Tag: when lines join

Quizzes (3) Questions (42)

Questions Related to when lines join

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

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

  1. $AD^{2}+BC^{2}-2AB.CD$
  2. $AD^{2}+BC^{2}+2AB.CD$
  3. $AD^{2}-BC^{2}+2AB.CD$
  4. $AD^{2}-BC^{2}-2AB.CD$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For any trapezium ABCD with AB parallel to CD, the sum of the squares of the diagonals AC and BD is given by the relation AC^2 + BD^2 = AD^2 + BC^2 + 2(AB)(CD). This can be derived using vector methods or the law of cosines on the triangles formed by the diagonals and sides.

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

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.

  1. $350$ $cm^2$
  2. $276$ $cm^2$
  3. $216$ $cm^2$
  4. $410$ $cm^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If the diagonals bisect the base angles of a trapezium, the non-parallel sides are equal to the segments of the base. With bases 23 and 13, the non-parallel sides are 13 each. The height can be calculated using the Pythagorean theorem, and then the area is found.

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

$ABCD$ 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

  1. $\dfrac{42}{13}$
  2. $\dfrac{13}{6}$
  3. $\dfrac{21}{13}$
  4. $\dfrac{13}{7}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given the midpoints P and Q, the area of the resulting shapes can be calculated using the properties of trapeziums and triangles. The ratio of the areas is derived from the geometric properties of the segments.

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

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

  1. $35 cm$
  2. $30 cm$
  3. $40 cm$
  4. $60 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the parallel sides be 3x and 4x. The area of a trapezium is given by (1/2) * (sum of parallel sides) * height. Thus, 385 = (1/2) * (3x + 4x) * 11. Solving this yields 385 = (7x / 2) * 11, so 385 = 38.5x, which means x = 10. The longer side is 4x = 4 * 10 = 40 cm.

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

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 :

  1. $\dfrac{a-b}{2}$
  2. $\dfrac{a+b}{2}$
  3. $\dfrac{ab}{2}$
  4. $\dfrac{ab}{a+b}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line segment joining the midpoints of the non-parallel sides of a trapezium is known as its median. The length of this line segment is always equal to half the sum of the lengths of the parallel sides, expressed as (a + b) / 2.

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