NCERT Exemplar Class 9 Maths Solutions Chapter 8 Quadrilaterals

NCERT Exemplar Class 9 Maths Solutions Chapter 8 Quadrilaterals

Edited By Safeer PP | Updated on Aug 31, 2022 12:33 PM IST

NCERT exemplar Class 9 Maths solutions chapter 8 deals with quadrilaterals and its properties. The NCERT exemplar Class 9 Maths chapter 8 solutions are prepared in such a way that students get an in-depth process flow to effectively approach the questions of NCERT Class 9 Maths. These NCERT exemplar Class 9 Maths chapter 8 solutions are curated by expert mathematicians at Careers360 and provide exhaustive solutions to the problems leading to a sturdy concept building of quadrilaterals. The NCERT exemplar Class 9 Maths solutions chapter 8 covers all the topics determined for CBSE Class 9 Syllabus.

Exercise 8.1

Question:1

Answer: [D] $120^{\circ}$
As we know that the sum of the angles of a quadrilateral is $360^{\circ}$.
i.e., $\angle 1+\angle 2+\angle 3+\angle 4=360^{\circ}$ …..(i)
Here $\angle 1=75^{\circ}$
$\angle 2=90^{\circ}$

$\angle 3=75^{\circ}$
Put the values in equation (i) we get
$75^{\circ}+90^{\circ}+75^{\circ}+\angle 4=360^{\circ}$
$\angle 4=360^{\circ}-240^{\circ}$
$\angle 4=120^{\circ}$
Hence option D is correct.

Question:2

As we know that, diagonals of a rectangle are equal in length.

$\therefore AC=BD$ { $\because$diagonals are equal}
$\frac{1}{2}AC=\frac{1}{2}BD$ {dividing both sides by 2}
$AO=BO$ { $\because$ O is midpoint of diagonal}
$\therefore \angle OBA=\angle OAB$
$\angle OAB=25^{\circ}$ {Given}
$\Rightarrow \angle OBA=25^{\circ}$
$\angle BOC=\angle OBA+\angle OAB$ {exterior angle is equal to the sum of two opposite interior angles}

$=25^{\circ}+25^{\circ}=50^{\circ}$
Hence the actual angle between the diagonals is 500.
Hence option B is correct

Question:3

Answer: [C] $50^{\circ}$

Given : ABCD is a rhombus such that $\angle ACB=40^{\circ}\Rightarrow \angle OCB=40^{\circ}$
To Find : $\angle ADB$
Since $AD \parallel BC$
$\Rightarrow \angle AOC=\angle ACB=40^{\circ}$ {alternate interior angles}
$\angle AOD=90^{\circ}$ {diagonals of a rhombus are perpendicular to each other}
We know that sum of angles of a triangle is $180^{\circ}$.
In $\triangle AOD$,
$\angle AOD+\angle OAD+\angle ADO=180^{\circ}$
$90^{\circ}+40^{\circ}+\angle ADO=180^{\circ}$
$\angle ADO=180^{\circ}-90^{\circ}-40^{\circ}$
$\angle ADO=50^{\circ}=\angle ADB$
Hence Option C is correct.

Question:4

The quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS taken in order is a rectangle. If
(A) PQRS is a rectangle
(B) PQRS is a parallelogram
(C) Diagonals of PQRS are perpendiculars
(D) Diagonals of PQRS are equal

According to question quadrilateral, ABCD is formed by joining the midpoints of PQRS
$\Rightarrow$ If PQRS is a rectangle

Here ABCD is not a rectangle because a rectangle is a four-sided polygon having all the internal angles $= 90^{\circ}$ and the opposite sides are equal in length.
$\Rightarrow$If PQRS is a parallelogram

Here ABCD is not a rectangle because a rectangle is a four-sided polygon having all the internal angles $= 90^{\circ}$ and the opposite sides are equal in length.
If diagonals of PQRS are perpendicular

Here ABCD is a rectangle because here
$\angle A=\angle B=\angle C=\angle D=90^{\circ}$ and opposite sides we equal that is AB = DC and AD = BD
If diagonals of PQRS are equal

ABCD is not a rectangle it is a square
Because here AB = BC = CD = AD and $\angle A=\angle B=\angle C=\angle D=90^{\circ}$

Here we saw that if diagonals of PQRS are perpendicular to each other then ABCD is a rectangle
option C is correct
(C) diagonals of PQRS are perpendicular

Question:5

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if

(A) PQRS is a rhombus

(B) PQRS is a parallelogram

(C) diagonals of PQRS are perpendicular

(D) diagonals of PQRS are equal.

According to question, the quadrilateral ABCD is formed by joining the midpoints of PQRS
If PQRS is a rhombus

Here ABCD is not a rhombus because in rhombus angles need not be right angles.
But here $\angle A=\angle B=\angle C=\angle D=90^{\circ}$

ABCD is a square not a rhombus.
If PQRS is a parallelogram

Here ABCD is not a rhombus because in rhombus all sides of if will be equal here sides of quadrilateral ABCD are not equal
it is not a rhombus
If diagonals of PQRS are perpendicular

Here AB D is not a rhombus because in rhombus angles are not right angles and all sides are equal but here AB = CD and BC = AD also $\angle A=\angle C=\angle D=90^{\circ}$
Hence ABCD is a rectangle not a square.
If diagonals of PQRS are equal

In ABCD quadrilateral here all sides are equal and angles of quadrilateral ABCD is not right angle, therefore, ABCD is a rhombus\
option D is correct
(D) diagonals of PQRS are equal

Question:6

If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3:7:6:4, then ABCD is a
(A) rhombus (B) parallelogram
(C) trapezium (D) kite

Given : 3 : 7 : 5 : 4 is the ratio of angles A, B, C, and D in a quadrilateral.
Let the angles be 3x, 7x, 6x and 4x.
As we know that the sum of angles is $360^{\circ}$
$3x+7x+6x+4x=360^{\circ}$
$20x=360^{\circ}$
$x=\frac{360^{\circ}}{20}=18^{\circ}$
$A=3x=3\times 18=54^{\circ}$
$B=7x=7\times 18=126^{\circ}$
$C=6x=6\times 18=108^{\circ}$
$D=4x=4\times 18=72^{\circ}$
Now draw the quadrilateral with given angles.

Hence, $BC\parallel AD$ and the sum of co-interior angles is $180^{\circ}$

Hence ABCD is a trapezium

Question:7

First of all, let us draw the figure according to the question.

