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Construction of triangle - class-VI

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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

  1. True

  2. False


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:

  1. $3-1-2$

  2. $1-2-3$

  3. $2-3-1$

  4. $2-1-3$


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.

  1. Right angled triangle

  2. Isosceles triangle

  3. Equilateral triangle

  4. Scalene triangle


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

  1. 4

  2. 2

  3. 1

  4. 0


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 $.

  1. $3$

  2. $4$

  3. $2$

  4. $5$

  5. $1$


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

  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


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?

  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


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 $QR$.

  1. True

  2. False


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

  1. $2\sqrt{10}$

  2. $6$

  3. $5$

  4. $\sqrt{145}$


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

  1. $220$

  2. $205$

  3. $190$

  4. $85$


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

  1. $0$

  2. $1$

  3. $3$

  4. $2$


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

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

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

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


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

  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 } } $


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

  1. $205$

  2. $210$

  3. $315$

  4. $216$


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?

  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


Correct Option: C

Mark the correct alternative of the following.
In which of the following cases, a right triangle cannot be constructed?

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

  2. $8$cm, $6$cm, $10$cm

  3. $5$cm, $9$cm, $11$cm

  4. None of these


Correct Option: C
Explanation:

Any triangle is right angled if it satisfies $a^2 + b^2 = c^2$, where c is the largest side from option (A)  we take 

$a = 12 cm$
$b = 5 cm$
$c = 13 cm$
we know $(12)^2 + (5)^2 = 169 = (13)^2$
So we can construct a right angled triangle.

Similarly in option (B) we can take 
$a = 8 cm$
$b = 6 cm$
$c = 10 cm$
with these sides also we can construct a right angled triangle.

Consider option (C) then 
$a = 5 cm$
$b = 9 cm$
$c = 11 cm$
$a^2 + b^2 = 25 + 81$   But  $c^2 = 121$
             $= 106$

So we cannot construct right angled triangle from $5 cm, 9 cm, 11 cm$.

Construct a $\triangle PQR$ in which $QR= 4.6\ cm., {\angle Q}={\angle R=50 ^{0}}$. Then the perimeter of the triangle is:

  1. $10.2\ cm$

  2. $13.2\ cm$

  3. $11.8\ cm$

  4. $12.4\ cm$


Correct Option: C

Construct a right angled $\triangle ABC$ with $\angle B = 90^\circ, BC = 5\ cm$ and $AC = 10\ cm$ and find the the length of side $AB$

  1. $6.2\ cm$

  2. $5\ cm$

  3. $8.7\ cm$

  4. $7.2\ cm$


Correct Option: C
Explanation:

Step 1. Draw a line segment $BC=5\ \ cm$

Step 2. At $B$ draw an angle of $90^{\circ}$ and extend the ray.
Step 3. Now taking $C$ and centre draw an arc of radius $10$ cm intersecting the previous ray at $A$.
Step 4. Join $C$ to $A$.
Now using a ruler measure $AB$
$AB=8.7$ cm

Construct a $\triangle PQR$ such that $\angle P = 30^\circ, \angle Q = 60^\circ$ and $PQ = 10\ cm$.Find the measure of $\angle R$
  1. $30^\circ$

  2. $45^\circ$

  3. $60^\circ$

  4. $90^\circ$


Correct Option: D

Length of two sides of a $\triangle ABC$ is $AB=6\ cm$ and $BC=7\ cm$. Then, which of the following can represent the third side of the triangle ? Also, construct the triangle formed by these three sides.

  1. $8\ cm$

  2. $13\ cm$

  3. $14\ cm$

  4. $15\ cm$


Correct Option: A
Explanation:

A triangle can be formed if sum of any two sides is greater then the third side.

