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., $\mathrm{\angle }1+\mathrm{\angle }2+\mathrm{\angle }3+\mathrm{\angle }4={360}^{\circ }$ …..(i)
Here $\mathrm{\angle }1={75}^{\circ }$
$\mathrm{\angle }2={90}^{\circ }$
$\mathrm{\angle }3={75}^{\circ }$
Putting the values in equation (i), we get
${75}^{\circ }+{90}^{\circ }+{75}^{\circ }+\mathrm{\angle }4={360}^{\circ }$
$\mathrm{\angle }4={360}^{\circ }-{240}^{\circ }$
$\mathrm{\angle }4={120}^{\circ }$
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}
$\therefore \mathrm{\angle }OBA=\mathrm{\angle }OAB$
$\mathrm{\angle }OAB={25}^{\circ }$ {Given}
$\Rightarrow \mathrm{\angle }OBA={25}^{\circ }$
$\mathrm{\angle }BOC=\mathrm{\angle }OBA+\mathrm{\angle }OAB$
[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 $\mathrm{\angle }ACB={40}^{\circ }$. Then $\mathrm{\angle }ADB$ is
(A) 40⁰ (B) 45⁰ (C) 50⁰ (D) 60⁰

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

Given : ABCD is a rhombus such that $\mathrm{\angle }ACB={40}^{\circ }\Rightarrow \mathrm{\angle }OCB={40}^{\circ }$
To Find : $\mathrm{\angle }ADB$
Since $AD\parallel BC$
$\Rightarrow \mathrm{\angle }AOC=\mathrm{\angle }ACB={40}^{\circ }$ {alternate interior angles}
$\mathrm{\angle }AOD={90}^{\circ }$ {diagonals of a rhombus are perpendicular to each other}
We know that the sum of angles of a triangle is ${180}^{\circ }$.
In $\mathrm{△}AOD$,
$\mathrm{\angle }AOD+\mathrm{\angle }OAD+\mathrm{\angle }ADO={180}^{\circ }$
${90}^{\circ }+{40}^{\circ }+\mathrm{\angle }ADO={180}^{\circ }$
$\mathrm{\angle }ADO={180}^{\circ }-{90}^{\circ }-{40}^{\circ }$
$\mathrm{\angle }ADO={50}^{\circ }=\mathrm{\angle }ADB$
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
$\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C=\mathrm{\angle }D={90}^{\circ }$ 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 $\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C=\mathrm{\angle }D={90}^{\circ }$
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 $\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C=\mathrm{\angle }D={90}^{\circ }$
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 $\mathrm{\angle }A=\mathrm{\angle }C=\mathrm{\angle }D={90}^{\circ }$
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 $\mathrm{\angle }A$ and $\mathrm{\angle }B$of a quadrilateral ABCD intersect each other at P, of $\mathrm{\angle }B$ and $\mathrm{\angle }C$ at Q, of $\mathrm{\angle }C$ and $\mathrm{\angle }D$ at R and of $\mathrm{\angle }D$and $\mathrm{\angle }A$ 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
$\mathrm{\angle }QPS=\mathrm{\angle }APB$ …..(1) (Vertically opposite angles)
In $\mathrm{△}APB=\mathrm{\angle }APB+\mathrm{\angle }PAB+\mathrm{\angle }ABP={180}^{\circ }$
$\mathrm{\angle }APB=\frac{1}{2}\mathrm{\angle }A+\frac{1}{2}\mathrm{\angle }B={180}^{\circ }$
$\mathrm{\angle }APB=180-\frac{1}{2}(\mathrm{\angle }A+\mathrm{\angle }B)$ …..(2)
From equations 1 & 2
$\mathrm{\angle }QPS=180-\frac{1}{2}(\mathrm{\angle }A+\mathrm{\angle }B)$ …..(3)
Similarly $\mathrm{\angle }QRS=180-\frac{1}{2}(\mathrm{\angle }C+\mathrm{\angle }D)$ …..(4)
Add equations 3 and 4
$\mathrm{\angle }QPS+\mathrm{\angle }QRS={360}^{\circ }-\frac{1}{2}(\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C+\mathrm{\angle }D)$
$={360}^{\circ }-\frac{1}{2}({360}^{\circ })$
$={360}^{\circ }-{180}^{\circ }$
$={180}^{\circ }$
$\therefore \mathrm{\angle }QPS+\mathrm{\angle }QRS={180}^{\circ }$
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 $APB\parallel CQD$
$\mathrm{\angle }APQ=\mathrm{\angle }PQD$ (Alternate angles)
and $\mathrm{\angle }MPQ=\mathrm{\angle }PQN$ (Alternate interior angles)
$\therefore PM\parallel QN$
Similarly $\mathrm{\angle }BPQ=\mathrm{\angle }PQC$ (Alternate angles)
$\Rightarrow PN\parallel QM$
So, quadrilateral PMQN is parallelogram
Since CQD is a line
$\therefore \mathrm{\angle }CQD={180}^{\circ }$
$\mathrm{\angle }CQP+\mathrm{\angle }PQD={180}^{\circ }$
$2\mathrm{\angle }MQP+2\mathrm{\angle }NQP={180}^{\circ }$
$2(\mathrm{\angle }MQP+\mathrm{\angle }NQP)={180}^{\circ }$
$\mathrm{\angle }MQP+\mathrm{\angle }NQP=\frac{180}{2}$
$\Rightarrow \mathrm{\angle }MQN=\frac{180}{2}$
$\Rightarrow \mathrm{\angle }MQN={90}^{\circ }$
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 $\mathrm{△}ABC$, 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 $\mathrm{△}ADC$
$AC\parallel SP$and $SP=\frac{1}{2}AC$…..(2)
From equations 1 and 2
$SP\parallel RQ$ and $SP=RQ$
Similarly, when we take $\mathrm{△}ABD$ and $\mathrm{△}BDC$ 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 $\mathrm{△}AOC$, 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 $\mathrm{△}ABO,PD\parallel AO$ 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 $\mathrm{△}ADB$ by mid-point theorem
$SP\parallel DB$
$SP=\frac{1}{2}DB$ …..(2)
Similarly in $\mathrm{△}ABC,AQ=\frac{1}{2}AC$ …..(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 $\mathrm{\angle }DAC={32}^{\circ }$ and $\mathrm{\angle }AOB={70}^{\circ }$, then $\mathrm{\angle }DBC$is equal to
(A) ${24}^{\circ }$ (B) ${86}^{\circ }$ (C) ${38}^{\circ }$ (D) ${32}^{\circ }$

