Understanding Quadrilaterals Test

Result

Understanding Quadrilaterals Test - 19

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$PQRS$$ is a parallelogram. Find $$y$$.

A
$$27^{\circ}$$

B
$$61^{\circ}$$

C
$$41^{\circ}$$

D
$$28^{\circ}$$

Solution

Given, $$PQRS$$ is a parallelogram.
Hence, $$\angle PSR = \angle PQR = 68^{\circ}$$ (Opposite angles of a parallelogram are equal)
Now, $$\angle PTS + \angle STQ = 180$$ (linear pair)
$$\angle PTS + 139 = 180$$
$$\angle PTS = 41^{\circ}$$
Now, $$\angle PTS = \angle RST = 41^{\circ}$$ (Alternate angles of parallel lines)
Now, $$\angle PSR = \angle PST + \angle TSR$$
$$68 = 41 + \angle TSP$$
$$\angle TSP = 27^{\circ}$$
Question 2
1 / -0

If ABCD is parallelogram. E is mid-point of AB and CE bisects $$ \angle BCD$$, then $$\angle DEC$$ is _____.

A
$$60^o$$

B
$$90^o$$

C
$$100^o$$

D
$$120^o$$

Solution

ABCD is parallelogram. E is mid-point of AB and CE bisects BCD
AB||CD and CE Traversal

EC Bisects $$\angle BCD$$
IN parallelogram. ABCD
$$\angle D$$+$$\angle C=180$$
$$\therefore \frac{1}{2}\angle D+\frac{1}{2}\angle C=\frac{1}{2}\times 180$$
$$\angle EDC+\angle ECD= 90^{0}$$
$$\bigtriangleup DEC$$
$$\angle DEC+\angle EDC+\angle ECD= 180^{0}$$
$$\therefore \angle DEC+90^{0}= 180^{0}$$
$$\therefore \angle DEC= 180^{0}- 90^{0}= 90^{0}$$
Question 3
1 / -0

The following figure GUNS is a parallelogram. Find $$x$$ and $$y.$$ (Lengths are in cm)

A
$$x= 6$$ and $$y= 9$$

B
$$x=5$$ and $$y=8$$

C
$$x= 9$$ and $$y= 6$$

D
$$x= 5$$ and $$y= 7$$

Solution

Given: $$GUNS$$ is a parallelogram

Hence, its opposite sides are equal.

i.e. $$GS = UN$$ and $$GU =SN$$

$$\Rightarrow 3x=18$$, i.e., $$x =6$$

and $$3y-1=26$$

$$\Rightarrow 3y=26 + 1 = 27$$

$$\Rightarrow y = 9$$

Hence, $$x= 6$$ and $$y= 9$$
Question 4
1 / -0

The following figure RUNS is a parallelogram. Find x and y. (Lengths are in cm)

A
$$x=3 $$ and $$y=13$$

B
$$x=13 $$ and $$y=3$$

C
$$x=13 $$ and $$y=13$$

D
$$x=3 $$ and $$y=3$$

Solution

$$RUNS$$ is a parallelogram and thus its diagonals bisect each other.
$$\therefore$$ $$x+y=16$$ ---- ( 1 )
and $$7+y=20$$
$$\therefore$$ $$y=20-7$$
$$\therefore$$ $$y=13\,cm$$
Substituting $$y=13$$ in equation ( 1 ) we get,
$$\Rightarrow$$ $$x+13=16$$
$$\Rightarrow$$ $$x=16-13$$
$$\therefore$$ $$x=3\,cm$$
$$\therefore$$ $$x=3\,cm$$ and $$y=13\,cm$$
Question 5
1 / -0

The adjacent angles of a parallelogram are as 2 : 3. Find the measures of all the angles

