Tag: construction of quadrilaterals

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Questions Related to construction of quadrilaterals

Multiple choice maths construction of quadrilaterals trapeziums and kites quadrilaterals and their properties closed figures

The ratio of the measures of the consecutive angles of a quadrilateral is 1:2:3:4. What type of quadrilateral is it

  1. Trapezium

  2. Kite

  3. Parallelogram

  4. Rectangle
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of angles in a quadrilateral is 360 degrees. If the ratio is 1:2:3:4, the angles are 36, 72, 108, and 144 degrees. Since one pair of consecutive angles (72+108) sums to 180, it is a trapezium.

Multiple choice maths construction of quadrilaterals trapeziums and kites quadrilaterals and their properties closed figures
In a kite shaped figure $ ABCD $, $ AB= AD $  and $ CB= CD $. Points $ P, Q $ and $ R $ are mid-points of sides $ AB, BC $ and $ CD $ respectively. Then, find $ \angle PQR  $
  1. $45^{\circ}$
  2. $30^{\circ}$
  3. $90^{\circ}$
  4. $60^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given: ABCD is a kite shaped figure. $AB = AD$, $BC = CD$, P, Q< R are mid points of AB, BC and CD respectively


To Prove: $\angle PQR = 90^{\circ}$

Construction: Join AC and BD

In $\triangle ABC$

P is mid point of AB and Q is mid point of BC

Then, $PQ \parallel AC$ (By Mid point theorem)

In $\triangle BCD$

Q is mid point of BC and R is mid point of CD

Then, $QR \parallel BD$ (mid point theorem)

We know, diagonals of the Kite shaped figure, bisect at right angles.
thus, $AC \perp BD$ 

hence, $PQ \perp QR$ (Angle made between two lines is same as the angle between their corresponding parallel sides)
or $\angle PQR = 90^{\circ}$

Multiple choice maths construction of quadrilaterals trapeziums and kites quadrilaterals and their properties closed figures

If angles P, Q, R and S of the quadrilateral PQRS taken in order are in the ratio 3 : 7 : 6 : 4 then PQRS is a

  1. rhombus

  2. parallelogram

  3. trapezium

  4. kite
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow$  $\angle P:\angle Q:\angle R:\angle S=3:7:6:4$    [ Given ]

$\Rightarrow$  Let $\angle P=3x,\,\angle Q=7x\,\angle R=6x$ and $\angle S=4x$.
$\Rightarrow$  In quadrilateral PQRS,
$\Rightarrow$  $\angle P+\angle Q+\angle R+\angle S=360^o$
$\therefore$   $3x+7x+6x+4x=360^o$
$\therefore$   $20x=360^o$
$\therefore$   $x=18^o$
$\Rightarrow$  $\angle P=3\times 18^o=54^o$
$\Rightarrow$  $\angle Q=7\times 18^o=126^o$
$\Rightarrow$  $\angle R=6\times 18^o=108^o$
$\Rightarrow$  $\angle S=4\times 18^o=72^o$
$\Rightarrow$  Now, $\angle P+\angle Q=54^o+126^o=180^o$
$\Rightarrow$  and $\angle R+\angle S=180^o+ 72^o=180^o$
$\Rightarrow$  If transversal intersects two lines in such a way that a pair of consecutive interior angles are supplementary, then the two lines are parallel.
$\therefore$    $PS\parallel QR$
$\therefore$  Quadrilateral $PQRS$ is a trapezium.

Multiple choice maths construction of quadrilaterals trapeziums and kites quadrilaterals and their properties closed figures

The area of a trapezium is $24{ cm }^{ 2 }$. The distance between its parallel sides is $4 cm$. If one of the parallel sides is $7 cm$. What is the measure of the other parallel side ?

  1. $5 cm$
  2. $8 cm$
  3. $12 cm$
  4. $7 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,


The area of the trapezium $A = 24c{m^2}$
Height $h = 4cm$

One of the side $a = 7cm$
Let the other side be $x$

SInce,
$A = \dfrac{1}{2} \times \left( {a + x} \right) \times h$
$24 = \dfrac{1}{2} \times \left( {7 + x} \right) \times 4$

$\dfrac{{24}}{2} = 7 + x$

$12 - 7 = x$

$x = 5$

Hence , the other side be the $5 \,cm$

Multiple choice maths construction of quadrilaterals trapeziums and kites quadrilaterals and their properties closed figures

If the diagonals of a quadrilateral are perpendicular to each other, the figure would always be included under the general classification:

  1. rhombus

  2. rectangle

  3. square

  4. isosceles trapezoid

  5. none of these
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Diagonals of a square, rhombus are perpendicular but a quadrilateral with perpendicular diagonal need not be a square, rhombus. 

Multiple choice maths construction of quadrilaterals trapeziums and kites quadrilaterals and their properties closed figures

ABCD is a kite in which AB=AD and CB=CD, if $\angle ABD=40^{0},$ $then find \angle A+\angle C.$

  1. $220^{0}$
  2. $200^{0}$
  3. $180^{0}$
  4. $210^{0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

we have,

$ABCD$ is a kite.

Then, Given that,

$ AB=AD\,\,\,\,\,.......\,\,\left( 1 \right) $

$ CB=CD\,\,\,\,\,.......\,\,\left( 2 \right) $

If $\angle ABD={{40}^{o}}$

Then, show Diagram,

$\angle ABD=\angle ADB={{40}^{o}}$

And

$ \angle ADB=\angle CBD={{40}^{o}}\,\,\left( \text{Alternate}\,\text{angle} \right) $

$ \angle ABD=\angle CDB={{40}^{o}}\,\,\,\left( \text{Alternate}\,\text{angle} \right) $

Then $\angle B+\angle D={{40}^{o}}+{{40}^{o}}+{{40}^{o}}+{{40}^{o}}={{160}^{o}}$

But, We know that,

In any quadrilateral

$ \angle A+\angle B+\angle C+\angle D={{360}^{o}} $

$ \angle A+\angle C={{360}^{o}}-{{160}^{o}} $

$ \angle A+\angle C={{200}^{o}} $

Hence, this is the answer.