Tag: maths

Questions Related to maths

Find the height of a trapezium whose area is  $68 { cm } ^ { 2 }$  and whose bases are  $13 { cm }$  and  $26 { cm }.$

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

  1. $\dfrac { 15 }{ 16 } { cm }$

  2. $\dfrac { 9 }{ 8 } { cm }$

  3. $\dfrac { 10 }{ 3 } { cm }$

  4. $\dfrac { 9 }{ 5 } { cm }$

Show answer
Correct Option: A

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 $

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

  1. True

  2. False

Show answer
Correct Option: A
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 $PQ \parallel AB \parallel DC$

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, 

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

  1. $ BE  $

  2. $ AB  $

  3. $ BC  $

  4. none of the above

Show answer
Correct Option: A
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

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:
$\displaystyle PA\times PD= PB\times PC.$

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

  1. True

  2. False

Show answer
Correct Option: A
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, $\frac{PA}{PC} = \frac{PB}{PD}$
$PA \times PD = PB \times PC$

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

Show answer
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

Show answer
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

Show answer
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. $300^{\circ}$

Show answer
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

Show answer
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

Show answer
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