Understanding Quadrilaterals Test - 16

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Understanding Quadrilaterals Test - 16

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Weekly Quiz Competition

Question 1
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In parallelogram ABCD, AP and AQ are perpendiculars from vertex of obtuse angle A as shown. If $$\angle x : \angle y = 2 : 1$$ ; find angles of the parallelogram.

A
$$\angle DAB = \angle C = 160^{\circ}$$ and $$\angle B = \angle D = 60^{\circ}$$

B
$$\angle DAB = \angle C = 120^{\circ}$$ and $$\angle B = \angle D = 60^{\circ}$$

C
$$\angle DAB = \angle C = 230^{\circ}$$ and $$\angle B = \angle D = 60^{\circ}$$

D
$$\angle DAB = \angle C = 100^{\circ}$$ and $$\angle B = \angle D = 60^{\circ}$$

Solution

Given, ABCD is a parallelogram and $$AQ\bot CD\quad and\quad AP\bot BC$$
$$\therefore \angle AQC=\angle APC=90$$

AQCP is forming a quadrilateral.
As we know that sum of angles of quadrilateral is 360,
$$\therefore \angle QAP+\angle AQC+\angle QCP+\angle APC=360$$
$$\\ or,\angle QAP+90+\angle QCP+90=360\quad \quad \quad \quad \quad \quad \quad \quad (\because \angle AQC=\angle APC=90)\\ or,\angle QAP+\angle QCP=360-180\\ or,\angle QAP+\angle QCP=180\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (1)$$
Given $$\angle QCP=x,\quad \angle QAP=y\\ x\quad :\quad y\quad =\quad 2\quad :\quad 1$$
Let angle be m.
$$\therefore x=2m\quad and\quad y=m$$
from (1),
$$\quad \angle QAP+\angle QCP=180\\ or,m+2m=180\\ or,\quad 3m=180\\ or,\quad m=60$$
$$\angle QCP=2m\\ or,\angle C=2\times 60=120\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (2)$$

In a parallelogram ABCD, Sum of any two consecutive angles is supplementary,
$$\quad \therefore \angle C+\angle D=180\\ or,\angle 120+\angle D=180\\ or,\angle D=180-120\\ or,\angle D=60\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (3)$$

In a paralleogram ABCD, opposite angles are equal,
$$\therefore \angle C=\angle A=120\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (from\quad 2)\\ \quad \angle D=\angle B=60\qquad \qquad \qquad \qquad \qquad \quad \quad (from\quad 3)$$
Question 2
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The perimeter of a parallelogram ABCD = 40 cm, AB = 3x cm, BC = 2x cm and CD = 2(y +1) cm. Find the values of x and y.

A
x$$ =$$ 4 and y $$=$$ 5

B
x$$ =$$ 2 and y $$= $$7

C
x $$=$$ 3 and y$$ =$$ 8

D
x $$=$$ 1 and y $$=$$ 12

Solution

Let the parallelogram be ABCD Perimeter= $$40 $$cm
AB $$= 3x$$ cm BC$$= 2x$$ cm(given)
Perimeter of parallelogram $$2(3x+2x)=40$$
solving equation we get $$ x=4$$
Also,AB=CD(opposite sides of parallelogram) $$3x=2(y+1)$$ (given)
Putting $$x=4$$
$$3(4)=2y+2$$
$$12=2y+2$$
$$10=2y$$
Hence $$ y=5$$ and $$x=4$$
Question 3
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$$ABCD$$ is parallelogram. If $$L$$ and $$M$$ are the middle points of $$BC$$ and $$ CD $$, then $$ \overrightarrow {AL}+ \overrightarrow {AM}= $$

A
$$ \dfrac { 1 }{ 2 } \overrightarrow{AC}$$

B
$$ \dfrac { 3 }{ 2 } \overrightarrow{AC}$$

C
$$\overrightarrow{AC}$$

D
None of the above

Solution

$$\displaystyle AL=AB+BL=AB+\frac { 1 }{ 2 } BC=AB+\frac { 1 }{ 2 } AD$$
$$\displaystyle AM=AD+DM=AD+\frac { 1 }{ 2 } DC=AD+\frac { 1 }{ 2 } AB$$
Adding, $$\displaystyle AL+AM=\frac { 3 }{ 2 } \left( AB+AD \right) =\frac { 3 }{ 2 } \left( AB+BC \right) =\frac { 3 }{ 2 } AC$$
Question 4
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The bisectors of interior angles of a parallelogram forms a ___________.