From the above diagram
$\angle QPS=\angle APB$ …..(1) (Vertically opposite angles)
In $\triangle APB=\angle APB+\angle PAB+\angle ABP=180^{\circ}$
$\angle APB=\frac{1}{2}\angle A+\frac{1}{2}\angle B=180^{\circ}$
$\angle APB=180-\frac{1}{2}(\angle A+\angle B)$ …..(2)
From equation 1 & 2
$\angle QPS=180-\frac{1}{2}(\angle A+\angle B)$ …..(3)
Similarly $\angle QRS=180-\frac{1}{2}(\angle C+\angle D)$ …..(4)
$\angle QPS+\angle QRS=360^{\circ}-\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)$
$=360^{\circ}-\frac{1}{2}(360^{\circ})$
$=360^{\circ}-180^{\circ}$
$=180^{\circ}$
$\therefore \angle QPS+\angle QRS= 180^{\circ}$
whose opposite angles are supplementary

Question:8

$\triangle ABC$, $AB = 5 cm$, $BC=8cm$ and$CA = 7 cm$. If D and E are respectively the mid-points of AB and BC, determine the length of DE.

Answer: $3.5cm$
Given: AB = 5cm, BC = 8cm and CA = 7cm and D and E are the mid points of AB and BC.
To Find: Length of DE

Using mid-point theorem $DE\parallel AC$
And
$DE=\frac{1}{2}AC$
$DE=\frac{1}{2}\times 7$
$DE=\frac{7}{2}cm$
$DE=3.5cm$

Question:9

Given : BDEF and FDCE are parallelograms.

If BDEF is a parallelogram then
$BD\parallel EF$ and BD = EF
…..(1)
Also if FDCE is a parallelogram then
$CD\parallel EF$ and CD = EF
…..(2)

From equation 1 and 2 we get
BD = CD = EF
Hence given statement is True.

Question:8

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form

(A) a square (B) a rhombus
(
C) a rectangle (D) any other parallelogram

Given : APB and COD are two parallel lines.
Construction : Let us draw the bisectors of the angles APQ, BPQ, CQP and PQD

Let the bisectors meet at point M and N
Since $APB\parallel CQD$
$\angle APQ=\angle PQD$ (Alternate angles)
and $\angle MPQ=\angle PQN$ (Alternate interior angles)
$\therefore PM\parallel QN$
Similarly $\angle BPQ=\angle PQC$ (Alternate angles)
$\Rightarrow PN\parallel QM$

Since CQD is a line
$\therefore \angle CQD=180^{\circ}$
$\angle CQP+\angle PQD=180^{\circ}$
$2\angle MQP+2\angle NQP=180^{\circ}$
$2\left (\angle MQP+\angle NQP \right )=180^{\circ}$
$\angle MQP+\angle NQP=\frac{180}{2}$
$\Rightarrow \angle MQN=\frac{180}{2}$
$\Rightarrow \angle MQN=90^{\circ}$

Hence, PMQN is a rectangle.
Therefore option (C) is correct.

Question:9

The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is:
(A) a rhombus (B) a rectangle
(C) a square (D) any parallelogram

Let ABCD is a rhombus and P, Q, R and S are its mid points.

In
$\triangle ABC$, R is the midpoint of AB and Q is the midpoint of BC.
$AC\parallel RQ$ and $RQ=\frac{1}{2}AC$ …..(1)
{using mid-point theorem}
Similarly in
$\triangle ADC$
$AC\parallel SP$and $SP=\frac{1}{2}AC$…..(2)
From equation 1 and 2
$SP\parallel RQ$ and $SP=RQ$

Similarly when we take $\triangle ABD$ and $\triangle BDC$ then we get $SR\parallel PQ$ and $SR=PQ$ .
Hence PQRS is a parallelogram
Therefore option (D) is correct.

Question:10

D and E are the mid-points of the sides AB and AC of DABC and O is any point on side BC. O is joined to A. If P and
Q are the mid-points of OB and OC respectively, then DEQP is

(A) a square (B) a rectangle
(C) a rhombus (D) a parallelogram

By midpoint theorem
$DE\parallel BC$ …..(1)
i.e., $DE=\frac{1}{2}(BC)$
$DE=\frac{1}{2}(BP+PO+OQ+QC)$
$DE=\frac{1}{2}(2PO+2OQ)$
{$\because$ P and Q are midpoints of OB and OC}
$DE=PO+OQ$
$DE=PQ$ …..(2)
Now in $\triangle AOC$, Q and E are the mid points of OC and AC.
$\therefore EQ\parallel AO$ and $EQ=\frac{1}{2}AO$ …..(3)
{Using mid-point theorem}
Similarly, in $\triangle ABO,PD\parallel AO$ and $PD=\frac{1}{2}AO$ …..(4)
From equation 3 and 4
$EQ\parallel PD$ and $EQ=PD$
From equation 1 and 3
$DE\parallel PQ$ and $DE= PQ$

Hence DEQP is a parallelogram.
Therefore option (D) is correct.

Question:11

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if,
(A) ABCD is a rhombus
(B) diagonals of ABCD are equal
(C) diagonals of ABCD are equal and perpendicular
(D) diagonals of ABCD are perpendicular.

Given : ABCD is a quadrilateral and P, Q, R and S are the midpoints of sides of AB, BC, CD and DA. Then, PQRS is a square
$\therefore PQ=QR=RS=PS$ …..(1)
$PR=SQ$
Also, $PR=BC$ and $SQ=AB$
$\therefore AB=BC$
Thus all sides are equal.
Hence ABCD is either a square or a rhombus.
In $\triangle ADB$ by mid-point theorem
$SP\parallel DB$
$SP=\frac{1}{2}DB$ …..(2)
Similarly in $\triangle ABC,AQ=\frac{1}{2}AC$ …..(3)
From equation 1
$PS=PQ$
$\frac{1}{2}DB=\frac{1}{2}AC$

Thus ABCD is a square so diagonals of a quadrilateral are also perpendicular.
Therefore option (C) is correct.

Question:12

The diagonals AC and BD of a parallelogram ABCD intersect each other at thepoint O. If $\angle DAC=32^{\circ}$ and $\angle AOB=70^{\circ}$, then $\angle DBC$is equal to
(A) $24^{\circ}$ (B) $86^{\circ}$ (C) $38^{\circ}$ (D) $32^{\circ}$

Answer: [C] $38^{\circ}$

Solution.
Given $\angle AOB=70^{\circ}$ and $\angle DAC=32^{\circ}$

$\angle ACB=32^{\circ}${$AD\parallel BC$ and AC is a transversal line}
$\angle AOB+\angle BOC=180^{\circ}$
$70^{\circ}+\angle BOC=180^{\circ}$
$\angle BOC=180^{\circ}-70^{\circ}=110^{\circ}$
In
$\triangle BOC$, we have
$\angle BOC+\angle OCB+\angle CBO=180^{\circ}$
$110^{\circ}+32^{\circ}+\angle CBO=180^{\circ}$
$\angle CBO=180^{\circ}-110^{\circ}-32^{\circ}$
$\angle CBO=38^{\circ}$
$\angle DBC=\angle CBD=38^{\circ}$
Hence option C is correct

Question:13

Which of the following is not true for a parallelogram?
(A) opposite sides are equal
(B) opposite angles are equal
(C) opposite angles are bisected by the diagonals
(D) diagonals bisect each other.

