Tag: maths

$Let\angle A,\angle B,\angle C\quad and\angle D\quad be\quad the\quad angles\quad of\quad quadilateral\ \angle A=3x,\angle B=4x,\angle C=5x\quad and\angle D=6x,where\quad x\quad is\quad a\quad constantIn\quad quadilateral\quad ABCD\quad \ \angle A+\angle B+\angle C+\angle D=360(Angle\quad sum\quad property)\ 3x+4x+5x+6x=360\ 18x=360\ x=20\ \angle A=3x=3\times 20=60\ \angle B=4x=4\times 20=80\ \angle C=5x=5\times 20=100\ \angle D=6x=6\times 20=120$
$In\quad quadilateral\quad ABCD\quad \ \angle A+\angle D=60+120=180\ \angle B+\angle C=80+100=180\ \therefore AB\parallel CD(Sum\quad of\quad consecutive\quad interior\quad angle\quad is\quad supplementary)\ But\quad \angle A+\angle B=60+80\neq 180\ \therefore AB\quad is\quad not\quad parallel\quad to\quad BC\ Hence\quad quadilateral\quad is\quad a\quad trapzium(as\quad it\quad has\quad only\quad one\quad pair\quad of\quad parallel\quad side)$\

Let angles be $3x, 7x, 6x$ and $4x$
Total sum of angles of a quadrilateral = $360^{\circ}$
$\Rightarrow 3x + 7x + 6x + 4x = 360^{\circ}$
$\Rightarrow 20x = 360^{\circ}$
$\Rightarrow x = 18^{\circ}$
$\therefore$ Angles are $ 3\, \times\, 18^{\circ} = 54^{\circ}$
 $ 7\, \times\, 18^{\circ} = 126^{\circ}$
 $ 6\, \times\, 18^{\circ} = 108^{\circ}$
 $ 4\, \times\, 18^{\circ} = 72^{\circ}$
All the angles of the figure ABCD are different, thus it is a trapezium.
Hence, option 'C' is correct.

Let $p$ and $q$ be the two parallel lines cut by two distinct transversals $l$ and $m$. The figure formed by the lines $AB, BC, CD, DA$ will always be a trapezium as at least one pair of given lines will be parallel.

An isosceles trapezium have equal opposite sides. Therefore diagonals of  trapezium are equal.

$Properties\  of \ an \ isosceles\  trapezium:  $

It has a pair of parallel sides.

It has a pair of opposite sides that are congruent.

Both pairs of opposite angles are supplementary, that is they sum to $180°$.

Consecutive angles along both bases are congruent.

Diagonals are congruent.

$\therefore $In an isosceles trapezium the base angles are always equal. Hence, it depends how we name an isosceles trapezium.

If the trapezium is an isosceles trapezoid it has equal angles according to the trapezoid theorem.
Therefore, B is the correct answer.

The pair of opposite sides of a parallelogram are equal and parallel but in the case of the trapezium, this is not true.