Tag: maths

Questions Related to maths

Suppose we have to cover the xy-plane with identical tiles such that no two tiles overlap and no gap is left between the tiles. Suppose that we can choose tiles of the following shapes: equilateral triangle, square, regular pentagon, regular hexagon. Then the tiling can be done with tiles of

maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

  1. All four shapes

  2. Exactly three of the four shapes

  3. Exactly two of the four shapes

  4. Exactly one of the four shapes

Show answer
Correct Option: B
Explanation:

For $ xy-plane $ to be completely covered with regular polygons of $n$ sides, integer number (say $p$) of polygons must meet at a vertex.
Interior angle of a polgon $ = 180^{\circ}\left ( \dfrac{n-2}{n} \right ) $

So, there must exist an integer $p$ such that $p \times 180^{\circ}\left ( \dfrac{n-2}{n} \right ) = 360^{\circ} $
$ \Rightarrow p\left ( \dfrac{n-2}{n} \right ) = 2 $

For $n = 3, p = 6$
For $n = 4, p = 4$
For $n = 5, p = \dfrac{10}{3}$
For $n=6, p = 3 $

Thus, apart from pentagon, rest polygons can cover the plane as stated.

The lengths of the sides of some triangles are given, which of them is not a right angled triangle?

maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

  1. $5$ cm , $12$ cm, $13$ cm

  2. $7$ cm, $24$ cm, $25$ cm

  3. $5$ cm, $8$ cm, 1$0$ cm

  4. $3$ cm, $4$ cm, $5$ cm

Show answer
Correct Option: C
Explanation:

For a right angled triangle.Sum of squares of two sides of a triangle=square of third i.e.,square of hypotenuse.

${ 5 }^{ 2 }+{ 12 }^{ 2 }=25+144=169$
$ { 13 }^{ 2 }=169$
$\implies\quad { 5 }^{ 2 }+{ 12 }^{ 2 }={ 13 }^{ 2 }$
 A)is a right angled triangle.
$ Now,{ 24 }^{ 2 }+{ 7 }^{ 2 }=576+49=625$
$\ implies\quad { 25 }^{ 2 }=625$
$\ implies\quad { 24 }^{ 2 }+{ 7 }^{ 2 }={ 25 }^{ 2 }$
B)is also a right angled triangle.
$ { 3 }^{ 2 }+{ 4 }^{ 2 }=9+16=25$
$ { 5 }^{ 2 }=25$
$\ implies\quad { 3 }^{ 2 }+{ 4 }^{ 2 }=5$
${ 5 }^{ 2 }=25$
 D)is also a right angled triangle.
$In\quad C)$
$ { 5 }^{ 2 }+{ 8 }^{ 2 }=25+64=89$
$ { 10 }^{ 2 }=100$
$ { 5 }^{ 2 }+{ 8 }^{ 2 }\neq { 10 }^{ 2 }$
$\therefore $C)is not a right angled triangle.

Construct a right angled triangle $PQR$, in which $\angle Q = 90^\circ $, hypotenuse $PR=8\,cm$ and $QR=4.5\,cm$. Draw bisector of angle $PQR$ and let it meet $PR$ at point $T$ then $T$ is equidistant from$PQ$ and $QR$.

maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

  1. True

  2. False

Show answer
Correct Option: A

Let $A(h, 0)$ & $B(0, k)$ be two given points and let $O$ be the origin. If area of $\Delta OAB$ is $6$ units & $h$ & $k$ are integers, then length(s) of $AB$ may be

maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

  1. $2\sqrt{10}$

  2. $6$

  3. $5$

  4. $\sqrt{145}$

Show answer
Correct Option: A,C,D

The sides $AB, BC, CA$ of a trinagle $ABC$ have $3, 4$ and $5$ interior point on them. The number of triangles that can be constructed using these points  as vertices are

maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

  1. $220$

  2. $205$

  3. $190$

  4. $85$

Show answer
Correct Option: B
Explanation:

Total number of points $12.$ If no three points are co-linear$,$

Th en$,$ total number of triangles would be $^{12}{C _3}.$ 
But $3$ points on $AB,4$ points on $BC$ and $5$ points on $CA$ are co-linear$.$
So$,$ total number of triangles formed should be$:$
${ = ^{12}}{C _3} - \left( {^3{C _3}{ + ^4}{C _3}{ + ^5}{C _3}} \right) = 220 - \left( {1 + 4 + 10} \right) = 205$  
Hence,
option $(B)$ is correct answer.