Answer: (C) opposite angles are bisected by the diagonals
Parallelogram: A parallelogram is a special type of quadrilateral whose opposite sides and angles are equal.
Hence we can say that the opposite sides of a parallelogram are of equal length (Option A) and the opposite angles of a parallelogram are of equal measure (Option B).
In parallelogram, diagonals bisect each other (Option D) but opposite angles are not bisected by diagonals (Option C).

Therefore option (C) is not true.

Question:14

D and E are the mid-points of the sides AB and AC respectively of $\triangle ABC$. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is
(A) $\angle DAE = \angle EFC$
(B) AE = EF
(C) DE = EF
(D) $\angle ADE = \angle ECF.$

Let us draw the figure according to question.

AE = EC
{E is the mid-point of AC}
Let DE = EF
$\angle AED=\angle FEC$ {vertically opposite angles}
$\therefore \triangle ADE\cong \triangle CFE$ {by SAS congruence rule}
$\therefore AD=CF$ {by CPCT rule}
$\therefore \angle ADE=\angle CFE$ {by CPCT}
Hence,$AD\parallel CF$

We need DE = EF.
Hence option C is correct.

Exercise 8.2

Question:1

Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If $OA = 3cm$ and $OD = 2 cm$, determine the lengths of AC and BD.

Given: ABCD is a parallelogram
$OA = 3 cm$ and $OD = 2 cm$

As we know that the diagonal of a parallelogram bisect each other

$AC = 2AO = 2 \times 3 = 6$
$AC = 6 cm$
$DB = 2DO = 2 \times 2 = 4$
$DB = 4 cm$

Hence the lengths of AC and BD is 6 cm and 4 cm respectively.

Question:2

Diagonals of a parallelogram are perpendicular to each other. Is this statement True or False and explain your answer.

A parallelogram is a special type of quadrilateral. It has equal and parallel opposite sides and equal opposite angles.
Diagonals of a parallelogram bisect each other but don’t bisect in such a way that the angles formed between the diagonals is 90 degrees. In order for the diagonals to bisect perpendicular to each other all the sides should be equal in length which is not true when it comes to parallelograms.
Hence the given statement is False.

Question:3

Can the angles $110^{\circ}, 80^{\circ}, 70^{\circ}$ and $95^{\circ}$ be the angles of a quadrilateral? Why or why Not? Explain

As we know that the sum of the angles of a quadrilateral, is $360^{\circ}$
Here
$110^{\circ} + 80^{\circ} + 70^{\circ} + 95^{\circ}= 355^{\circ}\neq 360^{\circ}$
These angles cannot be the angles of a quadrilateral because the sum of these angles is not equal to 3600.

Question:4

In quadrilateral ABCD, $\angle A + \angle D = 180^{\circ}$ .What special name can be given to this quadrilateral?

Let the given quadrilateral ABCD is,

It is given that $\angle A + \angle D = 180^{\circ}$
We can see the sum of co-interior angles is $180^{\circ}$ which is a property of trapezium.
Therefore the given quadrilateral ABCD is a trapezium.

Question:5

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

We know that sum of all angles of a quadrilateral is $360^{\circ}$
$\angle P+\angle Q+\angle R+\angle S=360^{\circ}$
…..(1)

It is given that all the angels are equal
i.e., $\angle P=\angle Q=\angle R=\angle S$
Put in equation 1 we get
$4\angle P=360^{\circ}$
$\angle P=\frac{360^{\circ}}{4}$
$P = 90^{\circ}$
Here all the angles of a quadrilateral are $90^{\circ}$.
Hence given quadrilateral is either a rectangle or square. We cannot say which one it is, as nothing is given about the length of the sides.

Question:6

Diagonals of a rectangle are equal and perpendicular. Is this statement True or False? Give reason for your answer.

Given that diagonals of a rectangle are equal and perpendicular.
Rectangle: A rectangle an equiangular quadrilateral, and all of its angles are equal.
Hence diagonals of a rectangle are equal but not necessary perpendicular to each other.
Let us consider a rectangle ABCD

Consider
$\triangle ACD$ and $\triangle BCD$
AC = BD (opposite sides of a rectangle are equal)
$\angle C=\angle D$ $(90^{\circ})$
AB = CD
(opposite sides of a rectangle are equal)
$\triangle ACD\cong \triangle BCD$ (SAS congruency)
Hence diagonals are equal.
Also, $\angle CAD=\angle DBC$ …(i)
Similarly, we can prove that $\triangle ACB$ and
$\triangle BDA$are congruent
Hence, $\angle ACB=\angle ADB$ …(ii)
Now, consider
$\triangle AOC$ and $\triangle BOD$
$\angle CAD=\angle DBC$From (i)
$\angle ACB=\angle ADB$ From (ii)
$\angle AOC=\angle BOD$ vertically opposite angles
$\triangle AOC$ and $\triangle BOD$
are also congruent.

But we cannot prove that $\angle AOC=\angle BOD=90^{\circ}$
Hence it is not necessary that diagonals will bisect each other at right angle, so they are not necessarily perpendicular to each other
Hence the given statement is False.

Question:7

An obtuse angle is an angle greater than $90^{\circ}$
Let all angles of a quadrilateral be obtuse angles. Now, the smallest obtuse angle will be equal to $91^{\circ}$
Hence the sum of all angles $= 91^{\circ} + 91^{\circ} + 91^{\circ} + 91^{\circ}= 364^{\circ}$
So we can conclude that, if all the angles are greater than 90o the sum of all the angles will be greater than $360^{\circ}$
But we know that sum of all the angles in a quadrilateral is $360^{\circ}$.
Therefore, a quadrilateral can have a maximum of three obtuse angles.
Hence given statement is False.

Question:10

Given: ABCD and AEFG are two parallelograms

Given $\angle C = 55^{\circ}$
then $\angle A$ is also $55^{\circ}$. {opposite angles of a parallelogram are equal}
Also AEFG is a parallelogram then
$\angle A=\angle F=55^{\circ}$
{opposite angels of a parallelogram is equal}

Hence, $\angle F=55^{\circ}$is the required answer.

