2 Marks Questions

Question 12 Marks

Express the linear equation $y -2=0$ in the form $ax + by + c =0$ and indicate the value of $a , b$ and c in case.

Answer

We need to express the linear equation $y -2=0$ in the form $ax + by + c =0$ and indicate the values of $a , b$ and c . $y -2=0$ can also be written as $0 \cdot x +1 \cdot y -2=0$.
We need to compare the equation $0 \cdot x+1 \cdot y -2=0$ with the general equation $ax + by + c =0$, to get the values of $a , b$ and c . Therefore, we can conclude that $a =0, b=1$ and $c =-2$

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Question 22 Marks

Write two solutions of the form $x=0, y=a$ and $x=b, y=0:-4 x+3 y=12$

Answer

Given, $-4 x+3 y=12$
Put value of $x=0$ in equation $(1),$ we get
$\Rightarrow 0+3 y=12$
$\Rightarrow y=4$
Thus, $x=0$ and $y=4$ is a solution
put value of $y=0$ in equation $(1),$ we get
$\Rightarrow-4 x+0=12$
$\Rightarrow x=-3$
Thus, $x =-3$ and $y = 0$ is a solution

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Question 32 Marks

In figure, OA and OB are respectively perpendiculars to chords CD and EF of a circle whose centre is O . If $OA =$ OB, prove that $\overline{E C} \cong \overline{D F}$.

Answer

Given: In figure, $O A$ and $O B$ are respectively perpendiculars to chords $C D$ and $E F$ of a circle whose centre is $O$. $O A=O B$.
To prove: $\overline{E C} \cong \overline{D F}$
Proof: $OA = OB \mid$ Given
$\therefore C D=E F \mid \because$ Chords equidistant from the centre are equal
$\Rightarrow \overline{C D} \cong \overline{E F}$. $\cdot$ If two chords of a circle are equal, then their corresponding arcs are congruent
$\Rightarrow \overline{C D}-\overline{E D} \cong \overline{E F}-\overline{E D}$
$\Rightarrow \overline{C E} \cong \overline{D F}$
$\Rightarrow \overline{E C} \cong \overline{D F}$

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Question 42 Marks

If $O$ is the centre of the circle, find the value of $x$ in the given figure:

Answer

We have, $\angle BAC =50^{\circ}$
$\angle DBC=70^{\circ}$
Therefore, $\angle BDC =\angle BAC =50^{\circ}......($Angles on same segment$)$
In triangle $BDC ,$ by angle sum property
$\angle BDC +\angle BCD +\angle DBC =180^{\circ}$
$50^{\circ}+ x +70^{\circ}=180^{\circ}$
$120^{\circ}+ x =180^{\circ}$
$x =60^{\circ}$

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Question 52 Marks

If a line intersects two concentric circles $($circles with the same centre$)$ with centre $O$ at $A , B , C$ and $D ,$ prove that $A B=C D$.

Answer

Given: $A$ line intersects two concentric circles $($circles with the same centre$)$ with centre $O$ at $A, B, C$ and $D.$
To prove : $AB = CD$
Construction : Draw $O M \perp B C$

Proof: $\because$ The perpendicular drawn from the centre of a circle to a chord bisects the chord.
$\therefore AM = DM$
and $ BM = CM $
Subtracting $(2)$ from $(1),$ we get
$ AM - BM = DM - CM$
$\Rightarrow AB = CD $

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Question 62 Marks

An isosceles triangle has perimeter $30 \ cm$ and each of the equal sides is $12 \ cm.$ Find the area of the triangle.

Answer

$a=12 \ cm, b =12 \ cm$
Perimeter $=30 \ cm$
$a + b + c =30$
$\Rightarrow 12+12+ c =30$
$\Rightarrow 24+ c =30$
$\Rightarrow c =30-24$
$\Rightarrow c =6 \ cm$
$s =\frac{30}{2} \ cm=15 \ cm$
$\therefore \text { Area of the triangle }=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{15(15-12)(15-12)(15-6)}$
$=\sqrt{15(3)(3)(9)}$
$=9 \sqrt{15} \ cm^2$

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Question 72 Marks

Prove the exterior angle formed by producing a side of a cyclic quadrilateral is equal to the interior opposite angle.

Answer

Given: ABCD is a cyclic quadrilateral whose side AB is produced to P to formed exterior $\angle CBP$.

To prove: $\angle CBP .=$ Interior opposite $\angle ADC$
Proof : $\because ABCD$ is a cyclic quadrilateral
$\therefore \angle ADC+\angle ABC=180^{\circ} 180^{\circ}$
$\because$ Opposite angles of a cyclic quadrilateral are supplementary
Also, $\angle ABC +\angle CBP =180^{\circ}$...(2) $\mid$ Linear Pair Axiom
From (1) and (2), we have
$\angle ABC+\angle CBP=\angle ABC+\angle ABC$

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Maths 2 Marks Questions

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