A
Square

B
Rectangle

C
Trapezium

D
Rhombus

Solution

$$ABCD$$ is a parallelogram

$$AE$$ bisects $$\angle BAD$$

$$BF$$ bisects $$\angle ABC$$

$$CG$$ bisects $$\angle BCD$$

$$DH$$ bisects $$\angle ADC$$

$$\angle BAD + \angle ABC = 180{^o}$$ (because adjacent angles of a parallelogram are supplementary)

$$ \angle BAJ = \cfrac{1}{2} \angle BAD $$ because $$AE$$ bisects $$\angle BAD$$

$$ \angle ABJ = \cfrac{1}{2} \angle ABC $$ because $$DH$$ bisects $$\angle ABC$$

$$\angle BAJ$$ + $$\angle ABJ$$ = $$90^{o}$$ [ halves of supplemetary angles are complementary]

$$\triangle ABJ$$ is a right triangle because its acute interior angles are complementary with $$\angle J = 90^o$$

Similarly, $$\angle K =\angle L= \angle I = 90^o$$

In quadrilateral $$IJLK$$, since all angles are $$90^o$$, it must be a rectangle.

Therefore, internal angle bisectors of a parallelogram forms a rectangle.
Question 5
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PQRS is a parallelogram and M, N are the mid-points of PQ and RS respectively. Which of the following is not true ?

A
RM trisects QS

B
PN trisects QS

C
$$\Delta$$PSN $$\sim$$ $$\Delta$$QMR

D
MS is not parallel to QN

Solution

In parallelogram $$PQRS,$$

We know that $$PQ= SR$$

Given, M is the mid-point of PQ and N is the mid-point of RS

$$\therefore PM=MQ=SN=NR$$ ...(1)

and we know that sides $$PQ$$ and $$RS$$ are parallel to each other.

$$\therefore QM$$ and $$NS$$ will also be parallel to each other.

$$\therefore$$ We can say that $$MSNQ$$ is a parallelogram

Hence its opposite sides will also be parallel

$$\Rightarrow MS \parallel QN$$

option D is incorrect so the correct answer according to the question is Option D.
Question 6
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Look at the pairs of shapes Which is a pair of rectangles?

A

B

C

D

Solution

Option B shows rectangle.
Hence, the answer is Option B.
Question 7
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If an angle of a parallelogram is 24$$^o$$ less than twice the smallest angle, then the value of largest angle of the parallelogram is

A
176$$^o$$

B
160$$^o$$

C
112$$^o$$

D
108$$^o$$

Solution

Let the smallest angle be $$x$$

then,the largest angle will be $$=180-x $$

but, the same equals to $$[2x-24] $$

so we have $$[2x-24]= 180-x$$

$$3x= 204 $$

$$x=68$$

thus the largest angle $$180-68= 112$$ degree
Question 8
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In the given figure, ABCD is a parallelogram in which $$ \angle DAB $$ = 75$$^o$$ and $$\angle$$DBC = 60$$^o$$ then the measure of $$\angle$$BDC is equal to ?

A
75$$^o$$

B
60$$^o$$

C
45$$^o$$

D
55$$^o$$

Solution

Opposite angles of parallelogram are equal.
$$\angle DAB = \angle BCD = {75}^{o}$$ and $$\angle DBC = {60}^{o}$$
$$ \angle BDC + \angle DBC + \angle BCD = {180}^{o}$$
$$ \angle BDC = {180}^{o} - \angle DBC - \angle BCD
= {180}^{o} - {75}^{o} - {60}^{o}
= {45}^{o} $$
Question 9
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A quadrilateral in which only one pair of opposite sides are parallel is called a ..........

A
square

B
rectangle

C
rhombus

D
trapezium

Solution

$$\Rightarrow$$ We know that in square, rectangle and rhombus two pairs of opposite sides are parallel, so answer can not be square, rectangle or rhombus.
$$\Rightarrow$$ We also know the properties of trapezium in which only one pair of opposite sides are equal.
$$\Rightarrow$$ So, correct answer is option D trapezium.
Question 10
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In a parallelogram ABCD, $$\angle$$D = 135$$^o$$, determine the measure of $$\angle$$A .

A
$$\angle$$A = 45$$^o$$

B
$$\angle$$A = 145$$^o$$

C
$$\angle$$A = 35$$^o$$

D
$$\angle$$A = 75$$^o$$

Solution

$$ABCD$$ is a parallelogram, $$\angle D=135$$
We know, in a parallelogram, opposite angles are equal.
Thus, $$\angle B =\angle D$$
$$=>\angle B=135$$
Also, in a parallelogram opposites sides are parallel. Thus,
$$AD\parallel BC$$
Now, $$AD\parallel BC$$ and $$AB$$ is the transversal, then
$$\angle A +\angle B=180$$
$$=>\angle A=180-135$$
$$=>\angle A=45$$