Question:11

Solution.
Acute angle is an angle less than $90^{\circ}$
Let all angles of a quadrilateral be acute angles. Now, the smallest acute angle will be equal to $89^{\circ}$
Hence sum of all angles $= 89^{\circ} + 89^{\circ} + 89^{\circ} + 89^{\circ} = 356^{\circ}$
So we can conclude that, if all the angles are less than 90o then the sum of all the angles will be less than $360^{\circ}$
But we know that sum of all the angles in a quadrilateral is $360^{\circ}$.
Therefore, a quadrilateral can have a maximum of three acute angles.
Hence given statement is False.

Question:12

Right angle is an angle equal to $90^{\circ}$
Let all angles of a quadrilateral be right angles.
Hence sum of all angles $= 90^{\circ} + 90^{\circ} + 90^{\circ} + 90^{\circ} = 360^{\circ}$
So we can conclude that, if all the angles are equal to $90^{\circ}$ then the sum of all the angles will be equal to $360^{\circ}$
Also, we know that sum of all the angles in a quadrilateral is $360^{\circ}$.
Therefore, a quadrilateral can have all angles as right angles.
Hence given statement is True.

Question:13

Diagonals of a quadrilateral ABCD bisect each other. If $\angle A = 35^{\circ}$, determine $\angle B$.

If the diagonals of a quadrilateral ABCD bisect each other then it parallelogram.

Sum of interior angles between two parallel lines is 1800 i.e.,
$\angle A+\angle B=180^{\circ}$
$\angle A=35^{\circ}$ (given)
$35^{\circ}+\angle B=180^{\circ}$
$\angle B=180^{\circ}-35^{\circ}$
$\angle B=145^{\circ}$

Question:14

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

Given: Opposite angles of a quadrilateral ABCD are equal and AB = 4
So ABCD is a parallelogram as we know that in a parallelogram opposite angles are equal

Also, we know that opposite sides in a parallelogram are equal
$AB = CD = 4 cm$
Hence $CD = 4cm$

Exercise 8.3

Question:1

One of the angle of a quadrilateral is of $108^{\circ}$ and the remaining three angles are equal. Find each of the three equal angles.

Answer: $84^{\circ}$
Solution.
Given that:
One of the angle of a quadrilateral is of $108^{\circ}$ and the remaining three angles are equal.
Let each of the three equal angles be $x ^{\circ}$

As we know that the sum of angles of a quadrilateral is $360 ^{\circ}$
$108^{\circ} + x^{\circ} + x^{\circ} + x^{\circ} = 360^{\circ}$
$3x^{\circ} + 108^{\circ}= 360^{\circ}$
$3x^{\circ} = 360^{\circ} %u2013 108^{\circ}$
$3x^{\circ}= 252^{\circ}$
$x=\frac{252^{\circ}}{3}$
$x=84^{\circ}$
Hence, each of the three equal angles is $84^{\circ}$.

Question:2

ABCD is a trapezium in which $AB \parallel DC$ and $\angle A = \angle B = 45^{\circ}$. Find angles C and D of the trapezium.

Answer: $\angle D=\angle C=135^{\circ}$
Solution.
Given: $AB\parallel DC$ and $\angle A=\angle E=45^{\circ}$

If $AB\parallel CD$ and BC is transversal then sum of co-interior angles is $180^{\circ}$
$\therefore \angle B+\angle C=180^{\circ}$
$45^{\circ}+\angle C=180^{\circ}$
$\angle C=180^{\circ}-45^{\circ}$
$\angle C=135^{\circ}$
Similarly
$\angle A+\angle D=180^{\circ}$
$45^{\circ}+\angle D=180^{\circ}$
$\angle D=180^{\circ}-45^{\circ}$
$\angle D=135^{\circ}$
Hence angles C and D are $135^{\circ}$ each.

Question:

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is $60^{\circ}$. Find the angles of the parallelogram.

Answer: $60^{\circ},120^{\circ}$
Solution.
Given that the angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is $60^{\circ}$.

In this figure
$\angle ADC$ and $\angle ABC$ are obtuse angles and DE and DF are altitudes.
$\angle BED$= $\angle BFD$ = $90^{\circ}$
$\angle FBE=360^{\circ}-(\angle FDE+\angle BED+\angle BFD)$
$=360^{\circ}-(60^{\circ}+90^{\circ}+90^{\circ})$

$=360^{\circ}-240^{\circ}$
$=120^{\circ}$
It is given that ABCD is a parallelogram
ADC = $120^{\circ}$
$\angle A+\angle B=180^{\circ}${Sum of interior angle $=180^{\circ}$}

$\angle A=180^{\circ}-B$
$\angle A=180^{\circ}-120^{\circ}$
$\angle A=60^{\circ}$
$\Rightarrow \angle C=\angle A=60^{\circ}$
Hence angles of the parallelogram are $60^{\circ}, 120^{\circ}$

Question:4

ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus.

Answer: $120^{\circ},120^{\circ},60^{\circ},60^{\circ}$
Solution.
Let sides of a rhombus be AB = BC = CD = DA = x
Join DB

Here $\angle DLA=\angle DLB=90^{\circ}$
{DL is perpendicular bisector of AB}

$AL=BL=\frac{X}{2}$
DL = DL {common side of $\triangle ADL$ and $\triangle BDL$}
$\triangle ALD\cong \triangle BLD${by SAS congruence}
$AD = BD$

In $\triangle ADB$, $AD = AB = DB = x$
$\triangle ADB$ is an equilateral triangle.
$\therefore \angle ADB=\angle ABD=\angle A=60^{\circ}$
Similarly, $\triangle DCB$ is an equilateral triangle
$\angle BDC=\angle DBC=\angle C=60^{\circ}$
Also,
$\angle A = \angle C$
$\angle D = \angle B$…..(1)

$\angle A+ \angle B+\angle C+\angle D=360^{\circ}$
$60^{\circ}+ B+60^{\circ}+B=360^{\circ}$
$2\angle B=\frac{240}{2}=120^{\circ}$
Hence answer is $120^{\circ},120^{\circ},60^{\circ},60^{\circ}$

Question:5

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram.

Given : ABCD is a parallelogram and AE = CF
To prove: OE = OF

Proof: Join BD and CA
Here $OA = OC$ and $OD = OB$ [ABCD is parallelogram]
$\Rightarrow$$OA = OC$…..(1)
And $AE = CF$…..(2) {given}
$OA -AE = OC - CF${by subtracting 2 from 1}
$OE = OF$
Hence, BFDE is a parallelogram.

