D \sim \triangle O B C$
$\angle O C B=\angle O D A=60^{\circ}$
Hence $\angle O C B=60^{\circ}$ and $\angle C O B=90^{\circ}$

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Question 42 Marks

In $\triangle A B C, \angle A=90^{\circ} A N \perp B C, B C=13 \ cm$ and $A C=5 Cm$. Find the ratio of $\operatorname{ar}(\triangle N A C): \operatorname{ar}(\triangle A B C)$

Answer

$\text { In right } \triangle B A C, A B^2+A C^2=B C^2$
$A B^2=13^2-5^2=144$
$A B=12 \ cm$
Now, in right $\triangle A N C$ Using Pythagoras theorem,
$A N^2+N C^2=A C^2$
$l^2+x^2=25 \ldots \ldots(i)$
Now, in right $\triangle A N B$, Using Pythagoras theorem
$A N^2+B N^2=A B^2$
$l^2+(13-x)^2=(12)^2$
$l^2+169+x^2-26 x=144$
By substituting the value of $l^2$ from eqn $(i)$ in eqn $(ii),$ we get.
$25-x^2+169+x^2-26 x=144 \ldots \ldots(ii)$
$x=\frac{25}{13}$
Put the value of $x$ in eqn $(i)$
$l^2+\left(\frac{25}{13}\right)^2=25$
$l=\frac{60}{13}$
Now, area of triangle is given by
$\frac{1}{2} \times \text { base } \times \text { height }$
$\frac{\text { area } \triangle A N C}{\text { area } \triangle A B C}=\frac{\frac{1}{2} \times \frac{25}{13} \times \frac{60}{13}}{\frac{1}{2} \times 5 \times 12}=\frac{25}{169}$
Hence $\operatorname{ar}(\triangle N A C): \operatorname{ar}(\triangle A B C)=25: 169$

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Question 52 Marks

In the given figure, if $A B \| DC$, find the value of $x$.

Answer

In $\triangle O A B$ and $\triangle O C D$,
$\angle A O B=\angle C O D ($vertically opposite angles$)$
$\angle O B A=\angle O D C ($alternate interior angles$)$
Hence, $\triangle O A B \sim \triangle O C D$
$\text{(AAA}$ similarity criterion$)$
Since, triangles $O A B$ and $O C D$ are similar,
Therefore, $\frac{O A}{O C}=\frac{O B}{O D}$
Substitutingvalues from the given figure, we get, $\frac{x+5}{x+3}=\frac{x-1}{x-2}$
Solving the equation,
$\Rightarrow(x+5)(x-2)=(x-1)(x+3)$
$\Rightarrow x^2+5 x-2 x-10=x^2+3 x-x-3$
Cancelling the same terms on both sides and solving, we get,
$\Rightarrow 3 x-10=2 x-3$
$\Rightarrow x=7$
Hence, the value of $x$ is $7 .$

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Question 62 Marks

In given figure $\text{ABCD}$ is a rectangle, in which $BC =2 AB$. A point $E$ lies on $C D$ produced such that $C E=2 B C$. Find $A C: B E$

Answer

$\text{ABCD}$ is a rectangle and each angle of rectangle is $90^{\circ}$.
Let length of $\quad A B=x\quad \quad \ldots \ldots(i)$
It is given that $B C=2 A B$,
$\Rightarrow B C=2 x ($Using$(i)).............(ii)$
$C E=2 B C=2(2 x)=4 x ($Using $(ii))$

Apply Py thagoras Theorem in right angled $\triangle A B C$,
$A C^2=A B^2+B C^2$
$\Rightarrow A C^2=x^2+(2 x)^2$
$\Rightarrow A C^2=5 x^2$
$\Rightarrow A C=\sqrt{5} x$
$($Neglecting the negative value of square root as length cannot be in negative$).....(iii)$
Apply Py thagoras Theorem in right angled $\triangle BCE$,
$B E^2=B C^2+C E^2$
$\Rightarrow B E^2=(2 x)^2+(4 x)^2$
$\Rightarrow B E^2=20 x^2$
$\Rightarrow B E=\sqrt{20} x=2 \sqrt{5} x ($Neglecting the negative value of square root as length cannot be in negative$)......(iv)$
From $(iii)$ and $(iv),$
$A C: B E=\sqrt{5} x: 2 \sqrt{5} x$
Hence, $A C: B E=1: 2$.

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Question 72 Marks

In $\triangle A B C A B=A C$, and $D$ is a point on side $A C$ such that $B C^2=A C \cdot C D$. Prove that $B D=B C$.

Answer

Consider $\triangle A B C$ where $A B=A C$ and $D$ is a point on side $A C$.

It is given that
$
\Rightarrow B C \times B C=A C \times C D
$
Rearranging the terms,
$
\Rightarrow \frac{A C}{B C}=\frac{B C}{C D}\quad \quad \ldots \ldots(i)
$
Which means that $\triangle A B C$ is similar to $\triangle B D C$.
$
\Rightarrow \frac{A C}{B C}=\frac{A B}{B D}
$
(Corresponding sides of similar triangles are proportional)
As it is given that $A B=A C$,
Therefore, from (i) $B D=B C$.
Hence proved.

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Question 82 Marks

In the given figure, ABC is a triangle in which $DE |\mid BC$. If $AD = x , DB = x -2, AE = x +2$ and $EC = x -1$, then find the value of x .

