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Construction of special quadrilaterals - class-IX

Description: construction of special quadrilaterals
Number of Questions: 21
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Tags: maths when lines join construction of quadrilaterals revision geometrical constructions construction of polygons quadrilateral practical geometry
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A square with side given can be constructed by using the property of its diagonals.

  1. True

  2. False


Correct Option: A
Explanation:

This statement is true 

We can use property that diagonals are at 45 degree with side and diagonals bisect each other at 90 degree.for construction of square.

Can we construct a rhombus $ABCD$ with $AB=4\ cm$? Its diagonal intersect at the point $O$ and $\angle OAB = 60^0$.

  1. Yes

  2. No

  3. Sometimes yes

  4. Can't say


Correct Option: A
Explanation:

Given : $AB=4$cm

Diagonal intersect at $O$ and $\angle OAB=60^{o}$ ....... $(1)$
Draw side $AB$ of $4$cm.
In a rhombus, all sides are equal and diagonals bisect the opposite angles
From $(1)$ we get, $\angle A=120^{o}$
$\implies \angle B=60^{o}$ ........... (Adjacent angles are supplementary)
Draw a side $AD$ from A of $4$cm such that $\angle BAD=120^{o}$
Now, from $D$, draw side $DC = 4$cm such that $\angle ADC=60^{o}$
And then join $B-C$ such that $BC=4$cm and $\angle DCB=120^{o}$.
At last we get a rhombus $ABCD$ with length of each side is $4$ cm and diagonals $AC$ and $BD$.
Hence, we can construct a rhombus with $AB=4\ cm$.

We cannot construct a square if:

  1. a side is given

  2. a diagonal is given

  3. one angle is $90^0$

  4. None of these


Correct Option: C
Explanation:

If a side is given then we can draw a square with the same side as given.

If diagonals are given, by joining the endpoints we can draw the square.
In square, all angles are of $90^{o}$.
If one angle is $90^{o}$ is given, we can't directly conclude that all the angles are $90^{o}$.

Hence, if one angle is $90^{o}$ then we cannot construct a square.

When given a square, the construction of an angle bisector at any vertex will create the diagonal of the square. 

  1. True

  2. False


Correct Option: A
Explanation:

This is statement is true

We know that diagonals of square bisects the angle. So angle bisector will be diagonal.

You are given the length of a diagonal of a rhombus and one of the angles of the rhombus. Which property of the rhombus will be used in the construction of this rhombus?

  1. The lengths of the sides of a rhombus are equal.

  2. The angles of a rhombus are $90^\circ$

  3. Diagonal of a rhombus bisects the opposite angles.

  4. Diagonals of a rhombus are perpendicular bisectors of each other.


Correct Option: C
Explanation:

$\Rightarrow$   We have given the length of diagonal of rhombus and one of angles of rhombus.

$\Rightarrow$  To construct an rhombus we will use the property that the diagonal of a rhombus bisect the opposite angle.
Because we know opposite angles of rhombus are equal, so it will be easier to construct rhombus.

If we have to construct a square $PQRS$ whose diagonal is $8 \sqrt 2$ cm then its side is equal to ?

  1. $8$ cm

  2. $4\sqrt2$ cm

  3. $4$ cm

  4. $8\sqrt2$ cm


Correct Option: A
Explanation:

If the diagonal of square is $a$, then its side $=\dfrac{a}{\sqrt2}$

If diagonal is $8\sqrt2 $ cm, then its side $=\dfrac{8\sqrt2}{\sqrt2}=8$ cm.

State the following statement is True or False
The side of a square is $\sqrt2$ times the diagonal of a square

  1. True

  2. False


Correct Option: B
Explanation:

The side of square is $\dfrac{1}{\sqrt2}$ times the diagonal of square.

State the following statement is True or False
We cannot construct the square if only diagonal is given

  1. True

  2. False


Correct Option: B
Explanation:

If $a$ is the diagonal of square then its side $=\dfrac{a}{\sqrt2}$

We can construct a square, with its side given.