Question:6

E is the mid-point of the side AD of the trapezium ABCD with $AB \parallel DC$. A line through E drawn parallel to AB intersect BC at F. Show that F is the mid-point of BC. [Hint: Join AC]

Given: ABCD is trapezium and $AB\parallel DC$
Constructions: Join AC which intersect EF at O

Proof: In $\triangle ADC$, E is the mid-point of line AD and $OE \parallel CD$.
By mid-point theorem, O is mid-point of AC
In $\triangle CBA$, as O is mid-point of AC
$OF\parallel AB$ (mid-point theorem)
Hence again by mid-point theorem, F is the mid-point of BC.
Hence proved

Question:7

Given : In $\triangle ABC$, $PQ\parallel AB$and $PR\parallel AC$ and $PQ\parallel BC$.
To Prove :
$BC=\frac{1}{2}QR$
Proof : In quadrilateral BCAR, $BR\parallel CA$ and $BC\parallel RA$

Hence it is a parallelogram
BC = AR ……(1)
In quadrilateral BCQA,$BC\parallel AQ$ and $AB\parallel QC$
Hence it is also a parallelogram
BC = AQ …..(2)|
Adding equations 1 and 2, we get
$2BC = AR + AQ$
$2BC = RQ$
$BC=\frac{1}{2}QR$
Hence proved

Question:8

D, E and F are the mid-points of the sides BC, CA and AB, respectively of an equilateral triangle ABC. Show that $\triangle DEF$ is also an equilateral triangle.

D and E are midpoints of BC and AC respectively, so using mid-point theorem:
$\Rightarrow DE=\frac{1}{2}AB$ …..(1)
E and F are midpoints of AC and AB, so using mid-point theorem:
$\Rightarrow EF=\frac{1}{2}BC$ …..(2)
F and D are midpoints of AB and BC, so using mid-point theorem:
$\Rightarrow FD=\frac{1}{2}AC$ …..(3)
It is given that $\triangle ABC$ is an equilateral triangle
$\Rightarrow AB=BC=CA$
$\frac{1}{2}AB=\frac{1}{2}BC=\frac{1}{2}CA$ {dividing by 2}

Using 1, 2 and 3 we get
DE = EF = FD
Hence DEF is an equilateral triangle.
Hence proved

Question:9

Given: ABCD is a parallelogram and AP = CQ
To Prove: AC and PQ bisect each other.
Proof :

Here ABCD is a parallelogram

$\Rightarrow AB\parallel DC$
$\Rightarrow AP\parallel QC$
It is given that $AP = CQ$
Thus APCQ is a parallelogram.
And we know that diagonals of a parallelogram bisect each other.
Hence AC and PQ bisect each other.
Hence proved

Question:10

Given : ABCD is a parallelogram, P is a mid-point of BC such that $\angle BAP = \angle DAP$.
To prove : $AD = 2CD$
Proof : Here ABCD is a parallelogram

$\therefore AD\parallel BC$ and AB is transversal
$\angle A+\angle B=180^{\circ}$ {sum of co-interior angles}

$\angle B=180-\angle A$
…..(1)
In
$\triangle ABP$,
$\angle PAB+\angle BPA+\angle B=180^{\circ}${using angle sum property of triangle}
Putting the values,
$\frac{1}{2}\angle A+\angle BPA+(180^{\circ}-\angle A)=180^{\circ}$
$\angle BPA-\frac{\angle A}{2}=0$
$\angle BPA=\frac{\angle A}{2}$ …..(2)
$\Rightarrow \angle BPA=\angle BAP$

$AB = BP$ {opposite sides of equal triangle}
$2AB = 2BP$ {multiply both sides by 2}
$2AB = BC$ { P is midpoint of BC}
$2CD = AD$
{ ABCD is parallelogram $AB = CD$ and $BC = AD$}

Hence proved

Exercise 8.4

Question:1

A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

Given : Here ABC is an isosceles triangle and ADEF is a square inscribed in $\triangle ABC$.
To prove : $CE = BE$

Proof: In isosceles $\triangle ABC$ and $AB = AC$ …..(1)
$\angle A=90^{\circ}$

$AD = AF$ …..(2)
Subtract equation 2 from 1
$AB - AD = AC - AF$
$BD = CF$ …..(3)
Now in
$\triangle BDE$and $\triangle CFE$
$BD = CF${from equation 3}
$\angle CFE = \angle EDB$ {$90^{\circ}$ each}
$DE = EF$ {side of a square}
$\triangle CFE\cong \triangle BDE$ {SAS congruence rule}

$CE = BE$ {by CPCT}

Hence vertex E of the square bisects the hypotenuse BC.
Hence proved

Question:2

In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of $\angle A$ meets DC in E. AE and BC produced meet at F. Find the length of CF.

Given: ABCD is a parallelogram in which $AB = 10 cm$ and $AD = 6 cm$.
Construction: In parallelogram ABCD, draw the bisector of $\angle A$ which meets DC in point E.
Produce AE and BC so that they meet at point F.
Also, produce AD to H and join H and F

Here ABFH is a parallelogram
$HF\parallel AB$
And $\angle AFH = \angle FAB$ …..(1) {alternate interior angles}

$AB = HF$ (opposite sides of a parallelogram)
$AF = AF$ (Common Side)
$\Rightarrow \triangle HAF\cong \triangle FAB$…..(2) { \ SAS Congruency}
Now, $\angle HAF=\angle FAB$ …..(3)
(EA is the bisector of $\angle A$)
$\Rightarrow \angle AFH=\angle HAF$ { using 1 and 3 }

So, $HF = AH$ {Sides opposite to equal angles are equal}
$HF = AB = 10 cm$

$AH = HF = 10 cm$
$AD + DH = 10 cm$
$DH = 10 -AD \Rightarrow 10 -6$
$DH=4cm$
Because FHDC is a parallelogram
opposite sides are equal
$DH = CF = 4 cm$
Hence the answer is $4 cm$

Question:3

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.

Given : ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA.

To prove : PQRS is a rhombus.
Proof :
To $\triangle ADC$, S and R are the mid-points of AD and DC respectively.
By mid point theorem we get
$SR\parallel AC$ and $SR=\frac{1}{2}AC$
…..(1)
In
$\triangle ABC$, P and Q are the mid-points of AB and BC respectively.
Then by mid-point theorem we get
$PQ\parallel AC$ and $PQ=\frac{1}{2}AC$
…..(2)
From equation 1 and 2 we get
$SR=PQ=\frac{1}{2}AC$ …..(3)
Similarly in
$\triangle BCD$we get
$RQ\parallel BD$ and $RQ=\frac{1}{2}BD$ …..(4)
And in
$\triangle BAD$
$SP\parallel BD$and $SP=\frac{1}{2}BD$ …..(5)
Using equation 4 and 5 we get
$RQ=SP=\frac{1}{2}BD$
It is given that AC = BD

$RQ=SP=\frac{1}{2}AC$......(6)
Now equate equation 3 and 6 we get
$SR = PQ = SP = RQ$

It shows that all sides of a quadrilateral PQRS are equal.
Hence PQRS is a rhombus.
Hence Proved.

