Quadrilaterals Test

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Quadrilaterals Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

ABCD is a quadrilateral whose diagonal AC divides it into two parts equal in area, then ABCD is :

A
always a rhombus

B
a parallelogram

C
a kite

D
a trapezium

Solution

A diagonal of a parallelogram divides it into two congruent triangles and we know congruent figures have equal area.
$$\therefore$$ Such a quadrilateral will be a parallelogram.
Hence, the answer is a parallelogram.
Question 2
1 / -0

Three angles of a quadrilateral are $$ 70^{\circ}, 120^{\circ}$$ and $$65^{\circ}$$. Then the fourth angle of the quadrilateral is:

A
$$ 95^{\circ}$$

B
$$ 75^{\circ}$$

C
$$ 105^{\circ}$$

D
$$ 90^{\circ}$$

Solution

We know, by angle sum property, the sum of all angles of a quadrilateral is $$360^o$$.

The given angles are, $$70^o, 120^o, 65^o$$.

Let the fourth angle be $$x^o$$.

Then, $$70^o+120^o+65^o+x^o=360^o$$ ...{Angle sum property of a quadrilateral}

$$\Rightarrow$$ $$360^o - (70^o + 120^o + 65^o ) =x^{\circ}$$

$$\Rightarrow$$ $$x^o=360^o - 255^o =105^{\circ}$$.

Therefore, the fourth angle is $$x^o=105^o$$.

Hence, option $$B$$ is correct.
Question 3
1 / -0

Angles of a quadrilateral are in the ratio $$3 : 6 : 8 : 13$$. The largest angle is :

A
$$178^{\circ}$$

B
$$90^{\circ}$$

C
$$156^{\circ}$$

D
$$36^{\circ}$$

Solution

Given ratio of angles of quadrilateral $$ABCD$$ is $$3:6:8:13$$
Let the angles of quadrilateral $$ABCD$$ be $$3x,6x,8x$$,and $$13x$$, respectively.

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
$$\Rightarrow 3x+6x+8x+13x=360^o$$
$$\Rightarrow 30x=360^o$$
$$\Rightarrow x=\dfrac{360^o}{12}$$
$$\therefore x=12^o$$.

$$\therefore \angle A =3x=3 \times 12^o =36^o$$,
$$ \angle B =6x=6\times 12^o =72^o$$,
$$ \angle C =8x=8 \times 12^o =96^o$$
and $$ \angle D =13x=13 \times 12^o =156^o$$.

$$\therefore$$ The largest angle $$=156^o$$.

Hence, option $$C$$ is correct.
Question 4
1 / -0

If three angles of a quadrilateral are $$70^{\circ}, 95^{\circ}$$ and $$105^{\circ}$$, then the fourth angle is :

A
$$90^{\circ}$$

B
$$100^{\circ}$$

C
$$110^{\circ}$$

D
$$85^{\circ}$$

Solution

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
The given angles are, $$70^o, 95^o, 105^o$$.
Let the fourth angle be $$x^o$$.
Then, $$70^o+95^o+105^o+x^o=360^o$$.

$$\Rightarrow$$ $$360^o - (70^o + 95^o + 105^o ) =x^{\circ}$$

$$\Rightarrow$$ $$x^o=360^o - 270^o =90^{\circ}$$.
Therefore, the fourth angle is $$x^o=90^o$$.

Hence, option $$A$$ is correct.
Question 5
1 / -0

In a quadrilateral $$ABCD$$, if $$\angle A = 80^{\circ},\angle B=70^{\circ}, \angle C=130^{\circ} ,$$ then $$\angle D$$ is:

A
$$80^{\circ}$$

B
$$70^{\circ}$$

C
$$130^{\circ}$$

D
$$150^{\circ}$$

Solution

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.

The given angles are $$\angle A=80^o, \angle B=70^o, \angle C=130^o, \angle D$$.

Then, $$\angle A+\angle B+\angle C+\angle D=360^o$$
$$\implies$$ $$80^o+70^o+130^o+\angle D=360^o$$.
$$\Rightarrow$$ $$\angle D=360^o - (80^o + 70^o + 130^o )$$

$$\Rightarrow$$ $$\angle D=360^o - 280^o =80^{\circ}$$.

Therefore, the unknown angle is $$\angle D=80^o$$.