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

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

$\mathrm{\angle }ACB={32}^{\circ }${$AD\parallel BC$ and AC is a transversal line}
$\mathrm{\angle }AOB+\mathrm{\angle }BOC={180}^{\circ }$
${70}^{\circ }+\mathrm{\angle }BOC={180}^{\circ }$
$\mathrm{\angle }BOC={180}^{\circ }-{70}^{\circ }={110}^{\circ }$
In $\mathrm{△}BOC$, we have
$\mathrm{\angle }BOC+\mathrm{\angle }OCB+\mathrm{\angle }CBO={180}^{\circ }$
${110}^{\circ }+{32}^{\circ }+\mathrm{\angle }CBO={180}^{\circ }$
$\mathrm{\angle }CBO={180}^{\circ }-{110}^{\circ }-{32}^{\circ }$
$\mathrm{\angle }CBO={38}^{\circ }$
$\mathrm{\angle }DBC=\mathrm{\angle }CBD={38}^{\circ }$
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 $\mathrm{△}ABC$. DE is produced to F. To prove that CF is equal and parallel to DA, we need additional information, which is
(A) $\mathrm{\angle }DAE=\mathrm{\angle }EFC$
(B) AE = EF
(C) DE = EF
(D) $\mathrm{\angle }ADE=\mathrm{\angle }ECF.$

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
$\mathrm{\angle }AED=\mathrm{\angle }FEC$ {vertically opposite angles}
$\therefore \mathrm{△}ADE\cong \mathrm{△}CFE$ {by SAS congruence rule}
$\therefore AD=CF$ {by CPCT rule}
$\therefore \mathrm{\angle }ADE=\mathrm{\angle }CFE$ {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 $OD=2cm$, determine the lengths of AC and BD.

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

As we know that the diagonals of a parallelogram bisect each other
$AC=2AO=2\times 3=6$
$AC=6cm$
$DB=2DO=2\times 2=4$
$DB=4cm$
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
${110}^{\circ }+{80}^{\circ }+{70}^{\circ }+{95}^{\circ }={355}^{\circ }\ne {360}^{\circ }$
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, $\mathrm{\angle }A+\mathrm{\angle }D={180}^{\circ }$ .What special name can be given to this quadrilateral?

Answer: Trapezium
Let the given quadrilateral be ABCD.

