SELINA Solution Class 9 Chapter 16 Area of Theorems [Proof and Use] Exercise 16A

Question 1

In the given figure, if the area of triangle ADE is 60 cm², state, given reason, the area of :
(i) Parallelogram ABED;
(ii) Rectangle ABCF;
(iii) Triangle ABE.

Sol:

(i) ΔADE and parallelogram ABED are on the same base AB and between the same parallels DE//AB, so an area of the triangle ΔADE is half the area of parallelogram ABED.

Area of ABED = 2 (Area of ADE) = 120 cm²

(ii)Area of the parallelogram is equal to the area of a rectangle on the same base and of the same altitude i.e, between the same parallels

Area of ABCF = Area of ABED = 120 cm²

(iii)We know that area of triangles on the same base and between same parallel lines are equal

Area of ABE = Area of ADE = 60 cm²

Question 2

The given figure shows a rectangle ABDC and a parallelogram ABEF; drawn on opposite sides of AB.
Prove that: 
(i) Quadrilateral CDEF is a parallelogram;
(ii) Area of the quad. CDEF
= Area of rect. ABDC + Area of // gm. ABEF.

Sol:

After drawing the opposite sides of AB, we get

Since from the figure, we get CD//FE, therefore, FC must parallel to DE. Therefore it is proved that the quadrilateral CDEF is a parallelogram.

The area of the parallelogram on the same base and between the same parallel lines is always equal and the area of the parallelogram is equal to the area of a rectangle on the same base and of the same altitude i.e, between the same parallel lines. 

So Area of CDEF= Area of ABDC + Area of ABEF 
Hence Proved

Question 3

In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS. Show that:

(i) 2 Area (POS) = Area (// gm PMLS)
(ii) Area (POS) + Area (QOR) = Area (// gm PQRS)
(iii) Area (POS) + Area (QOR) = Area (POQ) + Area (SOR).

Sol;

(i) Since POS and parallelogram, PMLS are on the same base PS and between the same parallels i.e. SP//LM.

As O is the center of LM and the Ratio of the area of triangles with the same vertex and bases along the same line is equal to the ratio of their respective bases.

The area of the parallelogram is twice the area of the triangle if they lie on the same base and in between the same parallels.

So 2(Area of PSO)=Area of PMLS
Hence Proved.

(ii) Consider the expression: Area ( ΔPOS) + Area ( QOR ):

LM is parallel to PS and PS is parallel to RQ, therefore, LM is
Since triangle POS lie on the base PS and in between the parallels PS and LM, we have,

Area ( ΔPOS ) = ${\displaystyle \frac{1}{2}\text{Area}(\square }$ PSLM )

Since triangle QOR lie on the base QR and in between the Parallels LM and RQ, we have,

Area ( ΔQOR ) = ${\displaystyle \frac{1}{2}\text{Area}(\square }$ LMQR )

Area ( ΔPOS ) + Area ( ΔQOR ) = ${\displaystyle \frac{1}{2}\text{Area}(\square }$ PSLM ) + ${\displaystyle \frac{1}{2}\text{Area}(\square }$ LMQR )

= ${\displaystyle \frac{1}{2}[\text{Area (PSLM )}+\text{Area}(\square }$ LMQR )]

= ${\displaystyle \frac{1}{2}[\text{Area}(\square }$ PQRS) ]

(iii) In a parallelogram, the diagonals bisect each other.
Therefore, OS = OQ

Consider the triangle PQS, since OS = OQ, OP is the median of the triangle PQS.
We know that the median of a triangle divides it into two triangles of equal area.

Therefore,
Area ( ΔPOS) = Area ( ΔPOQ )              ....(1)
Similarly, since OR is the median of the triangle QRS, we have, Area ( ΔQOR ) = Area ( ΔSOR )           ....(2)

Adding equations (1) and (2), we have,
Area ( ΔPOS ) + Area( ΔQOR) = Area ( ΔPOQ ) + Area( SOR)
Hence Proved.

Question 4

In parallelogram ABCD, P is a point on side AB and Q is a point on side BC.
Prove that:
(i) ΔCPD and ΔAQD are equal in the area.
(ii) Area (ΔAQD) = Area (ΔAPD) + Area (ΔCPB)

Sol:

Given ABCD is a parallelogram. P and Q are any points on the sides AB and BC respectively, join diagonals AC and BD.

proof:
(i) since triangles with the same base and between the same set of parallel lines have equal areas

area ( CPD ) =  area( BCD )                           …… (1)

again, diagonals of the parallelogram bisect area in two equal parts
area ( BCD ) = ( 1/2 ) area of parallelogram ABCD   …… (2)

from (1) and (2)
area( CPD ) = 1/2 area( ABCD )                           …… (3)
similarly area ( AQD ) = area( ABD ) = 1/2 area( ABCD )…… (4)
from (3) and (4)
area( CPD ) = area( AQD ),
hence proved.

