Tag: constructions of triangles

Questions Related to constructions of triangles

For construction of a $\triangle PQR$, where $\displaystyle QR=6\ cm, PR=10\ cm$ and $\angle Q=90^{\circ}$, its steps for construction is given below in jumbled form. Identify the third step from the following.

1. At point $ Q $, draw an angle of $ {90}^{\circ} $.
2. From $ R $ cut an arc of length $ PR = 10.0 \ cm $ using a compass. 
3. Name the point of intersection of the arm of the angle $ {90}^{\circ} $ and the arc drawn in step 3, as $ P $.
4. Join $P $ to $ Q $ . $ PQR $ is the required triangle. 
5. Draw the base side $ QR = 6\  cm $.

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

  1. $3$

  2. $4$

  3. $2$

  4. $5$

  5. $1$

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Correct Option: C
Explanation:

Step 1. Draw a line $QR=6\ \ cm$

Step 2. At point $Q$ ,draw an angle of $90^{\circ}$
Step 3. From $R$ cut an arc $PR=10\ \ cm$ using compass.
Step 4. Name the point of intersection of the arm of angle $90^{\circ}$ and the arc in step $3$ , as $P$
Step 5. Join $p$ to $Q$. $PQR$ is required triangle.
So the third step is $2$
Option $C$ is correct.

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

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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

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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

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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. $\sqrt{145}$

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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. $85$

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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. $2$

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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 } } $

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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 

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