Tag: construction of triangles - ii

Questions Related to construction of triangles - ii

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:

  1. $4$

  2. $7$

  3. $5$

  4. None of these


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:

  1. $4.5$

  2. $5.5$

  3. $6.5$

  4. $7.5$


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 $AB = 5 cm, \angle A = 30^{\circ}$?

  1. $5.5$

  2. $5$

  3. $2.5$

  4. None of these


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

The construction of $\Delta LMN$ when $MN=6$ $cm$ and $m\angle M=45^\circ$ is not possible when difference between $LM$ and $LN$ is equal to:

  1. $6.9$ $cm$

  2. $5.2$ $cm$

  3. $5$ $cm$

  4. $4$ $cm$


Correct Option: A
Explanation:

In a triangle sum of length of $2$ sides is $>$ third side.

or 

Difference of length of two sides is less than third side.

$LM,MN,NL$ are the length of sides.

$|LM-NL|<MN$

$|LM-NL|<6$

So construction of triangle is not possible as $|LM-NL|=6.9cm$

To construct a triangle similar to a given triangle ABC with its sides 6/5th of the corresponding sides of $\Delta$ABC. Correct order of steps of construction -
(a) Draw a ray AX inclined at certain angle with AB on opposite side of C.
(b) Starting from A, cut off six equal line segments AX$ _1$, X$ _1$X$ _2$, X$ _2$X$ _3$, X$ _3$X$ _4$, X$ _4$X$ _5$ and X$ _5$X$ _6$ on AX.
(c) Draw a line B'C' parallel to BC to intersect AC produced at C'
(d) Join X$ _5$B and draw a line X6B' parallel to X5B, to intersect AB produced at B'.

  1. abcd

  2. acbd

  3. abdc

  4. adcb


Correct Option: C
Explanation:

To construct a triangle like the one given in the following question, the given steps are to be followed:

1. Draw a ray AX inclined at a certain angle with AB on opposite side of C.
2. Starting from A, cut off six equal line segments $AX _1, X _1X _2, X _2X _3, X _3X _4, X _4X _5 and X _5X _6$ on $AX$.
3. Join $X _5B$ and draw a line $X _6B'$ parallel to $X _5B$, to intersect $AB$ produced at $B'$.
4.Draw a line $B'C'$ parallel to $BC$ to intersect $AC$ produced at $C'$.

Construct a $\Delta ABC$, whose perimeter is $10.5  cm$ and base angles are $60^o$ and $45^o$. Find the third angle.

  1. $75^o$

  2. $45^o$

  3. $90^o$

  4. $60^o$


Correct Option: A
Explanation:
  1. Draw a line segment PQ such that AB + BC + AC = PQ.
    2. Construct $\angle XPQ =\angle B = { 60 }^{ 0 }\ and\ \angle YQP =\angle C =45^{ 0 }$
    3. Bisect $\angle XPQ and \angle YQP$ and their bisectors will meet at A.
    4. Draw the perpendicular bisector DE of AP which meets PQ  at B. 
    5. Draw the perpendicular bisector FG of AQ which meets PQ  at C.
    6. Join AB & AC.
    7. ABC is the required triangle.
    Sum of interior angle of triangle= $180^o$
    so third angle = $180^o-45^o-60^o\ = 75^o $