Understanding Quadrilaterals Test

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Understanding Quadrilaterals Test - 23

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Question 1
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The diagonals of a parallelogram $$ABCD$$ intersect at $$O$$. A line through O intersect AB at X and DC at Y. Identify the correct option

A
$$OX = OY$$

B
$$OX > OY$$

C
$$OX < OY$$

D
$$OX - OY = 0$$

Solution

$$\Rightarrow$$ In given figure ABCD is a parallelogram.
$$\therefore$$ AB$$\parallel$$DC
$$\Rightarrow$$ Also AC is a transversal of AB$$\parallel$$DC.
$$\therefore$$ $$\angle1=\angle2$$ [Alternate interior angles]
$$\Rightarrow$$ Now in $$\triangle$$AXO and $$\triangle$$CYO, we have
$$\Rightarrow$$ $$\angle1=\angle2$$ [Alternate interior angles]
$$\Rightarrow$$ $$\angle3=\angle4$$ [Vertically opposite angles]
$$\Rightarrow$$ $$CO=OA$$ [Diagonals bisect each other]
$$\therefore$$ $$\triangle AXO \cong \triangle CYO$$ [ASA Criteria]
$$\therefore$$ $$OX=OY$$ [CPCT]
Question 2
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What do you call a parallelogram which has equal diagonals?

A
A trapezium

B
A rectangle

C
A rhombus

D
A kite

Solution

$$\Rightarrow$$ A parallelogram which has equal diagonals are called rectangle.
$$\Rightarrow$$ Trapezium, rhombus and kite does not contain equal diagonal.
$$\Rightarrow$$ Proof for rectangle contains equal diagonals are given below :-
$$\Rightarrow$$ In given figure ABCD is a rectangle.
$$\Rightarrow$$ In $$\triangle$$ABC and $$\triangle$$DCB
$$\Rightarrow$$ BC = CB [Common]
$$\Rightarrow$$ AB = DC [Opposite sides of rectangle]
$$\Rightarrow$$ $$\angle$$ABC = $$\angle$$DCA = 90$$^\circ$$ [Each angle of rectangle is 90$$^\circ$$]
$$\Rightarrow$$ $$\triangle$$ABC $$\cong$$ $$\triangle$$DCB [By SAS property]
$$\therefore$$ AC = BD [By CPCT]
Question 3
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ABCD is a parallelogram in which $$\angle A = (x+20)^\circ $$ and $$\angle C = (3x -10)^\circ $$. Find the value of $$x$$.

A
$$ 60^\circ $$

B
$$40^\circ$$

C
$$30^\circ $$

D
$$15^\circ$$

Solution

In given figure ABCD is a parallelogram.
We have given that, $$\angle A =(x+20)^\circ$$ and $$\angle C = (3x-10)^\circ$$
We know that opposite angles of parallelogram are equal.
So, $$\angle A = \angle C$$
$$(x+20)^\circ=(3x-10)^\circ$$
$$2x^\circ =30^\circ$$
$$\therefore$$ $$x = 15^\circ$$
Question 4
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$$ABCD$$ is a parallelogram and $$X, Y$$ are the mid-points of sides $$AB$$ and $$CD$$ respectively. What is the quadrilateral $$AXCY$$?

A
Parallelogram

B
Rhombus

C
Square

D
Rectangle
Question 5
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The figure below is a parallelogram with two angles given in terms of x. Determine the value of x.

A
$$9$$

B
$$10$$

C
$$20$$

D
$$22$$

E
$$24$$

Solution

We know that, adjacent angles of a parallelogram are supplementary; that is, they add to $$180^o$$.
Use this to set up an equation, and then solve it for $$x$$.
$$\therefore \ (4x + 12) + (3x + 14)=180$$
$$\Rightarrow 7x +26 =180$$
$$\Rightarrow 7x=154$$
$$\Rightarrow x=22$$
Hence, the value of $$x$$ is $$22$$.
Question 6
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Find the measure of each angle of a parallelogram, if largest angle is $$\displaystyle { 30 }^{ \circ }$$ less than twice the smallest angle.

A
$$\displaystyle { 60 }^{ \circ },{ 100 }^{ \circ },{ 90 }^{ \circ },{ 20 }^{ \circ }$$

B
$$\displaystyle { 80 }^{ \circ },{ 40 }^{ \circ },{ 120 }^{ \circ },{ 90 }^{ \circ }$$

C
$$\displaystyle { 100 }^{ \circ },{ 90 }^{ \circ },{ 90 }^{ \circ },{ 80 }^{ \circ }$$

D
$$\displaystyle { 70 }^{ \circ },{ 110 }^{ \circ },{ 70 }^{ \circ },{ 110 }^{ \circ }$$

Solution

Sum of angles of parallelogram is $$ 360^o$$
$$\Rightarrow x+(2x-30)+x+(2x-30)=360^o$$
$$ x^o+(2x-30)^o+x^o +(2x-30)^o=360^o$$
$$6x^o=420^o$$
$$\Rightarrow {x^o=70^o}$$
$$(2x-30)^o=2\times70^o$$ $$-30^o $$
$$=110^o$$
$$\Rightarrow\boxed {angles \,are 70^o,110^o,70^o,110}$$
Question 7
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The diagonals of a parallelogram divide the figure into four triangles that are

A
congruent

B
similar

C
equal in area

D
isosceles

E
none of these
Question 8
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Identify the true statement with respect to the given figure

A
$$ar (POQ) + ar(SOR) = ar(POS) + ar(QOR)$$

B
$$ar(POQ + ar(POS) = ar(ROQ) + ar(ROS)$$

C
Both the true (A) and (B)

D
Neither (A) nor (B) is true

Solution
Question 9
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Diagonal of a parallelogram separates it into

A
Two triangles of equal area

B
Two congruent triangles

C
Both a and b

D
None

Solution

In $$\Delta ABC$$ & $$\Delta ADC$$
$$BC=AD$$
$$DC=AB$$
$$\angle CBA=\angle ADC=\theta $$
$$\therefore$$ By SAS rule $$\Delta ABC$$ is congruent to $$\Delta ADC$$
$$\therefore $$ also have equal area.
Question 10
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In the parallelogram, what is the value of "x"?

A
$$30$$

B
$$50$$

C
$$70$$

D
$$80$$

Solution

By using properties of $$119m$$ $$\theta =80^{\circ}$$
Using exterior angle property $$\theta +x=150^{\circ}$$
$$\Rightarrow \boxed{x=70^{\circ}}$$