Hence, option $$A$$ is correct.
Question 6
1 / -0

Fill in the blank:

Line joining the mid-points of any two sides of a triangle is _____ to the third side.

A
perpendicular

B
parallel

C
Inclined at $$60^o$$

D
None of these

Solution

Line joining the mid-points of any two sides of a triangle is parallel to the third side.
Question 7
1 / -0

In a quadrilateral PQSR, with a diagonal PS, if $$QS = SR$$ and $$ \angle QSP = \angle RSP$$, then consider the following statements
Assertion (A) : $$\angle QPS = \angle RPS$$
Reason (R) : Triangles QPS and RPS are congruent.Of these statements

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true but R is not a correct explanation of A

C
A is true, but R is false

D
A is false, but R is true

Solution

In $$\triangle QPS$$ and $$\triangle RPS$$,
$$\Rightarrow$$ $$QS=SR$$ [ Given ]
$$\Rightarrow$$ $$QSP=\angle RSP$$ [ Given ]
$$\Rightarrow$$ $$SP=PS$$ [ Common side ]
$$\Rightarrow$$ $$\triangle QPS\cong\triangle RPS$$ [ By SAS Congruence rule ]
$$\Rightarrow$$ $$\angle QPS=\angle RPS$$ [ CPCT ]
$$\therefore$$ Both assertion and reason are correct and reason is correct explanation of assertion.
Question 8
1 / -0

Complete the following statement.
The straight lines which join the middle points of opposite sides of a quadrilateral ,

A
Are parallel to one another

B
Bisect one another

C
Trisect one another

D
None of these

Solution

On joining the midpoint of sides of a quadrilateral ,we get a parallelogram.
The diagonal of the parallelogram bisect each other.
Question 9
1 / -0

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the _______ side.

A
first

B
second

C
third

D
none of the above

Solution

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Question 10
1 / -0

The diagonals of a quadrilateral ABCD are perpendicular. The quadrilateral formed by joining the mid points of its sides, is ?

A
Trapezium

B
Rectangle

C
Square

D
Rhombus

Solution

From above figure,
AC, BD are diagonals of quadrilateral ABCD are perpendicular.
Points P, Q, R and S are mid-points of AB and BC respectively.
In $$\triangle$$ABC,
P and Q are mid points of AB and BC.
So, by Mid-point theorem,
$$\therefore$$ PQ$$\left|\right|$$AC; PQ = $$\dfrac{1}{2}$$AC -------- (1)
Similarly in $$\triangle$$ACD,
R and S are mid point of sides CD and AD.
$$\therefore$$ SR$$\left|\right|$$AC; SR = $$\dfrac{1}{2}$$AC ----------- (2)
From (1) and (2), we get
PQ$$\left|\right|$$SR and PQ = SR
Hence, PQRS is parallelogram -------( pair of opposite sides are equal and parallel)
Now,RS$$\left|\right|$$AC and QR$$\left|\right|$$BD.
Also, AC$$\perp$$BD ---- (Given)
$$\therefore$$ RS$$\perp$$QR.
Thus, PQRS is a rectangle.

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Re-Attempt Weekly Quiz Competition

Quadrilaterals Test - 22

Result

Quadrilaterals Test - 22

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SHARING IS CARING

Question 1

always a rhombus

a parallelogram

a kite

a trapezium

Solution

Question 2

$$ 95^{\circ}$$

$$ 75^{\circ}$$

$$ 105^{\circ}$$

$$ 90^{\circ}$$

Solution

Question 3

$$178^{\circ}$$

$$90^{\circ}$$

$$156^{\circ}$$

$$36^{\circ}$$

Solution

Question 4

$$90^{\circ}$$

$$100^{\circ}$$

$$110^{\circ}$$

$$85^{\circ}$$

Solution

Question 5

$$80^{\circ}$$

$$70^{\circ}$$

$$130^{\circ}$$

$$150^{\circ}$$

Solution

Question 6

perpendicular

parallel

Inclined at $$60^o$$

None of these

Solution

Question 7

Both A and R are true and R is the correct explanation of A

Both A and R are true but R is not a correct explanation of A

A is true, but R is false

A is false, but R is true

Solution

Question 8

Are parallel to one another

Bisect one another

Trisect one another

None of these

Solution

Question 9

first

second

third

none of the above

Solution

Question 10

Trapezium

Rectangle

Square

Rhombus

Solution

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