# Free Quadrilaterals 01 Practice Test - 9th Grade

### Question 1

If ABCD is a parallelogram, then which of the following is true?

AC=BD

AB=AC

∠A=130∘,∠C=130∘,∠B=50∘

AB⊥AD

#### SOLUTION

Solution :C

In a parallelogram, the adjacent sides and angles are not equal and no sides are perpendicular to other.

∠A+∠B=130∘+50∘=180∘

Here, ∠A and ∠B are co-interior angles and their sum is 180∘.

By the converse of co-interior angles theorem, AD || BC and AB is the transversal.

∠D+∠A=50∘+130∘=180∘

Here, ∠D and ∠A are co-interior angles and their sum is 180∘.

By the converse of co-interior angles theorem, AB || DC and AD is the transversal.

### Question 2

Opposite angles of a parallelogram are equal.

True

False

#### SOLUTION

Solution :A

In the given parallelogram, AD || BC and AB || DC

Consider the diagonal BD which acts as a transversal for BC and AD.

∠ADB=∠DBC=x∘ ( alternate angles)Consider the diagonal BD which acts as a transversal for AB and BC.

∠ABD=∠BDC=y∘ ( alternate angles)We can see that

∠B=∠ABD+∠DBC=x∘+y∘

Also, ∠D=∠ADB+∠BDC=x∘+y∘

Thus, ∠B=∠D.Hence, opposite angles of a parallelogram are equal.

### Question 3

Three angles of a quadrilateral measure 37∘, 130∘ and 53∘. Find the measure of the fourth angle.

40∘

50∘

90∘

140∘

#### SOLUTION

Solution :D

Let the fourth angle be x∘.

By angle sum property, we know that sum of interior angles of a quadrilateral is 360∘.

⇒37∘+130∘+53∘+x=360∘

⇒x=360∘−(37∘+130∘+53∘)=360∘ − 220∘=140∘

### Question 4

The sum of external angles of a quadrilateral is equal to the sum of its internal angles.

True

False

#### SOLUTION

Solution :A

The sum of internal angles of a quadrilateral is 360∘. The sum of external angles of any polygon, irrespective of the number of sides is 360∘. Hence, the given statement is true.

### Question 5

If a diagonal of a rectangle is inclined to one side of the rectangle at 35∘, then the acute angle between the diagonals is _____.

#### SOLUTION

Solution :C

Given: In rectangle ABCD,

∠ BAO=35°

To find: Angle between the diagonals ∠ AOD or ∠BOC

In ΔAOB

OA = OB

∵ Diagonals of a rectangle bisect each other

⇒∠OBA=∠OAB=35∘ (As the angles opposite to equal sides are equal)

Applying exterior angle property in ΔAOB

∠AOD=∠OBA+∠OAB

∠AOD=35°+35°

∠AOD=70°

### Question 6

If opposite angles of a parallelogram are supplementary, then the parallelogram is _____ .

#### SOLUTION

Solution :B

We know that,

the opposite angles of a parallelogram are equal.

ie. ∠ A=∠ C and ∠ B=∠ D ...(i)

If they are supplementary as well, then

∠ A+∠ C=180∘

from (i),

∠ A+∠ A=180∘

2∠ A=180∘

∠ A=90∘

Similarly,

∠ C=90∘

It means that each angle will be equal to 90∘.

A parallelogram whose internal angles are all 90∘ is a rectangle. Thus, the figure will be a rectangle.

### Question 7

If the ratio of sides (taken in order) of a quadrilateral is a:b:b:a, then it is a

kite

cyclic quadrilateral

parallelogram

square

#### SOLUTION

Solution :A

Here, the ratio is given to be a:b:b:a.

If KLMJ is a quadrilateral, the sides are in the ratio a:b:b:a.

Let KL = a, LM = b, MJ = b and JK = a.

Here, JK and LM are equal and LM and MJ are equal.

We know that, kite is a quadrilateral with two pairs of adjacent sides equal in length.

The data given implies that we have two disjoint pairs of adjacent sides which are equal i.e. there are two pairs of adjacent sides which are equal and the opposite sides are not equal.

This is the very definition of a kite.

### Question 8

In the given figure, ABCD is a parallelogram. From the given options, select the values of x and y.

#### SOLUTION

Solution :A and C

We know that the opposite angles of a parallelogram are equal.

So,

(3x−10)∘=(x+80)∘⇒3x−x=80∘+10∘⇒2x=90∘⇒x=45∘

Also, the consecutive angles of a parallelogram are supplementary.

So,

∠B+∠C=180∘(3x−10)∘+(y+25)∘=180∘(3×45∘)−10∘+y+25∘=180∘135∘−10∘+y+25∘=180∘⇒125∘+y+25∘=180∘⇒y=180∘–125∘−25∘⇒y=30∘

### Question 9

The quadrilateral formed by joining the mid-points of consecutive sides of a rectangle ABCD, taken in order, is a rhombus.

PQRS is a rectangle

PQRS is a parallelogram

diagonals of PQRS are perpendicular and bisect each other

diagonals of PQRS are equal and bisect each other

#### SOLUTION

Solution :C

Let ABCD be a rectangle such as AB = CD and BC = DA. P,Q,R and S are the midpoints of the sides AB, BC, CD and DA respectively.

Let us join AC and BD

In ΔABC,

P and Q are the mid-points of AB and BC respectively.

∴ PQ || AC and PQ = 12AC (Midpoint theorem)...........(1)

Similarly in Δ ADC,

SR || AC and SR=12AC (Midpoint theorem)..........(2)

Clearly, PQ || SR and PQ = SR

Since, in quadrilateral PQRS, one pair of opposite side is equal and parallel to each other, it is parallelogram

∴ PS || QR and PS = QR (opposite sides of parallelogram).........(3)

In Δ BCD, Q and R are the mid-points of sides BC and CD respectively.

∴ QR || BD and QR=12BD (Midpoint theorem)..........(4)

However, the diagonals of a rectangle are equal.

∴ AC = BD ...........(5)

By using equation (1), (2), (3), (4), (5), we obtain

PQ = QR = SR = PS

Therefore, PQRS is a rhombus and hence, the given statement is true.

### Question 10

If BDEF and FDCE are parallelograms, then

#### SOLUTION

Solution :

Since BDEF and FDCE are parallelograms,

BD = FE and DC = FE

Hence, we can see that BD is equal to DC and so, D is the mid-point of BC.