NCERT Exemplar Class 9 Maths Solutions Chapter 8 Quadrilaterals

Exercise 8.1

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

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

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

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

Answer:

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

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:

If diagonals of PQRS are perpendicular

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 ${180}^{\circ }$

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

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 equation 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 equation 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 quadrilateral whose opposite angles are supplementary

Question:8

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

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 $DE=\frac{1}{2}AC$
$DE=\frac{1}{2}\times 7$
$DE=\frac{7}{2}cm$
$DE=3.5cm$
Hence answer is 3.5 cm

Question:9

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

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)

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

(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 point 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)
$\Rightarrow PN\parallel QM$

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

(D) any parallelogram
Let ABCD is a rhombus and P, Q, R and S are its mid points.

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 equation 1 and 2
$SP\parallel RQ$ and $SP=RQ$

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

(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 mid points 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 equation 3 and 4
$EQ\parallel PD$ and $EQ=PD$
From equation 1 and 3
$DE\parallel PQ$ and $DE=PQ$

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.

Question:12

The diagonals AC and BD of a parallelogram ABCD intersect each other at thepoint 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) ${32}^{\circ }$

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.

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 $\mathrm{△}ABC$. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is
(A) $\mathrm{\angle }DAE=\mathrm{\angle }EFC$
(B) AE = EF
(C) DE = EF
(D) $\mathrm{\angle }ADE=\mathrm{\angle }ECF.$

(C) DE = EF
Let us draw the figure according to 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,$AD\parallel CF$

Exercise 8.2

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.

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

As we know that the diagonal of a parallelogram bisect each other
$AC=2AO=2\times 3=6$
$AC=6cm$
$DB=2DO=2\times 2=4$
$DB=4cm$

Question:2

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

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

Question:4

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

Let the given quadrilateral ABCD is,

Question:5

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

i.e., $\mathrm{\angle }P=\mathrm{\angle }Q=\mathrm{\angle }R=\mathrm{\angle }S$

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 $\mathrm{△}ACD$ 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.

Question:7

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

Question:10

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

Answer:

Question:11

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

Question:12

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

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.

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

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.

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.

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.

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

Question:4

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

Join DB

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

Question:5

E and F are points on 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

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]

Answer:

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

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:

Proof : In quadrilateral BCAR, $BR\parallel CA$ and $BC\parallel RA$

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 $\mathrm{△}DEF$ is also an equilateral triangle.

Answer:

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:

To Prove: AC and PQ bisect each other.
Proof :

Here ABCD is a parallelogram

Question:10

In 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

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.

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}

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.