It is given that $\mathrm{\angle }A+\mathrm{\angle }D={180}^{\circ }$
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,
$\mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R+\mathrm{\angle }S={360}^{\circ }$ …..(1)
It is given that all the angels are equal
i.e., $\mathrm{\angle }P=\mathrm{\angle }Q=\mathrm{\angle }R=\mathrm{\angle }S$
Put in equation 1, we get
$4\mathrm{\angle }P={360}^{\circ }$
$\mathrm{\angle }P=\frac{{360}^{\circ }}{4}$
$P={90}^{\circ }$
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 $\mathrm{△}ACD$ and $\mathrm{△}BCD$
AC = BD (opposite sides of a rectangle are equal)
$\mathrm{\angle }C=\mathrm{\angle }D$ $({90}^{\circ })$
AB = CD (opposite sides of a rectangle are equal)
$\mathrm{△}ACD\cong \mathrm{△}BCD$ (SAS congruency)
So, AD = BC
Hence, diagonals are equal.
Also, $\mathrm{\angle }CAD=\mathrm{\angle }DBC$ …(i)
Similarly, we can prove that $\mathrm{△}ACB$ and $\mathrm{△}BDA$are congruent
Hence, $\mathrm{\angle }ACB=\mathrm{\angle }ADB$ …(ii)
Now, consider $\mathrm{△}AOC$ and $\mathrm{△}BOD$
$\mathrm{\angle }CAD=\mathrm{\angle }DBC$From (i)
$\mathrm{\angle }ACB=\mathrm{\angle }ADB$ From (ii)
$\mathrm{\angle }AOC=\mathrm{\angle }BOD$ vertically opposite angles
$\mathrm{△}AOC$ and $\mathrm{△}BOD$ are also congruent.
But we cannot prove that $\mathrm{\angle }AOC=\mathrm{\angle }BOD={90}^{\circ }$
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

$\mathrm{△}ABC$, $AB=5cm$, $BC=8cm$ and$CA=7cm$. 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 $\mathrm{\angle }C={55}^{\circ }$, determine $\mathrm{\angle }F$ .

Answer:

Given: ABCD and AEFG are two parallelograms

Given $\mathrm{\angle }C={55}^{\circ }$
then $\mathrm{\angle }A$ is also ${55}^{\circ }$. {opposite angles of a parallelogram are equal}
Also AEFG is a parallelogram then
$\mathrm{\angle }A=\mathrm{\angle }F={55}^{\circ }$ {opposite angels of a parallelogram is equal}
Hence, $\mathrm{\angle }F={55}^{\circ }$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 $\mathrm{\angle }A={35}^{\circ }$, determine $\mathrm{\angle }B$.

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.,
$\mathrm{\angle }A+\mathrm{\angle }B={180}^{\circ }$
$\mathrm{\angle }A={35}^{\circ }$ (given)
${35}^{\circ }+\mathrm{\angle }B={180}^{\circ }$
$\mathrm{\angle }B={180}^{\circ }-{35}^{\circ }$
$\mathrm{\angle }B={145}^{\circ }$

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
$AB=CD=4cm$
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 $\mathrm{\angle }A=\mathrm{\angle }B={45}^{\circ }$. Find angles C and D of the trapezium.