Which of the following statements is true for a rhombus?

  1. It has only two pair of equal sides.

  2. Two of its angles are at right angles.

  3. Its diagonals bisect each other at right angles.

  4. It is always a square.


Correct Option: C
Explanation:

Rhombus is a flat shape with 4 equal straight sides.All sides have equal length.Opposite sides are parallel, and opposite angles are equal.The altitude is the distance at right angles to two sides.And the diagonals "p" and "q" of a rhombus bisect each other at right angles.
So (C) is correct.
Answer (C) 
Its diagonals bisect each other at right angles.

What would be the length of side $BC$ in Square $ABCD$ if the diagonal of the square given is $10$ cm?

  1. $5$ cm

  2. $5\sqrt2$ cm

  3. $10$ cm

  4. $10\sqrt2$ cm


Correct Option: B
Explanation:

The side of a square is $\dfrac{1}{\sqrt2}$ times of the diagonal.


If the length of diagonal $=10$ cm

Then length of side $=10\times \dfrac{1}{\sqrt2}=5\sqrt2$ cm.

If one diagonal of a square is the portion of the line $\frac { x }{ a } +\frac { y }{ b } =1$ intercepted by the axes, then the extremities of the other diagonal of the square are

  1. $\left( \frac { a+b }{ 2 } ,\frac { a+b }{ 2 } \right) $

  2. $\left( \frac { a-b }{ 2 } ,\frac { a+b }{ 2 } \right) $

  3. $\left( \frac { a-b }{ 2 } ,\frac { b-a }{ 2 } \right) $

  4. $\left( \frac { a+b }{ 2 } ,\frac { b-a }{ 2 } \right) $


Correct Option: C

The side of a regular hexagon is 'p' cm then its area is

  1. $ \displaystyle \frac{\sqrt{3}}{2}p^{2}cm^{2} $

  2. $ \displaystyle \frac{3\sqrt{3}}{2}p^{2}cm^{2} $

  3. $ \displaystyle 2\sqrt{3}p^{2}cm^{2} $

  4. $ \displaystyle 6p^{2}cm^{2} $


Correct Option: B
Explanation:

Given side of hexa gon is p cm 

Then area of hexagon =$\frac{(side)^{2}\times  n}{4tan\frac{180}{n}}=\frac{p^{2}\times 6}{4tan\frac{180}{6}}=\frac{6p^{2}}{4tan30^{0}}=\frac{3p^{2}}{2\times \frac{1}{\sqrt{3}}}=\frac{3\sqrt{3}p^{2}}{2} cm^{2}$

The diagonal of rectangle $ABCD$ intersect each other at $O$. If $\angle AOB = 30^0$, then we can construct a rectangle if _________ is given.

  1. diagonal

  2. one side

  3. both sides

  4. $\angle COD$


Correct Option: A,C
Explanation:

$ABCD$ is a rectangle

$\implies AB = CD$ and $AD = BC$ ... (1)
By knowing these, we can just draw the two pair of parallel lines but the length is not fixed.
So, to  draw a rectangle we need the length of the sides.
From (1), we need only the length of two adjacent sides.
Hence, we can construct a rectangle if both sides are given.

We can construct a parallelogram if:

  1. its adjacent sides and a diagonal are given

  2. its diagonal and one angle are given

  3. its four angles and a side are given

  4. None of these are given


Correct Option: A
Explanation:

Steps to create a parallelogram($ABCD$),

$(i)$ Draw one line segment of length $AB$.
$(ii)$ Make an arc of length $BC$ from $B$ and an arc of length $AC$ from $A$.
$(iii)$ Name the intersection point of both the arcs as $C$. Join $A-C$ and $B-C$.
$(iv)$ After completing this process we get $2$ adjacent sides and one diagonal of a parallelogram $ABCD$.
$(v)$ Since, opposite sides are equal and parallel in a parallelogram. So, draw an arc of length $AB$ from $C$ and an arc of length $BC$ from $A$ and name the intersection point as $D$.
$(vi)$) And then join $C-D$ and $A-D$.
At-last we get a parallelogram $ABCD$.
To construct $ABCD$, we need $2$ adjacent sides $AB$ and $BC$ and length of diagonal $AC$.
By knowing only these $3$ parameters we can construct a parallelogram.
Hence, option A is correct.

