ABCD is a parallelogram and X is the mid-point of AB. If \$Area(\Box AXCD) = \$ \$24\$ \${cm}^2\$, Find \$Area(\Delta ABC) \$.

We know that the area of a triangle is half of that of a parallelogram if they are on the same base and between the same parallels.In case \$\triangle XCB\$, base is half. \$\therefore\$ Area of \$\triangle XCB=\dfrac{1}{2}\times \dfrac{1}{2}\times ar(ABC)\$ \$=\dfrac{1}{4}\times ar(ABC)\$Now, \$ar(AXCD)=ar(ABCD)-ar(XCB)\$ \$=ar(ABCD)-\dfrac{1}{4}\times ar(ABCD)\$ \$=\dfrac{3}{4}\times ar(ABCD)\$\$\Rightarrow\$ \$ar(AXCD)=\dfrac{3}{4}\times ar(ABCD)\$\$\Rightarrow\$ \$24=\dfrac{3}{4}\times ar(ABCD)\$\$\Rightarrow\$ \$ar(ABCD)=32\,cm^2\$We know that a diagonal divides a parallelogram into two triangles of equal area.\$\Rightarrow\$ \$ar(ABC)=\dfrac{1}{2}\times ar(ABCD)\$ \$=\dfrac{1}{2}\times 32\$ \$=16\,cm^2\$

In Fig.8.2, \$ABCD\$ and \$AEFG\$ are two parallelograms. If \$\angle C = 55^o\$, determine \$\angle F\$.

\$ ABCD\$ and \$ AEFG\$ are two parallelograms sharing two sides \$AD\$ (or \$AG\$) and \$AE\$ (or \$AE\$) and the included angle. \$\angle C={ 55 }^{ o }\$.To find out- \$ \angle F\$Solution-We have \$ \angle F=\angle A\$ ......(opposite angles of a parallelogram).........(i)Again \$ \angle A=\angle C\$ ......(opposite angles of a parallelogram)...........(ii)\$ \therefore \angle F=\angle C\$ ......[From (i) and (ii)]And \$ \angle C={ 55 }^{ o }\$ .......(given)\$ \Rightarrow \angle F={ 55 }^{ o }\$

The diagonals \$AC\$ and \$BD\$ of a parallelogram \$ABCD\$ intersect each other at the point \$O\$.If \$\angle DAC = 32^o\$ and \$\angle AOB = 70^o\$, then \$\angle DBC\$ is equal to?

In given figure,Quadrilateral ABCD is a parallelogram.So, AD \$\left|\right|\$ BC\$\therefore\$ \$\angle\$DAC = \$\angle\$ACB --- ( Alternate angle)\$\therefore\$ \$\angle\$ACB = 32\$^\circ\$\$\angle\$AOB + \$\angle\$BOC = 180\$^\circ\$ --- (straight angle)\$\Rightarrow\$70\$^\circ\$ + \$\angle\$BOC = 180\$^\circ\$\$\therefore\$ \$\angle\$BOC = 110\$^\circ\$In \$\triangle\$BOC,\$\angle\$OBC + \$\angle\$BOC + \$\angle\$OCB = 180\$^\circ\$\$\Rightarrow\$\$\angle\$OBC + 110\$^\circ\$ + 32\$^\circ\$ = 180\$^\circ\$\$\Rightarrow\$ \$\angle\$OBC = 38\$^\circ\$\$\therefore\$ \$\angle\$DBC = 38\$^\circ\$

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>>Class 8
>>Maths
>>Understanding Quadrilaterals
>> $Medium Questions$

Understanding Quadrilaterals

Q: Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that)What can you say about the angle sum of a convex polygon with the number of sides (n) = 7?

