2 Marks Questions

Question 12 Marks

Cost of pen is two half times the cost of a pencil. Express this situation as a linear equation in two variable.

Answer

Let cost of pen $Rs. x$ and cost of a pencil be $Rs. y.$
According to statement of the question, we have
$x=2 \frac{1}{2} y$
$\Rightarrow 2 x =5 y$ or $2 x -5 y =0$

View full question & answer→

Question 22 Marks

The following values of x and y are thought to satisfy a linear equation :

X	1	2
y	1	3

Answer

From the table, we get two points $A(1,1)$ and $B(2,3)$ which lie on the graph of the linear equation Obviously, the graph will be a straight line so we first plot the points $A$ and $B$ and join them as shown in the fig from the fig we see that the graph cuts the x axis at the point $\left(\frac{1}{2}, 0\right)$ an and y - axis at the point $(0,-1)$

View full question & answer→

Question 32 Marks

Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Answer

Let two circles with respective centers $A$ and $B$ intersect each other at points $C$ and $D$.
We have to prove $\angle ACB =\angle ADB$
Proof: In triangles ABC and ABD,
AC = AD = r
BC = BD = r
AB = AB [Common]
$\therefore \Delta ABC \cong \Delta ABD$
[SSS rule of congruency]
$\Rightarrow \angle ACB =\angle ADB [ ByCPCT ]$

View full question & answer→

Question 42 Marks

In the given figure, $O$ is the centre of the circle and $\ce{\angle AOB=70^{\circ}}$. Calculate the values of $\ce{(i) \angle OCA, (ii) \angle OAC}$.

Answer

$i.$ The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
Thus, $\ce{\angle AOB =2 \angle OCA}$
$\ce{\Rightarrow \angle OCA=\left(\frac{\angle AOB}{2}\right)=\left(\frac{70^{\circ}}{2}\right)=35^{\circ}}$
$ii. \ce{OA = OC}($ Radius of a circle $)$
$\Rightarrow \ce{\angle OAC=\angle OCA}[$ Base angles of an isosceles triangle are equal$]$
$\Rightarrow \ce{\angle OAC}=35^{\circ}$

View full question & answer→

Question 52 Marks

Find an angle marked as x in given figure where O is the centre of the circle:

Answer

$x=\frac{1}{2} \times 110^{\circ}=55^{\circ}$
[ $\because$ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle.]

View full question & answer→

Question 62 Marks

Find the area of equilateral triangle whose side is $12 \ cm$ using Heron's formula.

Answer

in equilateral triangle, all sides are equal.
let`s sides are $a =12$
$S =\frac{12+12+12}{2} \ cm$
$=\frac{36}{2} \ cm$
$=18 \ cm$
$\therefore$ Area of equilateral $=\sqrt{ s ( s - a )( s - b )( s - c )}$
$=\sqrt{18(18-12)(18-12)(18-12)}$
$=\sqrt{18 \times 6 \times 6 \times 6}$
$=36 \sqrt{3} sq \ cm $

View full question & answer→

Question 72 Marks

ABC and ADC are two right triangles with common hypotenuse AC. Prove that $\angle CAD =\angle ABD.$

Answer

We have ABC and ADC two right triangles, right angled at B and D respectively.
$\Rightarrow \angle ABC=ADC\left[\operatorname{Each} 90^{\circ}\right]$
If we draw a circle with AC (the common hypotenuse) as diameter, this circle will definitely passes through of an arc AC, Because $B$ and $D$ are the points in the alternate segment of an arc $A C$.
Now we have $\widehat{ CD }$ subtending $\angle CBD$ and $\angle CAD$ in the same segment.
$\therefore \angle CAD=\angle CBD$
Hence proved.

View full question & answer→

Maths 2 Marks Questions

Test yourself on this topic