Tag: maths

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Questions Related to maths

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

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

  1. 2

  2. 4

  3. 8

  4. 16
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $0.00000004$

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

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

  1. $\dfrac{9}{16}$
  2. $\dfrac{16}{9}$
  3. $\dfrac{15}{9}$
  4. $\dfrac{9}{15}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
 In similar triangle,
         Rratio of areas of triangle = square of ratio of corresponding sides
       Ratio of areas of triangle=${\dfrac {3^2}{4^2}}$
                                                 =$\dfrac{9}{16}$
Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles
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 $\frac { | \overline { A P } + \overline { B Q } + \overline { C R } | } { | \overline { C S } | } =$
  1. $\frac { 1 } { 5 }$
  2. $\frac { 2 } { 5 }$
  3. $\frac { 5 } { 2 }$
  4. $\frac { 7 } { 10 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using vector geometry, the points P, Q, R divide the sides in given ratios. The sum of vectors AP, BQ, CR relates to the median CS. The ratio is 1/5 based on the geometric properties of the segments.

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

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:

  1. $4cm$
  2. $6.5cm$
  3. $4.5cm$
  4. $6cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $({s} _{1}=3 \ cm )$

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

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

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

State true or false:


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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.              

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

State true or false:

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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:

  1. $4.41$ $cm$
  2. $8.4$ $cm$
  3. $4.2$ $cm$
  4. $0.525$ $cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
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, $\dfrac{A _1}{A _2} = \dfrac{(h _1)^2}{(h _2)^2}$

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

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

$h _2 = 2 \times 2.1$

$h _2 = 4.2 cm$