\\( After\quad analysing\quad the\quad given\quad table\quad it\quad is\quad observed\quad that\quad when\\ (i)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=3\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 180 }^{ o }=\left( 3-2 \right) \times { 180 }^{ o }\\ (ii)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=4\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 360 }^{ o }=\left( 4-2 \right) \times { 180 }^{ o }\\ (iii)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=5\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 540 }^{ o }=\left( 5-2 \right) \times { 180 }^{ o }\\ (iv)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=6\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 720 }^{ o }=\left( 6-2 \right) \times { 180 }^{ o }\\ \therefore \quad By\quad induction,\quad it\quad can\quad be\quad concluded\quad that\\ When\quad the\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=n\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles=\left( n-2 \right) \times { 180 }^{ o }\\ \therefore \quad For\quad a\quad 7\quad sided\quad polygon\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \rightarrow sum\quad of\quad the\quad internal\quad angles=\left( 7-2 \right) \times { 180 }^{ o }={ 900 }^{ o }\\ \\ \\ \\ \\ \\ \\ \\)

Q: How many simple closed curves are there?

\$ A\quad simple\quad closed\quad curve\quad encloses\quad a\quad single\quad area \quad and \quad do \quad not\quad have \quad intersecting \quad lines.\\ Here\quad in\quad each\quad of\quad the\quad figures\quad 1,\quad 2,\quad 5,\quad 6,\quad 7\quad a\quad single\quad area\\ is\quad enclosed.\\ But\quad in\quad each\quad of\quad the\quad figures\quad 3,\quad 4,\quad 8\quad more\quad than\quad one\quad area\\ are\quad enclosed.\\ So\quad each\quad of\quad the\quad figures\quad 1,\quad 2,\quad 5,\quad 6,\quad 7\quad is\quad a\quad simple\quad closed \quad curve.\\ \$The number of simple closed curves is thus \$5\$

Q: Say true or false.It is possible to have a regular polygon with the measure of each exterior angle as \$22^{0}.\$

Here, let the number of sides of the polygon be \$n.\$So, the number of exterior angles is \$n\$.Also, it is a regular one. Therefore, the total number of exterior angles=\$n\$ and they are equal to each other.\$ \therefore\$ The sum of the exterior angles = \${ 360 }^{ o }\$\$ \therefore\$ Each angle=\$\theta =\dfrac { { 360 }^{ o } }{ n } \Longrightarrow n=\dfrac { { 360 }^{ o } }{ \theta }\$ when \$n \epsilon N\$i.e \$\theta\$ should be a complete divisor of \${ 360 }^{ o }.\$Here, \$\theta ={ 22 }^{ o }\$ which is not a complete divisor of \${ 360 }^{ o }.\$So any regular polygon can not have exterior angle =\${ 22 }^{ o } \$

Q: Opposite angles of a quadrilateral \$ABCD\$ are equal. If \$AB = 4\$ cm, determine \$CD\$.

Let \$ABCD\$ be a quadrilateral.Given, opposite angles of a quadrilateral \$ABCD\$ are equal.So, opposite sides of the quadrilateral \$ABCD\$ are equal.\$\Rightarrow AB=CD\$Given, \$AB=4\$ cmTherefore, \$CD=4\$ cm.

Q: Which of the following is not always true for a parallelogram?

Opposite angles are bisected by the diagonals.The above statement is not true for every parallelogram.It is true only for a rhombus (diamond) or square where the diagonals bisect the opposite angles they connect, and diagonals are perpendicular. In rectangles, the diagonals do not bisect the angles and are not perpendicular, but they do bisect each other.Hence, option C.

Q: All the angles of a quadrilateral are equal. What special name is given to this quadrilateral?

We know that the sum of the angles of the quadrilateral is \$360^{o}\$Now, the quadrilateral in given question has all angles equal.\$ \therefore\$ Each angle \$={ 360 }^{ o }\div 4={ 90 }^{ o }\$So, this quadrilateral would be rectangle or a square.

Q: In the above figure, it is given that \$BDEF\$ and \$FDCE\$ are parallelograms. Can you say that \$BD = CD\$? Why or why not?

Given:-\$ABC\$ is a triangle. \$D,\;E\;F\$ are points on \$BC,\;CA.\;AB\$ taken in order. \$BDEF\$ anf \$FDCE\$ are parallelograms.To Prove:-\$BD=CD\$Proof:-In parallelogram \$BDEF\$,\$BD=EF\quad\quad\quad\dots(i)\$[Opposite sides of a parallelogram are equal.]In parallelogram \$FDCE\$,\$CD=EF\quad\quad\quad\dots(ii)\$[Opposite sides of a parallelogram are equal.]From \$(i)\$ and \$(ii)\$,\$BD=CD\$Hence, proved.

