Tag: geometrical constructions

Questions Related to geometrical constructions

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'.

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

  1. abcd

  2. acbd

  3. abdc

  4. adcb

Show answer
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.

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

  1. $75^o$

  2. $45^o$

  3. $90^o$

  4. $60^o$

Show answer
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 $