Question:4

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that $AC \perp BD$. Prove that PQRS is a rectangle.

Given: ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA and $AC \perp BD$. To prove: PQRS is a rectangle
It is given that $AC \perp BD$

$\angle COD =\angle AOD = \angle AOB = \angle COB = 90^{\circ}$

In $\triangle ADC$, S and R are the mid-points of AD and DC so by mid-point theorem.
$SR \parallel AC$ and $SR=\frac{1}{2}AC$
…..(1)
Similarly in
$\triangle ABC$,
$PQ \parallel AC$ & $PQ =\frac{1}{2}AC$ …..(2)
Using 1 and 2
$SR=PQ=\frac{1}{2}AC$ …..(3)
Similarly, $SP\parallel RQ$ & $SP=RQ=\frac{1}{2}BD$ ……(4)
$OE \parallel FR, OF \parallel ER$
$\therefore \angle EOF = \angle ERF = 90^{\circ}$
{$\angle COD = 90^{\circ}$ because $AC \perp BD$ and opposite angles of a parallelogram are equal}
$\angle SRQ = \angle ERF = \angle SPQ = 90^{\circ}$
{Opposite angles in a parallelogram are equal}
$\angle RSP + \angle SRQ + \angle RQP + \angle QRS = 360^{\circ}$
$90^{\circ} + 90^{\circ} + \angle RSP + \angle RQP = 360^{\circ}$
$\angle RSP + \angle RQP = 180^{\circ}$
$\angle RSP =\angle RQP = 90^{\circ}$

If all the angles in parallelogram are $90^{\circ}$ then that parallelogram is a rectangle.
So, PQRS is a rectangle.
Hence Proved

Question:5

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and $AC \perp BD$. Prove that PQRS is a square.

Given: ABCD is a parallelogram and P, Q, R and S are the mid-points of sides AB, BC, CD and AD. Also AC = BD and $AC \perp BD$.
To prove: PQRS is a square

Proof: In $\triangle ADC$, S and R are the mid-points of sides AD and DC by mid-point theorem.
$SR \parallel AC$ and $SR =\frac{1}{2}AC$
…..(1)
In
$\triangle ABC$, P and Q are the mid-points of AB and BC respectively. Therefore, by mid-point theorem
$PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ …..(2)
From equation 1 and 2 we get
$PQ\parallel SR$ and $PQ=\frac{1}{2}AC$
…..(3)
Similarly, in
$\triangle BCD$, $RQ\parallel BD$ and R, Q are midpoints of CD, CB respectively, Therefore, by mid-point theorem
$RQ=\frac{1}{2}BD=\frac{1}{2}AC$ {Given BD = AC} …..(4)
And in $\triangle ABD$, $SP\parallel BD$ and S, P are midpoints of AD, AB respectively. Therefore, by mid-point theorem

$SP=\frac{1}{2}BD=\frac{1}{2}AC${Given AC = BD …..(5)
From equation 4 and 5 we get
$SP=RQ=\frac{1}{2}AC$ …..(6)
From equation 3 and 6 we get

$PQ = SR = SP = RQ$
Thus, all sides are equal
$OE \parallel FR, OF \parallel ER$
$\therefore \angle EOF =\angle ERF = 90^{\circ}$
{$\angle COD = 90^{\circ}$
because
$AC \perp BD$ and opposite angles of a parallelogram are equal}

$\angle SRQ = \angle ERF = \angle SPQ = 90^{\circ}$ {Opposite angles in a parallelogram are equal}
$\angle RSP + \angle SRQ + \angle RQP + \angle QRS = 360^{\circ}$
$90^{\circ} + 90^{\circ}+ \angle RSP +\angle RQP = 360^{\circ}$
$\angle RSP +\angle RQP = 180^{\circ}$
$\angle RSP = \angle RQP = 90^{\circ}$
All the sides are equal and all the interior angles of the quadrilateral are $90^{\circ}$

Hence, PQRS is a square.
Hence Proved

Question:6

If a diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

Let ABCD is a parallelogram and diagonal AC bisect the angle A and $\angle CAB = \angle CAD$

To Prove: ABCD is a rhombus
Proof: Here it is given that ABCD is a parallelogram.
$AB\parallel CD$ and AC is transversal
$\angle CAB=\angle ACD${alternate interior angles} …. (1)

Also $AD\parallel BC$ and AC is a transversal
$\angle CAD=\angle ACB${alternate interior angle} …. (2)
Now it is given that $\angle CAB = \angle CAD$

So from equations (1) and (2)
$\Rightarrow \angle ACD=\angle ACB$…. (3)
Also, $\angle A=\angle C${opposite angles of a parallelogram}
$\frac{\angle A}{2}=\frac{\angle C}{2}$ {dividing by 2}

$\angle DAC=\angle DCA${from 1 and 2}
$CD=AD$
{sides opposite of equal angles in $\triangle ADC$are equal}
$AB=CD$ & $AD=BC$
{$\because$ opposite sides of a parallelogram}

Hence,
$AB = BC = CD = AD$
Thus all sides are equal so ABCD is a rhombus.
Hence Proved.

Question:7

P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PRQS is a parallelogram.

Given: ABCD is a parallelogram, P and Q are the mid-points of AB and CD
To Prove: PRQS is a parallelogram.
Proof: Since ABCD is a parallelogram $AB\parallel CD \Rightarrow AP\parallel QC$

Also, AB = DC …..(1)

$\frac{1}{2}AB=\frac{1}{2}DC${dividing by 2}

$AP = QC$
{$\because$P and Q are the mid-points of AB and DC}
As,
$AP = QC$ and $AP\parallel QC$
Thus APCQ is a parallelogram

$AQ\parallel PC$ or $SQ\parallel PR$ …..(2)
$AB\parallel CD$ or $BP\parallel DQ$
$AB=DC$

$\frac{1}{2}AB=\frac{1}{2}DC${dividing both sides by 2}
$BP=QD$
{$\because$P and Q are the mid-points of AB and DC}
$BP\parallel QD$ and $BQ\parallel PD$

So, BPDQ is a parallelogram
$PD\parallel BQ$ or $PS\parallel QR$ …..(3)
From equation 2 and 3
$SQ\parallel RP$and $PS\parallel QR$
So, PRQS is a parallelogram.
Hence Proved

Question:9

Given: $AB \parallel DE$ and $AC \parallel DF$
Also, AB = DE and AC = DF
To prove: $BC\parallel EF$and BC = EF

Proof: $AB \parallel DE$ and AB = DE
$AC \parallel DF$ and AC = DF

$AC\parallel FD$ and AC = FD …..(1)
Thus ACFD is a parallelogram
$AD\parallel CF$ and AD = CF
…..(2)
$AB\parallel DE$ and AB = DE
…..(3)
Thus ABED is a parallelogram
$AD\parallel BE$ and AD = BE
…..(4)

From equation 2 and 4
AD = BE = CF and $AD \parallel CF \parallel BE$ …..(5)
In quadrilateral BCFE, BE = CF and $BE\parallel CF$ {from equation 5}
So, BCFE is a parallelogram BC = EF and $BC\parallel EF$

Hence Proved.