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

If $AB\parallel CD$ and BC is transversal then sum of co-interior angles is ${180}^{\circ }$
$\therefore \mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$
${45}^{\circ }+\mathrm{\angle }C={180}^{\circ }$
$\mathrm{\angle }C={180}^{\circ }-{45}^{\circ }$
$\mathrm{\angle }C={135}^{\circ }$
Similarly
$\mathrm{\angle }A+\mathrm{\angle }D={180}^{\circ }$
${45}^{\circ }+\mathrm{\angle }D={180}^{\circ }$
$\mathrm{\angle }D={180}^{\circ }-{45}^{\circ }$
$\mathrm{\angle }D={135}^{\circ }$
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, $\mathrm{\angle }ADC$ and $\mathrm{\angle }ABC$ are obtuse angles and DE and DF are altitudes.
In quadrilateral BEDF, $\mathrm{\angle }BED$= $\mathrm{\angle }BFD$ = ${90}^{\circ }$
$\mathrm{\angle }FBE={360}^{\circ }-(\mathrm{\angle }FDE+\mathrm{\angle }BED+\mathrm{\angle }BFD)$
$={360}^{\circ }-({60}^{\circ }+{90}^{\circ }+{90}^{\circ })$
$={360}^{\circ }-{240}^{\circ }$
$={120}^{\circ }$
It is given that ABCD is a parallelogram
ADC = ${120}^{\circ }$
$\mathrm{\angle }A+\mathrm{\angle }B={180}^{\circ }${Sum of interior angle $={180}^{\circ }$}
$\mathrm{\angle }A={180}^{\circ }-B$
$\mathrm{\angle }A={180}^{\circ }-{120}^{\circ }$
$\mathrm{\angle }A={60}^{\circ }$
$\Rightarrow \mathrm{\angle }C=\mathrm{\angle }A={60}^{\circ }$
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 $\mathrm{\angle }DLA=\mathrm{\angle }DLB={90}^{\circ }$ {DL is perpendicular bisector of AB}
$AL=BL=\frac{X}{2}$
DL = DL {common side of $\mathrm{△}ADL$ and $\mathrm{△}BDL$}
$\mathrm{△}ALD\cong \mathrm{△}BLD${by SAS congruence}
$AD=BD$
In $\mathrm{△}ADB$, $AD=AB=DB=x$
$\mathrm{△}ADB$ is an equilateral triangle.
$\therefore \mathrm{\angle }ADB=\mathrm{\angle }ABD=\mathrm{\angle }A={60}^{\circ }$
Similarly, $\mathrm{△}DCB$ is an equilateral triangle
$\mathrm{\angle }BDC=\mathrm{\angle }DBC=\mathrm{\angle }C={60}^{\circ }$
Also, $\mathrm{\angle }A=\mathrm{\angle }C$
$\mathrm{\angle }D=\mathrm{\angle }B$…..(1)
$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C+\mathrm{\angle }D={360}^{\circ }$
${60}^{\circ }+B+{60}^{\circ }+B={360}^{\circ }$
$2\mathrm{\angle }B=\frac{240}{2}={120}^{\circ }$
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 $\mathrm{△}ADC$, 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 $\mathrm{△}CBA$, 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 $\mathrm{△}ABC$ as shown in Figure. Show that $BC=\frac{1}{2}QR$

Answer:

Given : In $\mathrm{△}ABC$, $PQ\parallel AB$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 $\mathrm{△}DEF$ 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 $\mathrm{△}ABC$ 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 $\mathrm{\angle }BAP=\mathrm{\angle }DAP$. Prove that $AD=2CD$
.
Answer:

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

$\therefore AD\parallel BC$ and AB is transversal
$\mathrm{\angle }A+\mathrm{\angle }B={180}^{\circ }$ {sum of co-interior angles}
$\mathrm{\angle }B=180-\mathrm{\angle }A$ …..(1)
In $\mathrm{△}ABP$,
$\mathrm{\angle }PAB+\mathrm{\angle }BPA+\mathrm{\angle }B={180}^{\circ }${using angle sum property of triangle}
Putting the values,
$\frac{1}{2}\mathrm{\angle }A+\mathrm{\angle }BPA+({180}^{\circ }-\mathrm{\angle }A)={180}^{\circ }$
$\mathrm{\angle }BPA-\frac{\mathrm{\angle }A}{2}=0$
$\mathrm{\angle }BPA=\frac{\mathrm{\angle }A}{2}$ …..(2)
$\Rightarrow \mathrm{\angle }BPA=\mathrm{\angle }BAP$
$AB=BP$ {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 $\mathrm{△}ABC$.
To prove : $CE=BE$

Proof: In isosceles $\mathrm{△}ABC$ and $AB=AC$ …..(1)
$\mathrm{\angle }A={90}^{\circ }$
Here, ADEF is a square
$AD=AF$ …..(2)
Subtract equation 2 from 1
$AB-AD=AC-AF$
$BD=CF$ …..(3)
Now in $\mathrm{△}BDE$and $\mathrm{△}CFE$
$BD=CF${from equation 3}
$\mathrm{\angle }CFE=\mathrm{\angle }EDB$ {${90}^{\circ }$ each}
$DE=EF$ {side of a square}
$\mathrm{△}CFE\cong \mathrm{△}BDE$ {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 $\mathrm{\angle }A$ 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 $AB=10cm$ and $AD=6cm$.
Construction: In parallelogram ABCD, draw the bisector of $\mathrm{\angle }A$ 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 $\mathrm{\angle }AFH=\mathrm{\angle }FAB$ …..(1) {alternate interior angles}