Answer

In $\triangle ABC , DE \| BC$
$
\therefore \quad \frac{AD}{DB}=\frac{AE}{EC}
$
(By basic proportionality theorem)
$\begin{array}{cc}\Rightarrow & \frac{x}{x-2}=\frac{x+2}{x-1} \\ \Rightarrow & x(x-1)=(x+2)(x-2) \\ \Rightarrow & x^2-x=x^2-4 \\ \Rightarrow & x=4\end{array}$

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Question 92 Marks

In Figure-3, $\Delta ABC$ and $\Delta XYZ$ are shown. If $AB =3 \ cm, BC =6 \ cm, AC =2 \sqrt{3} \ cm, \angle A =80^{\circ}$, $\angle B=60^{\circ}, XY =4 \sqrt{3} \ cm, YZ =12 \ cm$ and $XZ =6 \ cm$ , then find the value of $\angle Y$.

Answer

$\frac{AB}{XZ}=\frac{3}{6}=\frac{1}{2}$
$\frac{BC}{YZ}=\frac{6}{12}=\frac{1}{2}$
$\frac{AC}{XY}=\frac{2 \sqrt{3}}{4 \sqrt{3}}=\frac{1}{2}$
We see that $\frac{ AB }{ XZ }=\frac{ BC }{ YZ }=\frac{ AC }{ XY }=\frac{1}{2}$
$\therefore \triangle ABC \sim \Delta XZY \quad(By SSS)$
$\Rightarrow \angle A=\angle X, \angle B=\angle Z, \angle C=\angle Y(CPST) \ldots(1)$
$\angle A+\angle B+\angle C=180^{\circ} \text { (Angle sum property) }$
$\Rightarrow \angle A+\angle B+\angle Y=180^{\circ}$
$\Rightarrow 80^{\circ}+60^{\circ}+\angle Y=180^{\circ}$
$\Rightarrow \angle Y=180^{\circ}-140^{\circ}$
$\Rightarrow \angle Y=40^{\circ}$

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Question 102 Marks

In figure, if $AD \perp BC,$ then prove that $AB ^2+ CD ^2= BD ^2+ AC ^2$.

Answer

By Pythagoras theorem
$\text { In } \triangle ABD, AB^2=AD^2+BD^2$
$\Rightarrow AD^2=AB^2-BD^2 \ldots \text { (i) }$
$\text { In } \triangle ADC, AC^2=AD^2+CD^2$
$\Rightarrow AD^2=AC^2-CD^2 \ldots \text { (ii) }$
From $(i)$ and $(ii),$
$AB^2-BD^2=AC^2-CD^2$
$\Rightarrow AB^2+CD^2=BD^2+AC^2$

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Question 112 Marks

In figure, ABC and DBC are two triangles on the same base $B C$. If $A D$ intersects $B C$ at $O$, show that
$
\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle DBC)}=\frac{AO}{DO}
$

Answer

To prove : $\frac{\operatorname{ar}(\triangle ABC )}{\operatorname{ar}(\triangle D B C)}=\frac{ AO }{ DO }$
We know that,
Area $=\frac{1}{2} \times$ base $\times$ height
Area of $\triangle ABC =\frac{1}{2} \times BC \times AE\quad \quad \ldots \ldots(i)$
In $\triangle DBC$,
Area of $\triangle DBC =\frac{1}{2} \times BC \times DF\quad \quad \ldots \ldots(ii)$
Dividing (ii) by (i),
$\Rightarrow \frac{\text { Area of } \triangle ABC }{\text { Area of } \triangle DBC }=\frac{\frac{1}{2} \times BC \times AE }{\frac{1}{2} \times BC \times DF }$
$\Rightarrow \frac{\text { Area of } \triangle ABC }{\text { Area of } \triangle DBC }=\frac{ AE }{ DF }\quad \quad \ldots \ldots(iii)$
In $\triangle AOE$ and $\triangle DOF$
$\Rightarrow \angle AEO =\angle DFO$ (both right angles)
$\angle AOE =\angle DOF$ (Vertically opposite)
$\therefore \quad \triangle AOE \sim \triangle DOF$ by AAsimilarity
Hence, $\frac{ AE }{ DF }=\frac{ AO }{ DO }$($\because$ corresponding sides will be in same ratio)
Putting in (iii), we get
$\Rightarrow \frac{\text { Area } \triangle ABC }{\text { Area } \triangle DBC }=\frac{ AE }{ DF }=\frac{ AO }{ DO }$
$\therefore \frac{\text { Area } \triangle ABC }{\text { Area } \triangle DBC }=\frac{ AO }{ DO }$, Hence proved

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Question 122 Marks

In the adjoining figure, $A , B$ and C are points on $OP , OQ$ and OR respectivelysuch that $AB \| PQ$ and $A C|\mid P R$ - Show that BC||QR.

Answer

In $\triangle O P Q$, we have
$
AB \| PQ
$
Therefore, by using basic proportionality theorem, we have
$
\frac{OA}{AP}=\frac{OB}{BQ}
$
IN $\triangle Q P R$, we have
$
AC \| PR
$
Therefore, by using basic proportionality theorem, we have
$
\frac{OC}{CR}=\frac{OA}{AP}
$
Comparing (i) \& (ii), we get
$
\frac{OB}{BQ}=\frac{OC}{CR}
$
Therefore, by using converse of basic proportionality theorem, we get $BC \| QR$

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