Question:10

E is the mid-point of a median AD of $\triangle ABC$ and BE is produced to meet AC at F. Show that $AF=\frac{1}{3}AC$ .

Given: In $\triangle ABC$, AD is a median and E is the mid-point of AD
To Prove:
$AF=\frac{1}{3}AC$.
Construction: Draw $DP\parallel EF$ and $\triangle ABC$ as given

Proof: In
$\triangle ADP$, E is mid-point of AD and $EF\parallel DB$

So, F is mid-point of AP {converse of midpoint theorem}
In $\triangle FBC$, D is mid-point of BC and $DP\parallel BF$

So, P is mid-point of FC {converse of midpoint theorem}
Thus, AF = FP = PC
$AF+FP+PC=AC=3AF$
$\therefore AF=\frac{1}{3}AC$
Hence Proved

Question:11

Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.
Hint: Prove all the sides are equal and diagonals are equal

Given: In a square ABCD; P, Q, R and S are the mid-points of AB, BC, CD and DA.
To Prove: PQRS is a square
Construction: Join AC and BD

Proof : Here ABCD is a square
AB = BC = CD = AD
P, Q, R, S are the mid-points of AB, BC, CD and DA.
In
$\triangle ADC$,
$SR\parallel AC$

$SR=\frac{1}{2}AC${using mid-point theorem} …..(1)
In
$\triangle ABC$,
$PQ\parallel AC$

$PQ=\frac{1}{2}AC${using mid-point theorem} …..(2)
From equation 1 and 2
$SR\parallel PQ$ and $SR= PQ=\frac{1}{2}AC$
…..(3)
Similarly, $SP\parallel BD$ and $BD\parallel RQ$

$SP=\frac{1}{2}BD$and $RQ=\frac{1}{2}BD$ {using mid-point theorem}
$SP=RQ=\frac{1}{2}BD$
Since diagonals of a square bisect each other at right angles.
AC = BD
$\Rightarrow SP=RQ=\frac{1}{2}AC$…..(4)
From equation 3 and 4
SP = PQ = SP = RQ
{All sides are equal}
$OE\parallel FR$ and $OF\parallel ER$

$\angle EOF=\angle ERF=90^{\circ}$
$\angle SRQ=\angle ERF=\angle SPQ=90^{\circ}$
{Opposite angles in a parallelogram are equal}

$\angle RSP + \angle SRQ + \angle RQP + \angle QRS = 360^{\circ}$
$90^{\circ} +90^{\circ}+ \angle RSP + \angle RQP = 360^{\circ}$
$\angle RSP + \angle RQP = 180^{\circ}$
$\angle RSP = \angle RQP = 90^{\circ}$
Since all the sides are equal and angles are also equal, so PQRS is a square.
Hence Proved

Question:12

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that $EF \parallel AB$ and EF $EF=\frac{1}{2}(AB+CD)$

Given: ABCD is a trapezium in which $AB\parallel CD$, E and F are the mid-points or sides AD and BC.

Constructions: Joint BE and produce it to meet CD at G.
Draw BOD which intersects EF at O
To Prove: $EF \parallel AB$ and $EF=\frac{1}{2}(AB+CD)$

Proof: In $\triangle GCB$, E and F are respectively the mid-points of BG and BC, then by mid-point theorem.
$EF \parallel GC$
But, $GC\parallel AB$ or $CD \parallel AB${given}
$\therefore EF\parallel AB$
In
$\triangle ADB$, $AB \parallel EO$ and E is the mid-point of AD. Then by mid-point theorem, O is mid-point of BD.
$EO=\frac{1}{2}AB$ ……(1)
In
$\triangle BDC$, $OF \parallel CD$and O is the mid-point of BD
$OF=\frac{1}{2}CD$ …..(2)
Adding 1 and 2, we get
$EO+OF=\frac{1}{2}AB+\frac{1}{2}CD$
$EF=\frac{1}{2}(AB+CD)$

Hence Proved

Question:13

Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.

Given: Let ABCD be a parallelogram and AP, BR, CQ, DS are the bisectors of

$\angle A, \angle B, \angle C$and $\angle D$ respectively.
To Prove: Quadrilateral PQRS is a rectangle.

Proof: Since ABCD is a parallelogram
Then $DC\parallel AB$ and DA is transversal.
$\angle A+\angle B=180^{\circ}${sum of co-interior angles of a parallelogram}

$\frac{1}{2}\angle A+\frac{1}{2}\angle B=90^{\circ}${Dividing both sides by 2}
$\angle PAD+\angle PDA=90^{\circ}$
$\Rightarrow \angle APD=90^{\circ}$ {$\because$ sum of all the angles of a triangle is 1800}
$\angle QRS=90^{\circ}$
Similarly,
$\angle RBC+\angle RCB=90^{\circ}$
$\Rightarrow \angle BRC=90^{\circ}$
$\angle QRS=90^{\circ}$
Similarly,
$\angle QAB+\angle QBA=90^{\circ}$
$\Rightarrow \angle AQB=90^{\circ}${$\because$ sum of all the angles of a triangle is 1800}
$\angle RQP=90^{\circ}${$\because$ vertically opposite angles}
Similarly,|
$\angle SDC+\angle SCD=90^{\circ}$
$\Rightarrow \angle DSC=90^{\circ}${$\because$ sum of all the angles of a triangle is 1800}
$\angle RSP=90^{\circ}${$\because$ vertically opposite angles}
Thus PQRS is a quadrilateral whose all angles are $90^{\circ}$
Hence PQRS is a rectangle.
Hence proved

Question:14

P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.

Given: ABCD is a parallelogram whose diagonals bisect each other at O.
To Prove: PQ is bisected at O.

Proof: In $\triangle ODP$ and $\triangle OBQ$
$\angle BOQ=\angle POD${Vertically opposite angles}
$\angle OBQ=\angle ODP${interior angles}
OB = OD {given}
$\triangle ODP\cong \triangle OBQ${by ASA congruence}
OP = OQ {by CPCT rule}
So, PQ is bisected at O

Question:15

ABCD is a rectangle in which diagonal BD bisects $\angle B$. Show that ABCD is a square.