(ii) We know that area of triangles on the same base and between same parallel lines are equal

So Area of AQD= Area of ACD= Area of PDC = Area of BDC = Area of ABC=Area of APD + Area of BPC
Hence Proved

Question 5

In the given figure, M and N are the mid-points of the sides DC and AB respectively of the parallelogram ABCD.

If the area of parallelogram ABCD is 48 cm²;
(i) State the area of the triangle BEC.
(ii) Name the parallelogram which is equal in area to the triangle BEC.

Sol:

(i) Since triangle BEC and parallelogram ABCD are on the same base BC and between the same parallels i.e. BC // AD.

So Area ( ΔBEC )= ${\displaystyle \frac{1}{2}\times \text{Area}(\square }$ABCD )

= ${\displaystyle \frac{1}{2}}$ x 48 = 24 cm²  

(ii) Area (${\displaystyle \square \text{ANMD})=\text{Area}(\square }$ BNMC )
= ${\displaystyle \frac{1}{2}\text{Area}(\square }$ ABCD)

= ${\displaystyle \frac{1}{2}}$ x 2 x Area ( ΔBEC )

= Area ( ΔBEC )

Therefore, Parallelograms ANMD and NBCM have areas equal to triangle BEC.

Question 6

In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E.
Prove that ΔADE and quadrilateral ABCD are equal in area.

Sol:

Since ΔDCB and ΔDEB are on the same base DB and between the same parallels i.e. DB // CE, therefore we get

Ar. ( ΔDCB) = Ar. ( ΔDEB )
Ar. ( ΔDCB + ΔADB ) = AR. (ΔDEB + ΔADB )
Ar. ( ABCD ) = Ar. ( ΔADE )
Hence proved.

Question 7

ABCD is a parallelogram a line through A cuts DC at point P and BC produced at Q. Prove that triangle BCP is equal in area to triangle DPQ.

Sol:

ΔAPB and parallelogram ABCD are on the same base AB and between the same parallel lines AB and CD.

∴ Ar. ( ΔAPB ) = ${\displaystyle \frac{1}{2}}$ Ar.( parallelogram ABCD ) ......(i)

ΔADQ and parallelogram ABCD are on the same base AD and between the same parallel lines AD and BQ.

∴ Ar.( ΔADQ ) = ${\displaystyle \frac{1}{2}}$ Ar.( parallelogram ABCD ) ......(ii)

Adding equation (i) and (ii), we get

∴ Ar.( ΔAPB ) + Ar.( ΔADQ ) = Ar.(parallelogram ABCD)
Ar.( quad ADQB ) - Ar.(Δ BPQ ) = Ar.(parallelogram ABCD)
Ar.( quad ADQB) - Ar.( ΔBPQ ) = Ar.(quad ADQB ) -Ar.( ΔDCQ )
                            Ar. ( ΔBPQ ) = Ar. ( ΔDCQ )

Subtracting Ar.ΔPCQ from both sides, we get

Ar. ( ΔBPQ ) - Ar.(ΔPCQ ) = Ar. ( ΔDCQ ) - Ar. ( ΔPCQ)
                    Ar. ( ΔBCP ) = Ar. ( ΔDPQ )
Hence proved.

Question 8

The given figure shows a pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF draw parallel to DB meets AB produced at F.
Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.

Sol:

Since triangle EDG and EGA are on the same base EG and between the same parallel lines EG and DA.
Therefore,
A( ΔEDG ) = A( ΔEGA )

Subtracting ΔEOG from both sides, we have
A( ΔEOG ) = A( ΔGOA )                      ......(i)

Similarly,
A( ΔDPC ) = A( ΔBPF )                       ........(ii)
Now
A( ΔGDF ) = A( ΔGOA ) + A( ΔBPF ) + A( pen. ABPDO ) 
                 = A( ΔEOD ) + A( ΔDPC ) + A( pen.ABPDO )
                 = A( pen. ABCDE )
Hence proved.

Question 9

In the given figure, AP is parallel to BC, BP is parallel to CQ.
Prove that the area of triangles ABC and BQP are equal.

SoL:

Joining PC we get,

ΔABC and ΔBPC are on the same base BC and between the same parallel lines AP and BC.
∴ A( ΔABC ) = A( ΔBPC )          ....(i)

ΔBPC and ΔBQP are on the same base BP and between the same parallel lines BP and CQ.
∴ A( ΔBPC ) = A( ΔBQP )          ....(ii)

From (i) and (ii), we get
∴A( ΔABC ) = A( ΔBQP ) 
Hence proved.

