Tag: when lines join

Questions Related to when lines join

$ABCD$ is a trapezium in which $AB\parallel CD$. Which of the following is equal to $AC^{2}+BD{2}$?

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

  1. $AD^{2}+BC^{2}-2AB.CD$

  2. $AD^{2}+BC^{2}+2AB.CD$

  3. $AD^{2}-BC^{2}+2AB.CD$

  4. $AD^{2}-BC^{2}-2AB.CD$

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Correct Option: A

In trapezium PQRS, PQ($23$cm) and RS($13$ cm) are the bases. Find the area of the trapezium if the diagonals bisect angles SPQ and PQR.

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

  1. $350$ $cm^2$

  2. $276$ $cm^2$

  3. $216$ $cm^2$

  4. $410$ $cm^2$

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Correct Option: B

$ABCD$ is a trapezium in which $BC \parallel AD, BC=20\ cm$ and $AD=45\ cm$. If $P$ and $Q$ are the midpoints of $AB$ and $CD$ respectively, then the ratio of $ar(\Box PBCQ)$ to $ar(\triangle PQD)$ is

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

  1. $\dfrac{42}{13}$

  2. $\dfrac{13}{6}$

  3. $\dfrac{21}{13}$

  4. $\dfrac{13}{7}$

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Correct Option: A

In trapezium $PQRS$, side $PQ\parallel SR$. Diagonals $PR$ and $QS$ intersect each other at point $M$.  $PM=3RM$ 

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

  1. True

  2. False

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Correct Option: B

The area of a trapezium is  $385 { cm } ^ { 2 } .$  Its parallel sides are in the ratio  $3 : 4$  and the perpendicular distance between them is  $11 { cm } .$  Its longer side is

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

  1. $35 cm$

  2. $30 cm$

  3. $40 cm$

  4. $60 cm$

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Correct Option: A

In a trapezium, the lengths of its parallel sides are $a$ and $b$. The length of the line joining the midpoints of its non-parallel sides is :

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

  1. $\dfrac{a-b}{2}$

  2. $\dfrac{a+b}{2}$

  3. $\dfrac{ab}{2}$

  4. $\dfrac{ab}{a+b}$

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Correct Option: A

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

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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

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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

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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

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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, $\frac{PA}{PC} = \frac{PB}{PD}$
$PA \times PD = PB \times PC$