Tag: closed figures

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

State true or false:

In trapezium $ ABCD  $, $ AB  $ is parallel to $ DC  $; $ P  $ and $ Q  $ are the mid-points of $ AD  $ and $ BC  $ respectively. $ BP $ produced meets $ CD $ produced at point $ E $. Hence, $ PQ $ is parallel to $ AB $

  1. True

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

Given: ABCD is a trapezium. $AB \parallel DC$. P and Q are mid points of AD and BC respectively.
BP produced meets CD at E

Construction: Join BD. Draw a parallel line from P which meets BD on M such that $PM \parallel AB$ and a parallel line from Q which meets BD on N such that $QN \parallel CD$

Now, In $\triangle ADB$
P is mid point of AD and $PM \parallel AB$. Thus, M is mid point of BD.

In $\triangle BDC$
Q is mid point of BC and $QN \parallel DC$. Thus, N is mid point of BD

Hence, M and N are same points. Thus, PM or QN is a straight line, PQ
and $PQ \parallel AB \parallel DC$

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

State true or false:

In trapezium $ ABCD  $, $ AB $ is parallel to $ DC $;  $ P $ and $ Q  $ are the mid-points of $ AD  $ and $ BC  $ respectively. $ BP $ produced meets $ CD $ produced at point $ E $. Hence, point $ P  $ bisects, 

  1. $ BE $
  2. $ AB $
  3. $ BC $

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: ABCD is a trapezium. $AB \parallel DC$. P and Q are mid points of AD and BC respectively.
BP produced meets CD at E

To prove: P is mid point of BE.
In $\triangle APB$ and $\triangle EPD$
$\angle APB = \angle EPD$ (Vertically opposite angles)
$\angle EDP = \angle PAB$ (Alternate angles)
$PA = PD$ (P is mid point of AD)
Thus, $\triangle APB \cong \triangle DPE$ (ASA rule)
Hence, $PE = PB$ (By cpct)
thus, P is mid point of BE

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

In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at a point
Such that:
$\displaystyle PA\times PD= PB\times PC.$

  1. True

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

In $\triangle$ APB and $\triangle$ CPD,
$\angle APB = \angle CPD$ (Vertically opposite angles)
$\angle ABP = \angle CDP$ (Alternate angles of parallel sides AB and CD)
$\angle BAP = \angle DCP$ (Alternate angles of parallel sides AB and CD)
Hence, $\triangle APB \sim \triangle CPD$ (AAA rule)
Thus, $\frac{PA}{PC} = \frac{PB}{PD}$
$PA \times PD = PB \times PC$

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)