Tag: when lines join

Questions Related to when lines join

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

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

  1. True

  2. False

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Correct Option: A
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.

State true or false:
In a trapezium $ABCD$ in which $AB$ is parallel to $DC$ and $AD= BC$, then $\angle ADC= \angle BCD$

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

  1. True

  2. False

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Correct Option: A
Explanation:

Draw $DM\perp AB$ and $CN\perp AB$.$\triangle DAM\cong \triangle CBN$

By R.H.S. $\therefore \angle DAB= \angle CBA$.$\Rightarrow \angle ADC= \angle BCD$

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

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

  1. rhombus

  2. parallelogram

  3. trapezium

  4. kite

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Correct Option: C
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.

In a trapezium ABCD, AB || CD. If angle A = $60^{\circ}$ then angle D =?

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

  1. $110^{\circ}$

  2. $120^{\circ}$

  3. $70^{\circ}$

  4. $300^{\circ}$

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Correct Option: B
Explanation:

AB I I CD, we have
$\angle A\, +\, \angle D\, =\, 180$ ($\because$ angles at the same side of transversal)
$\Rightarrow\, \angle D\, = 180^{\circ} - 60^{\circ} = 120^{\circ}$

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

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

  1. $14^{\circ}$

  2. $104^{\circ}$

  3. $76^{\circ}$

  4. None of these

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Correct Option: C
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)

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:

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

  1. $94$cm

  2. $92$cm

  3. $91$cm

  4. $90$cm

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Correct Option: C
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

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

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

  1. $\displaystyle 100^{\circ}$

  2. $\displaystyle 105^{\circ}$

  3. $\displaystyle 110^{\circ}$

  4. $\displaystyle 115^{\circ}$

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Correct Option: B
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.

In a trapezium. ABCD, $\angle ADC = 110^o$. Find $\angle A$.

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

  1. $50^o$

  2. $60^o$

  3. $70^o$

  4. $80^o$

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Correct Option: C
Explanation:

In a trapezium,     AB || CD
$\therefore$ sum of adjacent angles $= 180^o$
$\angle D + \angle A = 180^o$
$\angle A = 180^o - \angle D = 180^o - 110^o$
$\angle A= 70^o$

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 $\angle A$.

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

  1. $90^o$

  2. $100^o$

  3. $110^o$

  4. $120^o$

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Correct Option: B
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 = $100^o$

State whether true or false:

All trapeziums are parallelograms.

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

  1. True

  2. False

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Correct Option: B
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