Maths
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Easy Qs Med Qs Hard Qs

Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that)
What can you say about the angle sum of a convex polygon with the number of sides (n) = 7?

70 0^{0}

80 0^{0}

90 0^{0}

100 0^{0}

Medium

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How many simple closed curves are there?

Medium

View solution>

ABCD is a parallelogram and X is the mid-point of AB. If $A r e a (□ A X C D) =$ $24$ $c m^{2}$ , Find $A r e a (Δ A B C)$ .

16

c m^{2}

26

c m^{2}

12

c m^{2}

24

c m^{2}

Medium

View solution>

Say true or false.
It is possible to have a regular polygon with the measure of each exterior angle as $2 2^{0} .$

True

False

Medium

View solution>

In Fig.8.2, $A B C D$ and $A E F G$ are two parallelograms. If $∠ C = 5 5^{o}$ , determine $∠ F$ .

27 . 5^{o}

4 5^{o}

5 5^{o}

6 5^{o}

Medium

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Opposite angles of a quadrilateral $A B C D$ are equal. If $A B = 4$ cm, determine $C D$ .

3

4

5

6

Medium

View solution>

Which of the following is not always true for a parallelogram?

Opposite sides are equal.

Opposite angles are equal.

Opposite angles are bisected by the diagonals.

Diagonals bisect each other.

Medium

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The diagonals $A C$ and $B D$ of a parallelogram $A B C D$ intersect each other at the point $O$ .
If $∠ D A C = 3 2^{o}$ and $∠ A O B = 7 0^{o}$ , then $∠ D B C$ is equal to?

2 4^{o}

8 6^{o}

3 8^{o}

3 2^{o}

Medium

View solution>

All the angles of a quadrilateral are equal. What special name is given to this quadrilateral?

Rectangle

Trapezium

Rhombus

Parallelogram

Medium

View solution>

In the above figure, it is given that $B D E F$ and $F D C E$ are parallelograms. Can you say that $B D = C D$ ? Why or why not?

Yes

May be

Cannot be determined

Medium

View solution>

Understanding Quadrilaterals

Maths 6781 Views

Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that)What can you say about the angle sum of a convex polygon with the number of sides (n) = 7?

How many simple closed curves are there?

ABCD is a parallelogram and X is the mid-point of AB. If Area(□AXCD)= 24 cm2, Find Area(ΔABC).

Say true or false.It is possible to have a regular polygon with the measure of each exterior angle as 220.

In Fig.8.2, ABCD and AEFG are two parallelograms. If ∠C=55o, determine ∠F.

Opposite angles of a quadrilateral ABCD are equal. If AB=4 cm, determine CD.

Which of the following is not always true for a parallelogram?

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O.If ∠DAC=32o and ∠AOB=70o, then ∠DBC is equal to?

All the angles of a quadrilateral are equal. What special name is given to this quadrilateral?

In the above figure, it is given that BDEF and FDCE are parallelograms. Can you say that BD=CD? Why or why not?

Maths
6781 Views

Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that)
What can you say about the angle sum of a convex polygon with the number of sides (n) = 7?

ABCD is a parallelogram and X is the mid-point of AB. If $A r e a (□ A X C D) =$ $24$ $c m^{2}$ , Find $A r e a (Δ A B C)$ .

Say true or false.
It is possible to have a regular polygon with the measure of each exterior angle as $2 2^{0} .$

In Fig.8.2, $A B C D$ and $A E F G$ are two parallelograms. If $∠ C = 5 5^{o}$ , determine $∠ F$ .

Opposite angles of a quadrilateral $A B C D$ are equal. If $A B = 4$ cm, determine $C D$ .

The diagonals $A C$ and $B D$ of a parallelogram $A B C D$ intersect each other at the point $O$ .
If $∠ D A C = 3 2^{o}$ and $∠ A O B = 7 0^{o}$ , then $∠ D B C$ is equal to?

In the above figure, it is given that $B D E F$ and $F D C E$ are parallelograms. Can you say that $B D = C D$ ? Why or why not?