Tag: when lines join

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Questions Related to when lines join

Multiple choice maths when lines join trapeziums and kites quadrilaterals and their properties closed figures

State true or false:
In quadrilateral PQRS, $\angle P : \angle Q : \angle R : \angle S = 3 : 4 : 6 : 7$. The Quadrilateral PQRS is trapezium

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given in $\Box$ PQRS,$\angle P:\angle Q:\angle R:\angle S=3:4:6:7$

 Let $ \angle P=3x,\angle Q=4x,\angle R=6x,\angle S=7x$

Sum of interior angles of a quadrilateral$={ 360 }^{ o }$

So $\angle P+\angle Q+\angle R+\angle S=360^o$

$ \Rightarrow 3x+4x+6x+7x=360^o$

$ \Rightarrow 20x=360^o$

$ \Rightarrow x=\dfrac { 360 ^o}{ 20 } $

$ \Rightarrow x=18^o$

So $\angle P=3x=3\times 18^o={ 54 }^{ o }$

$\angle Q=4\times 18^o={ 72 }^{ o }$

$\angle R=6\times 18^o={ 108 }^{ o }$

$\angle S=7\times 18^o=126^{ o }$

Now  $\angle P+\angle S=54^o+126^o=180^o\quad \& \quad \angle Q+\angle R=72^o+108^o=180^o$

In  quadrilateral  PQRS,  $\angle P\& \angle S$ are supplementary  as  well  as  $\angle Q\& \angle R$  are supplementary.

This  is only possible when side PQ$\parallel$ SR ;  PS& QR are transversals &  the sum  of  interior  corresponding  angles  on  the  same side  of the  transversals  are  supplementary.

 So $PQ\parallel SR$.

Now $\angle P+\angle Q\neq 180 ^o\&  \angle S+\angle R\neq 180^o$

In quadrilateral PQRS, $\angle P\& \angle Q$ are not supplementary as well as $\angle S\& \angle R$ are not supplementary.

So QR is not parallel to SP.

So one pair of opposite sides are parallel.

None of the opposite angles are equal. 

None of the sides are given as equal.

The $\Box$ PQRS can only be a Trapezium.

Multiple choice maths when lines join trapeziums and kites quadrilaterals and their properties closed figures

If the angles $A, B, C, D$ of a quadrilateral , taken in order are in the ratio $7:13:12:8$, then $ABCD$ is:

  1. rhombus

  2. parallelogram

  3. trapezium

  4. kite
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the angles be $7x, 13x, 12x$ and $8x$
Then, $7x+13x+12x+8x={360}^{o}$
$\Rightarrow$ $40x={360}^{o}$ $\Rightarrow$ $x={9}^{o}$
$\therefore$ $40x={360}^{o}$
$\therefore$ The angles taken in order are ${63}^{o}, {117}^{o}, {108}^{o}, {72}^{o}$ 
This shows that tow pairs of adjacent angles are supplementary $({63}^{o}+{117}^{o}={108}^{o}$ and ${108}^{o}+{72}^{o}={180}^{o}$), but opposite angles are not equal.
Therefore, the given quadrilateral will be a trapezium.

Multiple choice maths when lines join trapeziums and kites quadrilaterals and their properties closed figures

Given a trapezium ABCD in which $AB||CD$ and $AD=BC$. If $\angle C=76^{\circ}$, then $\angle D$ equals

  1. $14^{\circ}$
  2. $104^{\circ}$
  3. $76^{\circ}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In trapezium ABCD, $AB \parallel CD$, $AD = BC$
Draw a perpendicular from A on CD to meet CD at M and a perendicular from B on CD to meet at N.
Now, in $\triangle ADM$ and $\triangle BNC$
$\angle AMD = \angle BNC$ (Each $90^o$)
$AM = BN$ (distance between parallel lines)
$AD = BC$ (Given)
Thus, $\triangle ADM \cong \triangle BCN$ (SAS rule)
Hence, $\angle D = \angle C = 76^{\circ}$ (by CPCT)

Multiple choice maths when lines join trapeziums and kites quadrilaterals and their properties closed figures

The line joining the mid points of the diagonals of a trapezium has length $3$cm. If the longer base is $97$cm then the shorter base is:

  1. $94$cm
  2. $92$cm
  3. $91$cm
  4. $90$cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The line joining the mid point of the diagonals of a trapezium is half the length of the difference between the two sides.
Let the smaller side be $x$
Then, $3 = \dfrac{97 -x}{2}$
$6= 97 - x$
$x = 91$ cm

Multiple choice maths when lines join trapeziums and kites quadrilaterals and their properties closed figures

The consecutive angles of a trapezium form an arithmetic sequence. If the smallest angle is $\displaystyle 75^{\circ}$, then the largest angle is

  1. $\displaystyle 100^{\circ}$
  2. $\displaystyle 105^{\circ}$
  3. $\displaystyle 110^{\circ}$
  4. $\displaystyle 115^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since, sum of all the four angles of a quadrilateral is $360^o$.

Angle 1 $=75^o$, Angle 2 $=75^o+x$, Angle 3 $=75^o+2x$, Angle 4 $=75^o+3x$
Angle 1 $+$ Angle 2 $+$ Angle 3 $+$ Angle 4 $=360^o$
$\therefore   75^o+75^o+x+75^o+2x+75^o+3x=360^o$
$\Rightarrow 300+6x=360\Rightarrow 6x=60 \Rightarrow x=10$
$\therefore$ Largest angle (Angle 4)$=75^o+3x=75^o+3\times 10=105^o$

Hence, option B.

Multiple choice maths when lines join trapeziums and kites quadrilaterals and their properties closed figures

In isoceles trapezoid ABCD, side CD is parallel to to side AB, line segment AC is congruent to line segment BD.The degree measure of angle BDC = $80^o$. Find the measures of the $\angle A$.

  1. $90^o$
  2. $100^o$
  3. $110^o$
  4. $120^o$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As per the property of isosceles trapezoid, 
Opposite sides of an isosceles trapezoid are the same length (congruent) and the angles on either side of the bases are the same size (congruent).
So, Angle C = $80^o$
Since the top and bottom angles are supplementary, we know that,
Angle A = $180 - 80$
Angle A = $100^o$
Similarly, the Angle of B = $100^o$

Multiple choice maths when lines join trapeziums and kites quadrilaterals and their properties closed figures

State whether true or false:

All trapeziums are parallelograms.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

There is some disagreement whether parallelograms, which have two pairs of parallel sides, should be regarded as trapezoids. Some define a trapezoid as a quadrilateral having only one pair of parallel sides (the exclusive definition), thereby excluding parallelograms