Given: In a rectangle ABCD, diagonal BD bisects B
To Prove: ABCD is a square
Construction: Join AC

Proof:
Given that ABCD is a rectangle. So all angles are equal to $90^{\circ}$
Now, BD bisects $\angle B$
$\angle DBA=\angle CBD$

Also,
$\angle DBA+\angle CBD=90^{\circ}$

So,
$2\angle DBA=90^{\circ}$

$\angle DBA=45^{\circ}$
In $\triangle ABD$,
$\angle ABD+\angle BDA+\angle DAB=180^{\circ}$
(Angle sum property)
$45^{\circ}+\angle BDA+90^{\circ}=180^{\circ}$
$\angle BDA=45^{\circ}$
In $\triangle ABD$,
AD = AB (sides opposite to equal angles in a triangle are equal)
Similarly, we can prove that BC = CD
So, AB = BC = CD = DA
So ABCD is a square.
Hence proved

Question:16

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. Prove that by joining these mid-points D, E and F, the triangles ABC is divided into four congruent triangles.

Given: In $\triangle ABC$, D, E and F are respectively the mid-points of the sides AB, BC and CA.
To prove: $\triangle ABC$ is divided into four congruent triangles.

Proof: Using given conditions we have
$AD=BD=\frac{1}{2}AB,BE=EC=\frac{1}{2}BC,AF=CF=\frac{1}{2}AC$
Using mid-point theorem
$EF\parallel AB$ and $EF=\frac{1}{2}AB=AD=BD$
$ED\parallel AC$ and $ED=\frac{1}{2}AC=AF=CF$
$DF\parallel BC$ and $DF=\frac{1}{2}BC=BE=EC$

In $\triangle ADF$ and $\triangle EFD$
AF = DE
DF = FD {common side}
$\triangle ADF\cong \triangle EFD$
{by SSS congruence}
Similarly we can prove that,
$\triangle DEF\cong \triangle EDB$
$\triangle DEF\cong \triangle CFE$

So, $\triangle ABC$ is divided into four congruent triangles
Hence proved

Question:17

Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.

Given: Let ABCD be a trapezium in which $AB\parallel DC$ and let M and N be the mid-points of diagonals AC and BD.

To Prove: $MN\parallel AB\parallel CD$
Proof: Join CN and produce it to meet AB at E
In $\triangle CDN$ and $\triangle EBN$we have
DN = BN {N is mid-point of BD}
$\angle DCN=\angle BEN${alternate interior angle}
$\angle CDN=\angle EBN${alternate interior angles}
$\triangle CDN\cong \triangle EBN$
DC = EB and CN = NE {by CPCT}
Thus in $\triangle CAE$, the points M and N are the mid-points of AC and CE, respectively.
$MN\parallel AE$ {By mid-point theorem}
$MN\parallel AB\parallel CD$

Hence Proved

Question:18

P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and CQ = QR.

Given: In a parallelogram ABCD, P is the mid-point of DC
To Prove: DA = AR and CQ = QR

Proof: ABCD is a parallelogram
BC = AD and $BC\parallel AD$
Also,
DC = AB and $DC\parallel AB$

P is mid-point of DC
$DP=PC=\frac{1}{2}DC$
Now $QC\parallel AP$ and $PC\parallel AQ$
So APCQ is a parallelogram
$AQ=PC=\frac{1}{2}DC$
$\frac{1}{2}AB=BQ$…..(1) {$\because$ DC = AB}
In $\triangle AQR$ & $\triangle BQC$AQ = BQ {from equation 1}
$\angle AQR=\angle BQR${vertically opposite angles}
$\angle ARQ=\angle BCQ${alternate angles of transversal}
$\triangle AQR\cong \triangle BQC${AAS congruence}
AR = BC {by CPCT}
BC = DA
AR = DA Also, CQ = QR {by CPCT} Hence Proved .

Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 8:

The chapter on Quadrilaterals in NCERT exemplar Class 9 Maths solutions chapter 8 covers the below-mentioned topics:

◊ Properties of angles within the quadrilateral and to prove that sum of all angles will be 360°.

◊ Sum of angles of any polygon of side length more than three.

◊ Different types of quadrilaterals such as rectangles, squares, parallelogram, trapezium, and rhombus.

◊ Condition for which any quadrilateral to be parallelogram trapezium rhombus et cetera.

◊ NCERT exemplar Class 9 Maths chapter 8 solutions discuss the midpoint theorem of the parallelogram, which will help us solve many geometry problems related to parallelograms.

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

 Chapter 1 Number System Chapter 2 Polynomials Chapter 3 Coordinate geometry Chapter 4 Linear equations in Two Variable Chapter 5 Introduction to Euclid’s Geometry Chapter 6 Lines and Angles Chapter 7 Triangles Chapter 9 Area of Parallelograms and Triangles Chapter 10 Circles Chapter 11 Constructions Chapter 12 Heron’s Formula Chapter 13 Surface Areas and Volumes Chapter 14 Statistics and Probability

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 8:

These Class 9 Maths NCERT exemplar chapter 8 solutions give basic ideas of quadrilaterals and it is useful in higher classes. Students can use the Class 9 Maths NCERT exemplar solutions chapter 8 Quadrilaterals in the form of reference content to practice a variety of problems based on quadrilaterals. The detailed solutions are adequate for students to build a strong concept base and attempt books such as NCERT Class 9 Maths, RD Sharma Class 9 Maths, RS Aggarwal Class 9 Maths etcetera.

Check the Solutions of Questions Given in the Book

 Chapter No. Chapter Name Chapter 1 Number Systems Chapter 2 Polynomials Chapter 3 Coordinate Geometry Chapter 4 Linear Equations In Two Variables Chapter 5 Introduction to Euclid's Geometry Chapter 6 Lines And Angles Chapter 7 Triangles Chapter 8 Quadrilaterals Chapter 9 Areas of Parallelograms and Triangles Chapter 10 Circles Chapter 11 Constructions Chapter 12 Heron’s Formula Chapter 13 Surface Area and Volumes Chapter 14 Statistics Chapter 15 Probability

Also Check NCERT Books and NCERT Syllabus here

1. Is it necessary that rhombus will be a parallelogram?

Yes, the rhombus is a special kind of parallelogram which diagonally intersects each other perpendicularly. All sides of the rhombus are equal, and the square is a special case of a rhombus.

2. Is Diamond in playing cards a parallelogram?

The diamond in the playing card is the rhombus and sometimes rhombus is also called a diamond. We know that rhombus is a special case of parallelogram therefore diamond shape will be a parallelogram.

3. What is the sum of interior angles of any polygon of n sides?

For any polygon of n sides, the sum of interior angle will be equal (n -2 )180°. For triangle n is equal to 3 hence the sum of interior angles will be 180°. For quadrilaterals n is equal to 4 hence the sum of interior angle will be 360°

4. Are these solutions of the chapter Quadrilaterals available in an offline mode?

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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9