2 Marks Questions

Question 12 Marks

If the radius and slant height of a cone are in the ratio $7: 13$ and its curved surface area is $286 \ cm^2,$ find its radius.

Answer

We are given that,Two ratio in radius and slant height of a cone $= 7 : 13$
Let radius $(r) = 7x$
and slant height $(1) = 3x$
Curved surface area $=\pi r l$
$=\frac{22}{7} \times 7 x \times 13 x=286$
$286 x ^2=286$
$x^2=\frac{286}{286}=1$
$\therefore\ x=\sqrt{1}=1$
Therefore Radius $=7 x =7 \times 1=7 \ cm$

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Question 22 Marks

Curved surface area of a cone is $308 \ cm^2$ and its slant height is $14 \ cm$. Find the radius of the base.

Answer

Slant height of cone $=14 \ cm$
Let radius of circular end of cone be r.
Curved surface area of cone $=\pi r l$
$308 \ cm^2=\left(\frac{22}{7} \times r \times 14\right) \ cm$
$\Rightarrow r =\left(\frac{308}{44}\right) \ cm =7 \ cm$
Thus, the radius of circular end of the cone is $7 \ cm.$

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Question 32 Marks

Express the decimal $0 . \overline{235}$ in the form $\frac{p}{q}$, where $p,q$ are integers and $q \neq 0$.

Answer

Let $x=0 . \overline{235}$
i.e. $x=0.235235\ldots\ldots (i)$
Multiply both sides by $1000 ,$ we get
$\Rightarrow 1000 x=235.235235\ldots\ldots (ii)$
On subtracting $(i)$ from $(ii),$ we get
$999 x =235$
$\Rightarrow x=\frac{235}{999}$
$\therefore 0 . \overline{235}=\frac{235}{999}$

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Question 42 Marks

Find two rational and two irrrational numbers between $0.5$ and $0.55.$

Answer

We know that, $0.5<0.55$
Consider $x =0.5, y =0.55$ and $n =2$
$d=\frac{y-x}{n+1}=\frac{0.55-0.5}{2+1}=\frac{0.05}{3}$
Two rational and two irrational numbers which lies between $0.5$ and $0.55$ are $x + d$ and $x +2 d$
so we get,
$=0.5+\frac{0.05}{3}$ and $0.5+2 \times \frac{0.05}{3}$
$=\frac{1.5+0.05}{3}$ and$\frac{1.5+0.1}{3}$
$=\frac{1.55}{3}$ and $\frac{1.6}{3}$
By division
$= 0.51$ and $0.53$
Two irrational numbers which lies between $0.5$ and $0.55$ are $0.5151151115 ....$ And $0.5353553555$

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Question 52 Marks

Name the quadrant in which the point lies :(i) A(1, 1) (ii) (–2, –4) (iii) C(1, –2).

Answer

(i) $(+,+)$ are the signs of the co-ordinates of points in the I quadrant.
$\therefore A (1,1)$ lies in the I quadrant.
(ii) $(-,-)$ are the signs of the co-ordinates of points in the III quadrant.
$\therefore B (-2,-4)$ lies in the III quadrant.
(iii) (+,-) are the signs of the co-ordinates of points in the IV quadrant.
$\therefore C (1,-2)$ lies in the IV quadrant.

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Question 62 Marks

In the given figure, we have $B X=\frac{1}{2} A B$ and $B Y=\frac{1}{2} B C$ and $A B=B C$. Show that $B X=B Y$.

Answer

We have AB = BC [Given]
Now, by Euclid’s axiom 7, we have things which are halves of the same things are equal to one another.
$\therefore \frac{1}{2} AB=\frac{1}{2} BC$
Hence, $BX = BY .\left[\because BX =\frac{1}{2} AB\right.$ and $BY =\frac{1}{2} BC$ (Given) $]$

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Question 72 Marks

Point C is called a mid point of line segment AB, prove that every line segment has one and only one mid-point.

Answer

Let a line AB have two mid-points, say, C and D. Then
AB = AC + CB = 2AC . . . . (i) . . . [As C is the mid-point of AB]
and AB = AD + DB = 2AD . . . . (ii) [As D is the mid-point of AB]
From equation (i) and (ii)
AC = AD and CB = DB
But this will possible only when D lies on point C. So every line segment has one and only one mid-point.

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