Here $AB=6$ cm and $BC=7$ cm
Now $AB+BC>AC$
$6+7>AC$
$AC<13$ cm
So only option $A$ is possibel.
Steps of construction

Step 1. Draw a line segment $AB=6\ \ cm$
Step 2. Assuming $A$ a centre draw an arc of radius $8 \ \ cm$
Step 3. Now assuming $B$ as centre draw an arc of $7 \ \ cm$ intersecting the previous arc at $C$.
Step 4. Now join $A$ to $C$ and $B$ to $C$.

The perimeter of a triangle is $45\ cm$. Length of the second side is twice the length of first side. The third side is $5$ more than the first side. Find the length of each sides and construct the triangle made by these three sides.

  1. $11,19,15$

  2. $10,20,15$

  3. $10,16,19$

  4. $13,15,17$


Correct Option: B
Explanation:

Let the length of first side $=x$

Length of second side $=2x$
Length of third side $=x+5$
Perimeter $=45cm$
$x+2x+x+5=45\ 4x=45-5\ 4x=40\ x=10$
So the sides are
$x=10$
$2x=2\times 10=20\ x+5=10+5=15$
Option $B$ is correct.

Construct a triangle $ABC$ in which $AB = 5 cm$ and $BC = 4.6 cm$ and $AC = 3.7 cm$
Steps for the construction is given in jumbled form.Choose the appropriate sequence for the above
1) With radius as $5\ cm$ from $C$, cut an arc.
2)They arcs will intersect at point $A$. Join $AB$ and $AC$. $ABC$ is the required triangle.
3)Draw a line segment $BC = 4\ cm.$
4)With radius as $3$ cm from $B$, cut the arc. 

  1. $1,4,3,2$

  2. $4,3,2,1$

  3. $3,1,4,2$

  4. $4,1,3,2$


Correct Option: C
Explanation:

Correct sequence is:

1. Draw a line segment $AB=4$ cm.
2. With radius $5$ cm from $C$ , cut an arc.
3. With radius $3$ cm from $B$ , cut an arc.
4. The arc will intersect at point $A$, Join $AB$ and $AC$ .$ABC$ is required triangle.
So correct sequence is $3,1,4,2$
So option $C$ is correct.

Construct an isosceles $\triangle  XYZ,$ where $YZ=5$ units and $\angle XYZ=35^{o}$. Also, find the measure of $\angle YXZ$.

  1. $35^{o}$

  2. $70^{o}$

  3. $110^{o}$

  4. $140^{o}$


Correct Option: C
Explanation:

$YZ=5$ CM $,\angle XYZ=35^{\circ}$

As the triangle is isosceles therefore $\angle XZY=50^{\circ}$
Steps of construction:
1. Draw a line segment $XY=5$ cm.
2. At $Y$ draw an angle of $35^{\circ}$ and extend the arm.
3. At $Z$ draw an angle of $35^{\circ}$ and extend the ray such that it intersect the previous ray at $X$
4. Join $Y$ to $X$ and $Z$ to $X$
Now measure $\angle YXZ$
$\angle YXZ=110^{\circ}$

Construct an isosceles $\triangle  ABC,$ where base $AB=7\ cm$ and $\angle ABC=50^{o}$. Also, find the measure of $\angle ACB$.

  1. $50^{0}$

  2. $80^{o}$

  3. $100^{o}$

  4. $120^{o}$


Correct Option: B
Explanation:

$AB=7$ cm $,\angle ABC=50^{\circ}$

As the triangle is isosceles therefore $\angle CAB=50^{\circ}$
Steps of construction:
1. Draw a line segment $AB=7$ cm.
2. At $A$ draw an angle of $50^{\circ}$ and extend the arm.
3. At $B$ draw an angle of $50^{\circ}$ and extend the ray such that it intersect the previous ray at $C$
4. Join $A$ to $C$ and $B$ to $C$
Now measure $\angle ACB$
$\angle ACB=80^{\circ}$

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 fourth 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. $5$

  2. $1$

  3. $2$

  4. $3$

  5. $4$


Correct Option: D
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 fourth step is $3$
Option $D$ is correct.

