Tag: maths

Questions Related to maths

Choose the correct statement:

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. Of all the line segments that can be drawn from a point outside a line, the perpendicular is the shortest.

  2. The difference of two sides of a triangle is equal to the third side.

  3. The sum of the three sides of a triangle is less than the sum of its three medians.

  4. If two sides of a triangle are unequal then the larger side has the smaller angle opposite to it.

Show answer
Correct Option: A
Explanation:

XY is a fixed line.

$LX=MO=NT=FP=a$
$In\triangle PXL$
we can see that
${ PX }^{ 2 }={ PL }^{ 2 }+{ LX }^{ 2 }$
$\implies\quad { PX }^{ 2 }={ PL }^{ 2 }+{ a }^{ 2 }\quad $-(1)
$ In\quad \triangle PMO:$
$ { PO }^{ 2 }+{ OM }^{ 2 }={ PM }^{ 2 }$
$ \implies\quad { PM }^{ 2 }={ PO }^{ 2 }+{ a }^{ 2 }\quad$ -(2)
$ In\quad \triangle PNT:$
$ { PT }^{ 2 }+{ TN }^{ 2 }={ PN }^{ 2 }$
$\implies\quad { PN }^{ 2 }={ PT }^{ 2 }+{ a }^{ 2 }\quad$ -(3)
$ In\quad \triangle PFN:$
$ { PN }^{ 2 }={ PF }^{ 2 }+{ FN }^{ 2 }$
$\implies\quad { PF }^{ 2 }={ PN }^{ 2 }-{ FN }^{ 2 }$
$ \because In\quad \triangle PFN,PN\quad is\quad $hypotenuse,
$ \therefore PN>PF\quad -(4)$
 comparing$(1)(2)(3)& (4)$
$ PL>PO>PT$
$\implies\quad PX>PM>PN$
$comparing \quad with(4)$
 $PN>PF$
$\implies\quad PX>PM>PN>PF$
Similarly,we can prove for right side of $PF.$
$\therefore PF$ is the shortest distance on $XY$ from $P.$
$\therefore A)$is correct.

For constructing a triangle when the base, one base angle and the difference between lengths of other two sides are given, the base length is equals to:

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. The difference between lengths of other two sides

  2. The given base length

  3. The largest side

  4. None of these

Show answer
Correct Option: B
Explanation:

The length of the base is exactly given. Hence there is no need to extend the same.

The construction of a $\Delta ABC$ in which $BC=6$ $cm$ and $\angle B=50^\circ$, is not possible when $(AB-AC)$ is equal to:

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. $5.6\ cm$

  2. $5\ cm$

  3. $6\ cm$

  4. $4.8\ cm$

Show answer
Correct Option: C
Explanation:

For construction of triangle sum of any two sides must be greater then third side

$\Rightarrow AC+BC>AB$
$AC+6>AB$
$6>AB-AC$
$AB-AC<6$
So $AB-AC\neg 6$
Option $C$ is correct.

The construction of $\Delta EFG$ when $FG=3$ $cm$ and m$\angle G=60^\circ$ is possible when difference of $EF$ and $EG$ is equal to:

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. $3.2$ $cm$

  2. $3.1$ $cm$

  3. $3$ $cm$

  4. $2.8$ $cm$

Show answer
Correct Option: D
Explanation:
The difference possible isn't a constant

It could range from $0$ to less than $3\ cm$

It says less than $3\ cm$, and not equal to $3\ cm$ . This is so because sum of 2 sides is always greater than the third side in a triangle.

$\therefore EF-EG < 3 \ cm$

From the given options, D is the only possible solution as the rest are $\geq 3 \ cm$

For constructing a triangle whose perimeter and both base angles are given, the first step is to:

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. Draw a base of any length

  2. Draw the base of length $=$ perimeter

  3. Draw the base angles from a random line.

  4. Draw a base of length $=\dfrac13 \times$ perimeter.

Show answer
Correct Option: B
Explanation:
Below are the steps for constructing a triangle whose perimeter and base angles are given.

Step 1 : Draw a line segment/base equal to perimeter 

Step 2 : From any point X draw ray at one of the given base angles. From any point Y draw ray at second of the given base angles

Step 3 : draw angle bisector of X and Y, two angle bisectors intersect each other at point A

Step 4 : Draw line bisector of XA and AY respectively these two line bisectors intersect XY at point B and C

Step 5 : Join A to B and A to C 

Step 6 : Triangle ABC is the required triangle.


Hence, the answer is 'Draw the base of length $=$ perimeter'.

The construction of $\triangle ABC$ in which $AB = 6\ cm, \angle A = 30^\circ$, is not possible when $AC+BC = $

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. $6.3\ cm$

  2. $7.2\ cm$

  3. $5.6\ cm$

  4. $6.9\ cm$

Show answer
Correct Option: C
Explanation:

For constructing a triangle the sum of any two sides must be greater then third side.

Hence one side is $AB=6cm$
$\Rightarrow AC+BC>6cm$
So when $AC+BC=5.6cm$ it is not possible to construct a triangle.

The construction of $\triangle ABC$ in which $AB = 5\ cm, \angle A = 45^\circ$, is possible when $AC+BC = $

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. $4.8\ cm$

  2. $5.6\ cm$

  3. $3.2\ cm$

  4. $2.8\ cm$

Show answer
Correct Option: B
Explanation:

For constructing a triangle the sum of any two sides must be greater then third side.

Hence one side is $AB=5cm$
$\Rightarrow AC+BC>5cm$
Here we have $AC+BC=5.6cm$ in option $C$
So it is possible to construct a triangle.

Construct a $\triangle ABC$ in which:
$AB= 5.4\ cm$, $\angle CAB= 45^{0}$ and $AC\, +\, BC= 9\ cm$. Then the length of $AC$ (in $cm.$) is:

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. $4$

  2. $7$

  3. $5$

  4. None of these

Show answer
Correct Option: C

The construction of $\Delta LMN$ when $MN=7$ $cm$ and $m\angle M=45^\circ$ is not possible when difference of $LM$ and $LN$ is equal to:

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. $4.5$

  2. $5.5$

  3. $6.5$

  4. $7.5$

Show answer
Correct Option: D
Explanation:

The triangle inequality rule states that the length of a side of a triangle is less than the sum of the lengths of the other two sides and greater than the difference of the lengths of the other two sides.


In $\triangle LMN$, if $MN=7 \ cm$ then $LM-LN<7 \ cm$

This is not possible, from the given options, if $LM-LN=7.5 \ cm$

Option D.

Which of the following could be the value of $AC-BC$ in the construction of a triangle $ABC$ in which base $AB = 5 cm, \angle A = 30^{\circ}$?

maths construction of triangles constructions of triangles construction of triangles - ii constructions

  1. $5.5$

  2. $5$

  3. $2.5$

  4. None of these

Show answer
Correct Option: C
Explanation:

The triangle inequality rule states that the length of a side of a triangle is less than the sum of the lengths of the other two sides and greater than the difference of the lengths of the other two sides.


In $\triangle ABC$, if $AB=5 \ cm$ then $AC-BC<5 \ cm$

This is not, from the given options, if $AC-BC=2.5 \ cm$

Option C