Tag: maths

Questions Related to maths

State true or false
Square numbers can only have even number of zeros at the end.

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$If\quad a\quad number\quad has\quad 0\quad in\quad the\quad unit’s\quad place,then\quad its\quad square\quad end\quad with\quad two\quad 0's.$

The square root of sum of the digits in the square of $121$ is 

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. $4$

  2. $3$

  3. $6$

  4. $9$

Show answer
Correct Option: A
Explanation:
The square root of sum of digits in the square of 121 

$ Sol^{n} $ square of $121 = 14641 $

sum of digits $ \Rightarrow 1+4+6+4+1 = 16 $

and square root of 16; 
$ \sqrt{16} = 4 $

$10^8$ is expanded by writing ............ number of zeros after 1

maths powers and exponents decimal fracfions measurements again understanding tenths and hundredths

  1. 8

  2. 0

  3. 10

  4. None of these

Show answer
Correct Option: A
Explanation:

8 zeros after 1 for $10^8$

A decimal number has 16 decimal places The number of decimal places in the square root of this number will be

maths powers and exponents decimal fracfions measurements again understanding tenths and hundredths

  1. 2

  2. 4

  3. 8

  4. 16

Show answer
Correct Option: C
Explanation:

The square root will have half the number of decimal places as the number it self has.
Hence square root of $16$ decimal places has $8$ decimal places.

For e.g. the square root of $0.0000000000000016$ will be $0.00000004$

If $\triangle ABC\sim \triangle DEF$ and $AB:DE=3:4$, then the ratio of area of triangles taken in order is 

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $\dfrac{9}{16}$

  2. $\dfrac{16}{9}$

  3. $\dfrac{15}{9}$

  4. $\dfrac{9}{15}$

Show answer
Correct Option: A
Explanation:
 In similar triangle,
         Rratio of areas of triangle = square of ratio of corresponding sides
       Ratio of areas of triangle=${\dfrac {3^2}{4^2}}$
                                                 =$\dfrac{9}{16}$
In $\Delta A B C$, $P,Q,R$ are points on $\overline { B C } , \overline { C A } , \overline { A B }$ respectively, dividing them in the ratio $1 : 4,3 : 2$ and $3 : 7$. The points $S$ divides $AB$ in the ratio $1 : 3$.
Then $\frac { | \overline { A P } + \overline { B Q } + \overline { C R } | } { | \overline { C S } | } =$
maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $\frac { 1 } { 5 }$

  2. $\frac { 2 } { 5 }$

  3. $\frac { 5 } { 2 }$

  4. $\frac { 7 } { 10 }$

Show answer
Correct Option: A

The areas of two similar triangles are $16cm^2$ and $36cm^2$ respectively. If the altitude of the first triangle is $3cm$, then the corresponding altitude of the other triangle is:

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $4cm$

  2. $6.5cm$

  3. $4.5cm$

  4. $6cm$

Show answer
Correct Option: C
Explanation:
Let ${A} _{1}$ and ${A} _{2}$ be the areas of the similar triangles.

$\Rightarrow \dfrac{{A} _{1}}{{A} _{2}}=\dfrac{{s} _{1}^{2}}{{s} _{2}^{2}}$

$\Rightarrow \dfrac{16}{36}=\dfrac{{\left(3\right)}^{2}}{{s} _{2}^{2}}$ given $({s} _{1}=3 \ cm )$

$\Rightarrow {s} _{2}^{2}=\dfrac{36\times 9}{16}$

$\Rightarrow {s} _{2}=\dfrac{6\times 3}{4}=4.5 \ cm$  

State true or false:


The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

To prove: Ratio of the areas of two triangles of the same bases is equal to the ratio of their heights.
Proof:
Let us take triangle ${ T } _{ 1 }$ with height  ${ h } _{ 1 }$ and base ${ b } _{ 1 }$ and Triangle ${ T } _{ 2 }$ with height  ${ h } _{ 2 }$ and base ${ b } _{ 2 }$.

Area of Triangle ${ T } _{ 1 }$ $=\dfrac { 1 }{ 2 } \times \text{base}\times \text{height}\ =\dfrac { 1 }{ 2 } \times { b } _{ 1 }\times { h } _{ 1 }$

Area of Triangle ${ T } _{ 2 }$ $=\dfrac { 1 }{ 2 } \times \text{base}\times \text{height}\ =\dfrac { 1 }{ 2 } \times { b } _{ 2 }\times { h } _{ 2 }$

Ratio of area of two triangles,
$\dfrac { \text{Area of triangle} \ { T } _{ 1 } }{\text{ Area of triangle} \ { T } _{ 2 } } =\frac { \frac { 1 }{ 2 } \times { b } _{ 1 }\times { h } _{ 1 } }{ \frac { 1 }{ 2 } \times { b } _{ 2 }\times { h } _{ 2 } }$      
$ =\dfrac { h _{ 1 } }{ { h } _{ 2 } } $                                      (Base of two triangles are equal. So, $ { b } _{ 1 } = { b } _{ 2 }$) 
Proved.              

State true or false:

The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

To prove: Ratio of the areas of two triangles of the same height is equal to the ratio of their bases.
Proof:
Let us take triangle ${ T } _{ 1 }$ with height  ${ h } _{ 1 }$ and base ${ b } _{ 1 }$ and Triangle ${ T } _{ 2 }$ with height  ${ h } _{ 2 }$ and base ${ b } _{ 2 }$.

Area of Triangle ${ T } _{ 1 }$ $=\dfrac { 1 }{ 2 } \times\ \text{ base}\times \text{height}\ =\dfrac { 1 }{ 2 } \times { b } _{ 1 }\times { h } _{ 1 }$

Area of Triangle ${ T } _{ 2 }$ $=\dfrac { 1 }{ 2 } \times\ \text{ base}\times \text{height}\ =\dfrac { 1 }{ 2 } \times { b } _{ 2 }\times { h } _{ 2 }$

Ratio of area of two triangles,
$\dfrac { \text{Area of triangle} \  { T } _{ 1 } }{\text{ Area of triangle} \  { T } _{ 2 } } =\dfrac { \frac { 1 }{ 2 } \times { b } _{ 1 }\times { h } _{ 1 } }{ \frac { 1 }{ 2 } \times { b } _{ 2 }\times { h } _{ 2 } } \ $
$ =\dfrac { { b } _{ 1 } }{ { b } _{ 2 } } $                      (Height of two triangles are equal So,$ { h } _{ 1 } = { h } _{ 2 }$) 
Hence, proved.

The areas of two similar triangles are $12$ ${cm}^{2}$ and $48$ ${cm}^{2}$. If the height of the smaller one is $2.1$ $cm$, then the corresponding height of the bigger one is:

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $4.41$ $cm$

  2. $8.4$ $cm$

  3. $4.2$ $cm$

  4. $0.525$ $cm$

Show answer
Correct Option: C
Explanation:

Areas of two similar triangles are $12 cm^2$ and $48 cm^2$
For similar triangles the ratio of areas is equal to the ratio of square of corresponding heights
Hence, $\dfrac{A _1}{A _2} = \dfrac{(h _1)^2}{(h _2)^2}$

$\dfrac{12}{48} = \dfrac{(2.1)^2}{(h _2)^2}$

$(h _2)^2= 4 \times (2.1)^2$

$h _2 = 2 \times 2.1$

$h _2 = 4.2 cm$