NCERT Exemplar Class 9 Maths Solutions Chapter 8 Quadrilaterals

Question:1
If three angles of a quadrilateral are $75^{\circ}, 90^{\circ}$ and $75^{\circ}$ then find the fourth angle
(A) 90^o (B) 95^o (C) 105^o (D) 120^o

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 $\angle 1=75^{\circ}$
$\angle 2=90^{\circ}$
$\angle 3=75^{\circ}$
Putting 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
A diagonal of a rectangle is inclined to one side of the rectangle at $25 ^{\circ}$. The acute angle between the diagonals is
(A) 55⁰ (B) 50⁰ (C) 40⁰ (D) 25⁰

Answer: [B] 50^o
As we know, 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

ABCD is a rhombus such that $\angle ACB = 40^{\circ}$. Then $\angle ADB$ is
(A) 40⁰ (B) 45⁰ (C) 50⁰ (D) 60⁰

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 the 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 midpoints 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

Answer:

According to the 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.

Answer:

According to the 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 a 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 a rhombus, all sides 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 the ABCD quadrilateral here, all sides are equal, and the angles of quadrilateral ABCD are not right angles, 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

Answer: [C]
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 the given angles.

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

Question:7

If bisectors of $\angle A$ and $\angle B$of a quadrilateral ABCD intersect each other at P, of $\angle B$ and $\angle C$ at Q, of $\angle C$ and $\angle D$ at R and of $\angle D$and $\angle A$ at S, then PQRS is a
(A) rectangle
(B) rhombus
(C) parallelogram
(D) quadrilateral whose opposite angles are supplementary

Answer: [D] quadrilateral whose opposite angles are supplementary
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 equations 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)
Add equations 3 and 4
$\angle QPS+\angle QRS=360^{\circ}-\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)$
$=360^{\circ}-\frac{1}{2}(360^{\circ})$
$=360^{\circ}-180^{\circ}$
$=180^{\circ}$
$\therefore \angle QPS+\angle QRS= 180^{\circ}$
Hence, PQRS is a quadrilateral whose opposite angles are supplementary.

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

Answer: (C) a rectangle
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 points 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)
$\Rightarrow PN\parallel QM$
So, quadrilateral PMQN is parallelogram
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

Answer: (D) Any parallelogram
Let ABCD be a rhombus and P, Q, R, and S be its midpoints.

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 equations 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 midpoints 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

Answer: (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 midpoints 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 equations 3 and 4
$EQ\parallel PD$ and $EQ=PD$
From equations 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.

Answer: [C]

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 the 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 the point 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 a 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 midpoints 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 additional information, which is
(A) $\angle DAE = \angle EFC$
(B) AE = EF
(C) DE = EF
(D) $\angle ADE = \angle ECF.$

Answer: (C) DE = EF
Let us draw the figure according to the 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.

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.

Answer: 6 and 4
Given: ABCD is a parallelogram
$OA = 3 cm$ and $OD = 2 cm$

As we know that the diagonals 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 are 6 cm and 4 cm, respectively.

Question:2

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

Answer: False
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

Answer: No
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?

Answer: Trapezium
Let the given quadrilateral be ABCD.

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 a 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?

Answer: Rectangle or Square
We know that the sum of all angles of a quadrilateral is $360^{\circ}$
If PQRS is a quadrilateral,
$\angle P+\angle Q+\angle R+\angle S=360^{\circ}$ …..(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 a reason for your answer.

Answer: False
Given that diagonals of a rectangle are equal and perpendicular.
Rectangle: A rectangle is an equiangular quadrilateral, and all of its angles are equal.
Hence, diagonals of a rectangle are equal but not necessarily 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)
So, AD = BC
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 a right angle, so they are not necessarily perpendicular to each other
Hence, the given statement is False.

Question:7

Can all four angles of a quadrilateral be obtuse angles? Give a reason for your answer.

Answer: False
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 90^o, the sum of all the angles will be greater than $360^{\circ}$
But we know that the 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:8

$\triangle ABC$, $AB = 5 cm$, $BC=8cm$ and$CA = 7 cm$. If D and E are respectively the midpoints 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 midpoints of AB and BC.
To Find: Length of DE

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

Question:9

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

Answer: True
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 equations 1 and 2, we get,
BD = CD = EF
Hence given statement is True.

Question:10

In Figure, ABCD and AEFG are two parallelograms. If $\angle C = 55^{\circ}$, determine $\angle F$ .

Answer:

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

Can all the angles of a quadrilateral be acute angles? Give a reason for your answer.

Answer: False
Solution.
An 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 90^o, then the sum of all the angles will be less than $360^{\circ}$
But we know that the 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

Can all the angles of a quadrilateral be right angles? Give a reason for your answer.

Answer: Yes
A 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 the 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 $\angle B$.

Answer:

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.

Answer:
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 $CD = 4cm$

Question:1

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

Answer: $84^{\circ}$
Solution.
Given that:
One of the angles of a quadrilateral is $108^{\circ}$, 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 $360 ^{\circ}$
$108^{\circ} + x^{\circ} + x^{\circ} + x^{\circ} = 360^{\circ}$
$3x^{\circ} + 108^{\circ}= 360^{\circ}$
$3x^{\circ} = 360^{\circ} � 108^{\circ}$
$3x^{\circ}= 252^{\circ}$
$x=\frac{252^{\circ}}{3}$
$x=84^{\circ}$
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.
In quadrilateral BEDF, $\angle BED$= $\angle BFD$ = $90^{\circ}$
$\angle FBE=360^{\circ}-(\angle FDE+\angle BED+\angle BFD)$
$=360^{\circ}-(60^{\circ}+90^{\circ}+90^{\circ})$
$=360^{\circ}-240^{\circ}$
$=120^{\circ}$
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 the 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 the diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram.