Construct a parallelogram $ABCD$ with $AB=24$ cm and $AD=16$ cm. The distance between AB and DC is $10$ cm. Find the area of parallelogram $ABCD$ in sq. cm.

  1. $240$

  2. $235$

  3. $270$

  4. None of these


Correct Option: A
Explanation:

Area of parallelogram ABCD

$=AB\times \left( altitude\quad associated\quad with\quad AB \right) \ =24\times 10\ =240 sq. cm$
So, correct answer is option A. 

Construct a parallelogram $ABCD$, with adjacent sides $AB=4$ cm, $BC = 5$ cm and height corresponding to  (base) $BC = 3.5$ cm. Find the area of parallelogram ABCD in sq. cm.

  1. $21.5$

  2. $14.5$

  3. $17.5$

  4. None of these


Correct Option: C
Explanation:

Area of parallelogram ABCD

$=BC\times \left( altitude\quad associated\quad with\quad BC \right) \ =5\times 3.5\ =17.5$
So, correct answer is option C.

State whether the following statement is True or False.
The length of diagonal of rectangle is more than any side of rectangle.

  1. True

  2. False


Correct Option: A
Explanation:

Length diagonal having sides $a$ and $b$ $=\sqrt{a^2+b^2}$ 

Which is greater than any of its side that is $a$ and $b$.

Construct a rectangle $ABCD$, where $AB=10$ cm and $BC=8$ cm.Steps for its construction is given in a jumbled form. Identify its correct sequence.
1) Join these cuts with a line $CD$ and rectangle $ABCD$ is formed
2) Draw a straight line $AB$ of length $10$ cm
3) Draw perpendicular lines at $A$ and $B$ using protractor.
4) Using compass cut arc at the perpendicular from $A$ and $B$ of lengths $8$ cm

  1. $2,4,3,1$

  2. $2,3,4,1$

  3. $3,2,4,1$

  4. $3,4,2,1$


Correct Option: B
Explanation:

Correct sequence for constructing rectangle $ABCD$ is:

Draw a straight line $AB$ of $10 $ cm.
Draw perpendicular lines at $A$ and $B$ using proctor.
Using compass cut arc at the perpendicular from $A$ and $B$ of lengths $8$ cm.
Join these cuts with a line $CD$ and rectangle $ABCD$ is formed.
Correct sequence is $2,3,4,1$.

Let $ABCD$ be a square in which $A$ lies on the positive y-axis and $B$ lies on the positive x-axis. If $D$ is the point $(12, 17)$ the coordinates of $C$ are.

  1. $(17, 12)$

  2. $(17, 5)$

  3. $(14, 16)$

  4. $(15, 3)$


Correct Option: A

Construct a parallelogram $ABCD$ with $AB=24$ cm and $AD=16$ cm. The distance between AB and DC is $10$ cm. Find the distance between AD and BC.

  1. $25 $ cm

  2. $15 $ cm

  3. $13 $ cm

  4. None of these


Correct Option: B
Explanation:

The distance between AB and DC is the height (DE).
Area of the parallelogram ABCD = base x height = $24\times 10$ = $240$ sq cm
This is also the area of the same parallelogram with AD as the base and AH as the height.
height= $13$ width= $15$
Area of ABCD = ADx AH = $16 \times AH$
But area = $240$sqcm
I.e. AH = $\dfrac{240}{16} = 15$ cm

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