$AB=HF$ (opposite sides of a parallelogram)
$AF=AF$ (Common Side)
$\Rightarrow \mathrm{△}HAF\cong \mathrm{△}FAB$…..(2) { \ SAS Congruency}
Now, $\mathrm{\angle }HAF=\mathrm{\angle }FAB$ …..(3) (EA is the bisector of $\mathrm{\angle }A$)
$\Rightarrow \mathrm{\angle }AFH=\mathrm{\angle }HAF$ { using 1 and 3 }
So, $HF=AH$ {Sides opposite to equal angles are equal}
$HF=AB=10cm$
$AH=HF=10cm$
$AD+DH=10cm$
$DH=10-AD\Rightarrow 10-6$
$DH=4cm$
Because FHDC is a parallelogram
opposite sides are equal
$DH=CF=4cm$
Hence, the answer is $4cm$

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 $\mathrm{△}ADC$, 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 $\mathrm{△}ABC$, 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 $\mathrm{△}BCD$we get
$RQ\parallel BD$ and $RQ=\frac{1}{2}BD$ …..(4)
And in $\mathrm{△}BAD$
$SP\parallel BD$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 $AC\perp BD$
$\mathrm{\angle }COD=\mathrm{\angle }AOD=\mathrm{\angle }AOB=\mathrm{\angle }COB={90}^{\circ }$

In $\mathrm{△}ADC$, 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 $\mathrm{△}ABC$,
$PQ\parallel AC$ & $PQ=\frac{1}{2}AC$ …..(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,
$OE\parallel FR,OF\parallel ER$
$\therefore \mathrm{\angle }EOF=\mathrm{\angle }ERF={90}^{\circ }$
{$\mathrm{\angle }COD={90}^{\circ }$ because $AC\perp BD$ and opposite angles of a parallelogram are equal}
In quadrilateral RSPQ
$\mathrm{\angle }SRQ=\mathrm{\angle }ERF=\mathrm{\angle }SPQ={90}^{\circ }$ {Opposite angles in a parallelogram are equal}
$\mathrm{\angle }RSP+\mathrm{\angle }SRQ+\mathrm{\angle }RQP+\mathrm{\angle }QRS={360}^{\circ }$
${90}^{\circ }+{90}^{\circ }+\mathrm{\angle }RSP+\mathrm{\angle }RQP={360}^{\circ }$
$\mathrm{\angle }RSP+\mathrm{\angle }RQP={180}^{\circ }$
$\mathrm{\angle }RSP=\mathrm{\angle }RQP={90}^{\circ }$
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 $\mathrm{△}ADC$, 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 $\mathrm{△}ABC$, 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 $\mathrm{△}BCD$, $RQ\parallel BD$ 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 $\mathrm{△}ABD$, $SP\parallel BD$ 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,
$OE\parallel FR,OF\parallel ER$
$\therefore \mathrm{\angle }EOF=\mathrm{\angle }ERF={90}^{\circ }$
{$\mathrm{\angle }COD={90}^{\circ }$ because $AC\perp BD$ and opposite angles of a parallelogram are equal}
In quadrilateral RSPQ
$\mathrm{\angle }SRQ=\mathrm{\angle }ERF=\mathrm{\angle }SPQ={90}^{\circ }$ {Opposite angles in a parallelogram are equal}
$\mathrm{\angle }RSP+\mathrm{\angle }SRQ+\mathrm{\angle }RQP+\mathrm{\angle }QRS={360}^{\circ }$
${90}^{\circ }+{90}^{\circ }+\mathrm{\angle }RSP+\mathrm{\angle }RQP={360}^{\circ }$
$\mathrm{\angle }RSP+\mathrm{\angle }RQP={180}^{\circ }$
$\mathrm{\angle }RSP=\mathrm{\angle }RQP={90}^{\circ }$
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 $\mathrm{\angle }CAB=\mathrm{\angle }CAD$

To Prove: ABCD is a rhombus
Proof: Here, it is given that ABCD is a parallelogram.
$AB\parallel CD$ and AC is transversal
$\mathrm{\angle }CAB=\mathrm{\angle }ACD${alternate interior angles} …. (1)
Also, $AD\parallel BC$ and AC is a transversal
$\mathrm{\angle }CAD=\mathrm{\angle }ACB${alternate interior angle} …. (2)
Now it is given that $\mathrm{\angle }CAB=\mathrm{\angle }CAD$
So from equations (1) and (2)
$\Rightarrow \mathrm{\angle }ACD=\mathrm{\angle }ACB$…. (3)
Also, $\mathrm{\angle }A=\mathrm{\angle }C${opposite angles of a parallelogram}
$\frac{\mathrm{\angle }A}{2}=\frac{\mathrm{\angle }C}{2}$ {dividing by 2}
$\mathrm{\angle }DAC=\mathrm{\angle }DCA${from 1 and 2}
$CD=AD$ {sides opposite of equal angles in $\mathrm{△}ADC$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.