State the following statement is True or False
In a right angle triangle $ABC$ such as $AC=5 cm ,BC=2 cm$ , $\angle B=90^o$
Then the length of $AB$ after construction is $7$cm

  1. True

  2. False


Correct Option: B
Explanation:

In the given triangle $\Delta ABC$:


$AC=5$ and $BC=2$.

So by the property of triangle(sum of two sides are always greater than the third side):

$AB<(AC+BC)\implies AB<7$.

But in the given question it is given that $AB=7$, which is not possible.
So given statement is incorrect.

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 second 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. $2$

  2. $1$

  3. $4$

  4. $5$

  5. $3$


Correct Option: B
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 second step is $1$
Option $B$ is correct.

For construction of a $\triangle PQR$, when $\displaystyle QR=6\ cm, PR=10\ cm$ and $\angle Q=90^{\circ}$, its steps for construction is given below in jumbled form. Identify the fifth 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. $2$

  2. $3$

  3. $1$

  4. $5$

  5. $4$


Correct Option: E
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 fifth step is $4$
Option $E$ is correct.

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 first 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. $2$

  2. $1$

  3. $3$

  4. $5$

  5. $4$


Correct Option: D
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 first step is $5$
Option $D$ is correct.

Construct a triangle $ABC$, in which $AB = 5.5 cm, AC = 6.5 cm$ and $\angle BAC = 70^{\circ}$.
Steps for its construction is given in a jumbled form.Identify its correct sequence.
1) At $A$, construct a line segment $AE$, sufficiently large, such that $\angle BAC$ at $70^\circ$, use protractor to measure $70^\circ$
2) Draw a line segment which is sufficiently long using ruler.
3) With $A$ as centre and radius $6.5cm$, draw the line cutting $AE$ at C, join $BC$, then $ABC$ is the required triangle.
4) Locate points $A$ and $B$ on it such that $AB = 5.5cm$.

  1. $2,4,1,3$

  2. $2,1,4,3$

  3. $1,2,4,3$

  4. $4,2,1,3$


Correct Option: A
Explanation:

Below are the correct steps.

i) Draw a line segment which is sufficiently long using ruler.
ii) Locate points $A$ and $B$ on it such that $AB=5.5 \ cm$
iii) At $A$ construct a line segment $AE$ , sufficiently large, such that $\angle BAC=70^\circ$, use protractor to measure
iv) With $A$ as centre and radius $6.5 \ cm$ draw the line cutting $AE$ at $C$, join $BC$ then $ABC$ is the required triangle.

So, the correct sequence of given steps is $2,4,1,3$.

Which of the following steps is INCORRECT, while constructing $\triangle$LMN, right angled at M, given that LN = 5 cm and MN = 3 cm?
Step 1. Draw MN of length 3 cm.
Step 2. At M, draw MX $\perp$ MN. (L should be some where on this perpendicular).
Step 3. With N as centre, draw an arc of radius 5 cm. (L must be on this arc, since it is at a distance of 5 cm from N).
Step 4. L has to be on the perpendicular line MX as well as on the arc drawn with centre N. Therefore, L is the meeting point of these two and $\triangle$LMN is obtained.

  1. Only Step 4

  2. Both Step 2 and Step 3

  3. Only Step 2

  4. None of these


Correct Option: A

In a right-angled triangle, the square of the hypotenuse is equal to twice the product of the other two sides. One of the acute angles of the triangle is

  1. $40^{\circ}$

  2. $42^{\circ}$

  3. $44^{\circ}$

  4. $45^{\circ}$


Correct Option: D
Explanation:

In a right angled triangle, if the square of the hypotenuse is  equal to twice the product of  the other two sides, then the two angles are equal.
Since, one of the angle is 90, the sum of other two will be 90. 
Thus, each angle should be $45^{\circ}$

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