Tag: constructions of triangles

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Questions Related to constructions of triangles

Multiple choice maths construction of triangle construction of triangles - i construction of triangles 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 $.

  1. $3$
  2. $4$
  3. $2$
  4. $5$
  5. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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 $QR$.

  1. True

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

The angle bisector theorem states that any point on the angle bisector of an angle is equidistant from the two sides forming the angle. Since T lies on the bisector of angle PQR, it is by definition equidistant from PQ and QR.

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

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

  1. $220$
  2. $205$
  3. $190$
  4. $85$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $0$
  2. $1$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles
Which of the following pair of sides can form triangle?
  1. $5\ cm, 6\ cm , 4\ cm$
  2. $13\ cm , 12\ cm , 24\ cm$
  3. $2\ cm , 7\ cm , 9\ cm$
  4. $5.6\ cm , 6.5\ cm , 12\ cm$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
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.

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

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

  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 } } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given AB = A1B1 and M, N, M1, N1 are midpoints, the triangles are congruent. The vector relationship between the midpoints of sides in congruent triangles preserves the distance between corresponding points. Specifically, M and M1 are midpoints of AB and A1B1, and N and N1 are midpoints of BC and B1C1; the displacement MM1 is equal to BB1.

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

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

  1. $205$
  2. $210$
  3. $315$
  4. $216$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 

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