A
$$108^o$$

B
$$60^o$$

C
$$116^o$$

D
$$90^o$$

Solution

Let ABCD be a parallelogram. Let $$\angle A$$ and $$\angle B$$ be its two adjacent angles such that $$\angle A:\angle B=2:3$$
Let $$\angle A = 2x^o$$ and $$\angle B = 3x^o$$
Since adjacent interior angles are supplementary.
$$\therefore \angle A + \angle B =180^o$$
$$\Rightarrow 2x^o + 3x^o =180^o$$
$$\Rightarrow 5x^o=180^o\Rightarrow x^o=\frac {180}{5}=36^o$$
$$\therefore \angle A=2x^o=72^o, \angle B=3x^o=108^o$$
Since opposite angles of parallelogram are equal.
$$\therefore \angle C=\angle A=72^o$$ and $$\angle D=\angle B=108^o$$
Question 6
1 / -0

Two adjacent angles of a parallelogram are in the ratio $$4:5$$, Find all the angles of the parallelogram

A
$$\angle C = \angle A = 90^o, \angle D =\angle B = 100^o$$

B
$$\angle C = \angle A = 80^o, \angle D =\angle B = 100^o$$

C
$$\angle C = \angle A = \angle D =\angle B = 90^o$$

D
None of above

Solution

Let $$ABCD$$ be a parallelogram.
Given that, $$\angle A:\angle B=4:5$$
Sum of the ratios $$=4+5=9$$
But, $$\angle A+\angle B=180^o$$
(Adjacent angles of the parallelogram, $$ABCD$$)
$$\therefore \angle A=\dfrac {4}{9}\times 180^o=80^o$$
$$\angle B=\dfrac {5}{9}\times 180^o=100^o$$
So $$\angle C = \angle A = 80^o, \angle D =\angle B = 100^o$$
(Opposite angles of a parallelogram are equal).
Question 7
1 / -0

In the given figure $$AE = BC$$ and $$AE || BC$$ and the three sides $$AB, CD$$ and $$ED$$ are equal in length If $$\displaystyle \angle A=102^{\circ}$$ find the measure of $$\displaystyle \angle BCD$$

A
$$\displaystyle 138^{\circ}$$

B
$$\displaystyle 162^{\circ}$$

C
$$\displaystyle 88^{\circ}$$

D
None of these

Solution

As $$AE = BC$$ and $$AE || BC$$
So if a line will be drawn parallel to $$AB$$ from $$E$$, the line $$EC$$ will be pass through $$C$$ and $$AB=EC$$
And parallelogram will be formed.
and in parallelogram opposite angels are congruent.
So,
$$\displaystyle \angle BAE=\angle BCE=102^{\circ}$$
And AB=EC=CD=ED
So, ECD is equilateral triangle and all the three angle will be equal.
$$\displaystyle \angle ECD=60^{\circ}$$
So, $$\displaystyle \angle BCD=\angle BCE+\angle ECD=102^{\circ}+60^{\circ}=162^{\circ}$$
Answer - option $$B$$ is correct
Question 8
1 / -0

Given here are some figures:
Which one is convex polygon among $$2$$ and $$3$$?

A
$$2$$

B
$$3$$

C
Both

D
None of these

Solution

A convex polygon is a simple polygon whose interior is a convex set. In a convex polygon, all interior angles are less than or equal to $$180$$ degrees.
So, Image $$2$$ is a convex polygon.
Question 9
1 / -0

In a parallelogram PQRS, X is the mid point of PS and Y is the mid point of QR, then XY divides QS in the ratio of

A
$$1:4$$

B
$$1:1$$

C
$$1:2$$

D
$$1:3$$

Solution

Suppose XY and diagonal QS intersect each other at point O.
Since X and Y are the midpoints of PS and QR. This means XY is parallel to PQ.
Now in triangle PQS, XO is parallel to PQ And X is midpoint of PS. By applying midpoint theorem on triangle PQS, O becomes midpoint of QS.
So, XY devides QS in the ration 1:1.
So, correct answer is option B.
Question 10
1 / -0

Which of the following is/are an example(s) of simple closed curve?

A
Circle

B
Rectangle

C
Square

D
All of the above

Solution

A curve that does not cross itself and ends at the same point where it begins, is called a simple closed curve.
Therefore, $$D$$ is the correct answer.