Answer:

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 intersects BC at F. Show that F is the midpoint of BC. [Hint: Join AC]

Answer:

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

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

Question:7

Through A, B and C, lines RQ, PR and QP have been drawn, respectively, parallel to sides BC, CA and AB of a $\triangle ABC$ as shown in Figure. Show that $BC=\frac{1}{2}QR$

Answer:

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 midpoints of the sides BC, CA and AB, respectively, of an equilateral triangle ABC. Show that $\triangle DEF$ is also an equilateral triangle.

Answer:

D and E are midpoints of BC and AC, respectively, so using the mid-point theorem:
$\Rightarrow DE=\frac{1}{2}AB$ …..(1)
E and F are midpoints of AC and AB, so using the mid-point theorem:
$\Rightarrow EF=\frac{1}{2}BC$ …..(2)
F and D are midpoints of AB and BC, so using the 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

Points P and Q have been taken on opposite sides AB and CD, respectively, of a parallelogram ABCD such that AP = CQ (Figure). Show that AC and PQ bisect each other.

Answer:

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

In the Figure, P is the mid-point of side BC of a parallelogram ABCD such that $\angle BAP=\angle DAP$. Prove that $AD = 2CD$
.
Answer:

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

Question:1

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

Answer:

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)
$\angle A=90^{\circ}$
Here, ADEF is a square
$AD = AF$ …..(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.

Answer: 4 cm

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.

Answer:

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 midpoints of AD and DC, respectively.
By the mid-point theorem, we get
$SR\parallel AC$ and $SR=\frac{1}{2}AC$ …..(1)
In $\triangle ABC$, P and Q are the midpoints of AB and BC, respectively.
Then, by the mid-point theorem, we get
$PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ …..(2)
From equations 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 equations 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 equations 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.

Answer:

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 midpoints of AD and DC, so by the 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)
In quadrilateral EOFR,
$OE \parallel FR, OF \parallel ER$
$\therefore \angle EOF = \angle ERF = 90^{\circ}$
{$\angle COD = 90^{\circ}$ because $AC \perp BD$ and opposite angles of a parallelogram are equal}
In quadrilateral RSPQ
$\angle SRQ = \angle ERF = \angle SPQ = 90^{\circ}$ {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 a 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.

Answer:

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 midpoints of sides AD and DC by the mid-point theorem.
$SR \parallel AC$ and $SR =\frac{1}{2}AC$ …..(1)
In $\triangle ABC$, P and Q are the midpoints of AB and BC, respectively. Therefore, by the mid-point theorem
$PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ …..(2)
From equations 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 the mid-point theorem
$SP=\frac{1}{2}BD=\frac{1}{2}AC${Given AC = BD …..(5)
From equations 4 and 5, we get
$SP=RQ=\frac{1}{2}AC$ …..(6)
From equations 3 and 6, we get
$PQ = SR = SP = RQ$
Thus, all sides are equal
In quadrilateral EOFR,
$OE \parallel FR, OF \parallel ER$
$\therefore \angle EOF =\angle ERF = 90^{\circ}$
{$\angle COD = 90^{\circ}$ because $AC \perp BD$ and opposite angles of a parallelogram are equal}
In quadrilateral RSPQ
$\angle SRQ = \angle ERF = \angle SPQ = 90^{\circ}$ {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.

Answer:

Let ABCD be 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 midpoints 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.

Answer:

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 equations 2 and 3
$SQ\parallel RP$and $PS\parallel QR$
So, PRQS is a parallelogram.
Hence Proved

Question:9

In Figure, $AB \parallel DE$, AB = DE, $AC \parallel DF$ and AC = DF. Prove that $BC \parallel EF$ and BC = EF.

Answer:

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
In ACFD quadrilateral
$AC\parallel FD$ and AC = FD …..(1)
Thus ACFD is a parallelogram
$AD\parallel CF$ and AD = CF …..(2)
In ABED quadrilateral
$AB\parallel DE$ 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 $AF=\frac{1}{3}AC$.

Answer:

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 midpoints of the consecutive sides of a square is also a square.
Hint: Prove all the sides are equal and diagonals are equal

Answer:

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, and S are the midpoints 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 equations 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 equations 3 and 4
SP = PQ = SP = RQ {All sides are equal}
In quadrilateral OERF
$OE\parallel FR$ and $OF\parallel ER$
$\angle EOF=\angle ERF=90^{\circ}$
In quadrilateral RSPQ
$\angle SRQ=\angle ERF=\angle SPQ=90^{\circ}$ {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 the angles are also equal, 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 $EF=\frac{1}{2}(AB+CD)$

Answer:

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 the mid-point theorem, O is the 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.

Answer:

Given: Let ABCD be a parallelogram and AP, BR, CQ, DS be 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.

Answer:

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.

Answer:

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 triangle ABC is divided into four congruent triangles.

Answer:

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 the 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 $\triangle EFD$
AD = EF
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 midpoints of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.

Answer:

Given: Let ABCD be a trapezium in which $AB\parallel DC$ and let M and N be the midpoints 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 midpoints 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.

Answer:

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 .