If $b=3, c=4, \angle B=\dfrac{\pi}{3}$, then the number of triangles that can be constructed is

maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

  1. $0$

  2. $1$

  3. $3$

  4. $2$

Show answer
Correct Option: A
Explanation:
Given $b=3,c=4,\angle B=\dfrac{\pi}{3}$

We know that as per Sine rule

$\dfrac{\sin \dfrac{\pi}{3}}{3}=\dfrac{\sin C}{4}$

$\dfrac{\sqrt 3}{2\cdot3}=\dfrac{\sin C}{4}$

$\sin C = 2\sqrt{3}$  which is greater than 1

and sin lies between -1 and 1 that means angle C is not possible

Thus Zero triangles can be construct.

Hence A is the correct option

Which of the following pair of sides can form triangle?
maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

  1. $5\ cm, 6\ cm , 4\ cm$

  2. $13\ cm , 12\ cm , 24\ cm$

  3. $2\ cm , 7\ cm , 9\ cm$

  4. $5.6\ cm , 6.5\ cm , 12\ cm$

Show answer
Correct Option: A,B,D
Explanation:

As in a triangle, the sum of any two sides should be strictly greater than the third side.

 
In $(A), 5+6 > 4$   $\therefore$ This pair will form triangle.


In $(B), 13+12 > 24$ This pair will form triangle.

In $(C), 2+7\ \ngtr 9$  This pair will not form triangle.

In $(D), 5.6+6.5 > 12$  This pair will form triangle.

Consider $\triangle ABC$ and $\triangle { A } _{ 1 }{ B } _{ 1 }{ C } _{ 1 }$ in such a way that $\overline { AB } =\overline { { A } _{ 1 }{ B } _{ 1 } } $ and M, N, ${ M } _{ 1 }$, ${ N } _{ 1 }$ be the mid points of AB, BC, ${ A } _{ 1 }{ B } _{ 1 }$ and ${ B } _{ 1 }{ C } _{ 1 }$ respectively, then

maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

  1. $\overline { M{ M } _{ 1 } } =\overline { N{ N } _{ 1 } } $

  2. $\overline { C{ C } _{ 1 } } =\overline { M{ M } _{ 1 } } $

  3. $\overline { C{ C } _{ 1 } } =\overline { N{ N } _{ 1 } } $

  4. $\overline { M{ M } _{ 1 } } =\overline { B{ B } _{ 1 } } $

Show answer
Correct Option: D

The sides  $A B , B C , C A$  of a triangle  $A B C$  have  $3,4$  and  $5$  interior points respectively on them. The number of triangles that can be constructed using these points as vertices is

maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

  1. $205$

  2. $210$

  3. $315$

  4. $216$

Show answer
Correct Option: A
Explanation:

Total number of points $12$. If no three points are co-linear then total number of the   triangles would be  $^{12}{C _3}$.

But $3$ points on AB , $4$ points on BC and $5$ points on CA are co-linear.

So, total number of triangles formed should be 

$\begin{array}{l} { =^{ 12 } }{ C _{ 3 } }-\left( { ^{ 3 }{ C _{ 3 } }{ +^{ 4 } }{ C _{ 3 } }{ +^{ 5 } }{ C _{ 3 } } } \right)  \\ =220-\left( { 1+4+10 } \right)  \\ =205 \end{array}$



Mark the correct alternative of the following.
In which of the following cases can a right triangle ABC be constructed?

maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

  1. $AB=5$cm, $BC=7$cm, $AC=10$cm

  2. $AB=7$cm, $BC=8$cm, $AC=12$cm

  3. $AB=8$cm, $BC=17$cm, $AC=15$cm

  4. None of these

Show answer
Correct Option: C