Tag: construction of triangle

Questions Related to construction of triangle

State true or false:

Whether it is possible to construct a triangle or not with its sides equal to $5$ cm, $7$ cm, and $4$ cm
Ans: Yes

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
Explanation:

A triangle with three sides is possible, if sum of two smaller sides is greater than the third side.
For given sides, $5$ cm, $7$ cm and $4$ cm
$5 + 4 > 7$
$9 > 7$
Thus, a triangle is possible.

The steps for construction of $\triangle DEF$ with $DE = 4\ cm, EF=6.5\ cm$ and $DF = 8.6\ cm$ are given below in jumbled order:
1. Draw arcs of length $4\ cm$ from $4\ cm$ from $D$ and $6.5\ cm$ from $F$ and mark the intersection point as $E$.
2. Join $D-E$ and $F-E$.
3. Draw a line segment of length $DF = 8.6\ cm$.

The correct order of the steps is:

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

  1. $3-1-2$

  2. $1-2-3$

  3. $2-3-1$

  4. $2-1-3$

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Correct Option: A
Explanation:

Correct sequence is

Step 1: Draw a line segment of length $DF=8.6 cm$.
Step2 :Draw arcs of length $4 cm$ from $4 cm$ from $D$ and $6.5 cm$ from $F$ and mark the intersection point as $E$
Step 3: Join $D-E$ and $F-E$.
So the sequence is $3-1-2$

In $\triangle ABC$, $AB=5\ cm, BC= 6\ cm ,AC=4\ cm$. Identify the type of triangle.

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

  1. Right angled triangle

  2. Isosceles triangle

  3. Equilateral triangle

  4. Scalene triangle

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Correct Option: D
Explanation:
$5cm+6cm>4cm$
$\Rightarrow AB+BC>AC$

$6cm+4cm>5cm$
$\Rightarrow BC+AC>AB$

$4cm+5cm>6cm$
$\Rightarrow AC+AB>BC$

Sum of any two sides taken in pair is greater than the third side. So a triangle can be formed.

Now all the sides of triangle are unequal . So the triangle is Scalene triangle.

Further ${(LargestSide)}^{2}$ is not equal to $({Side1})^{2} + ({Side2})^{2}$. Hence, not right angled.

Option $D$ is correct

The number of triangles with any three of the length 1, 4, 6 and 8 cms, as sides is

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

  1. 4

  2. 2

  3. 1

  4. 0

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Correct Option: C
Explanation:

A triangle is formed if and only if sum of two sides is greater than the third side. Thus with given lengths, there is only one triangle possible with sides 1 cm, 4 cm , 6 cm.

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.