Question 10

In the figure given alongside, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC.

If BH is perpendicular to FG
prove that:
(i) ΔEAC ≅ ΔBAF.
(ii) Area of the square ABDE
⇒ Area of the rectangle ARHF.

SOl:

(i) ∠EAC = ∠EAB + ∠BAC
∠EAC = 90° + ∠BAC            ....(i)
∠BAF = ∠FAC + ∠BAC
∠BAF = 90° + ∠BAC            .....(ii)

From (i) and (ii), we get
∠EAC = ∠BAF
In ΔEAC and ΔBAF, we have, EA = AB
∠EAC = ∠BAF and AC = AF
∴ ΔEAC ≅ ΔBAF                ....( SAS axiom of congruency )

(ii) Since ΔABC is a right triangle, We have,
AC² = AB² + BC²              ....( Using pythagoras theorm in ΔABC )
⇒ AB² = AC² - BC² 
⇒ AB² = ( AR + RC )² - ( BR² + RC² )  ....( Since AC = AR + RC and Using Pythagoras Theorem in ΔBRC )
⇒ AB² = AR² + 2AR x RC + RC² - ( BR² + RC² ) ....( Using the identity ) 
⇒ AB² = AR² + 2AR x RC + RC² - ( AB² - AR² + RC² )  ...( Using Pythagoras Theorem in ΔABR )

⇒ 2AB² = 2AR² + 2AR x RC
⇒ AB² = AR( AR + RC )
⇒ AB² =  AR + AC
⇒ AB² = AR x AF
⇒ Area (${\displaystyle \square }$ABDE ) = Area( rectangle ARHF ).

Question 11

In the following figure, DE is parallel to BC.
Show that: 
(i) Area ( ΔADC ) = Area( ΔAEB ).
(ii) Area ( ΔBOD ) = Area( ΔCOE ).

SoL:

(i) In ΔABC, D is the midpoint of AB and E is the midpoint of AC.
${\displaystyle \frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}}$

DE is parallel to BC.
∴ A( ΔADC ) = A( ΔBDC ) = ${\displaystyle \frac{1}{2}}$ A( ΔABC )
Again,
∴ A( ΔAEB ) = A( ΔBEC ) = ${\displaystyle \frac{1}{2}}$ A( ΔABC )

From the above two equations, we have
Area( ΔADC ) = Area( ΔAEB ).
Hence Proved.

(ii) We know that the area of triangles on the same base and between the same parallel lines are equal.
Area( ΔDBC )= Area( ΔBCE )
Area( ΔDOB ) + Area( ΔBOC ) = Area( ΔBOC ) + Area( ΔCOE )
So, Area( ΔDOB ) = Area( ΔCOE ).

Question 12

ABCD and BCFE are parallelograms. If area of triangle EBC = 480 cm²; AB = 30 cm and BC = 40 cm.

Calculate : 
(i) Area of parallelogram ABCD;
(ii) Area of the parallelogram BCFE;
(iii) Length of altitude from A on CD;
(iv) Area of triangle ECF.

Sol:

(i) Since ΔEBC and parallelogram ABCD are on the same base BC and between the same parallels i.e. BC // AD.

∴ A( ΔEBC ) = ${\displaystyle \frac{1}{2}}$ x A( parallelogram ABCD )

parallelogram ABCD = 2 x A( ΔEBC )
                                  = 2 x 480 cm²
                                  = 960 cm²
(ii) Parallelograms on same base and between same parallels are equal in area.
Area of BCFE = Area of ABCD = 960 cm²

(iii) Area of triangle ACD=480 = ${\displaystyle \frac{1}{2}}$ x 30 x Altitude
Altitude = 32 cm

(iv) The area of a triangle is half that of a parallelogram on the same base and between the same parallels.
Therefore,
Area( ΔECF ) = ${\displaystyle \frac{1}{2}}$ Area(${\displaystyle \square }$CBEF )
Similarly, Area( ΔBCE ) = ${\displaystyle \frac{1}{2}}$Area(${\displaystyle \square }$CBEF )

⇒ Area( ΔECF ) = Area( ΔBCE ) = 480 cm².

Question 13

In the given figure, D is mid-point of side AB of ΔABC and BDEC is a parallelogram.

Prove that: Area of ABC = Area of // gm BDEC.

Sol:

Here AD = DB and EC = DB, therefore EC = AD
Again, 
∠EFC = ∠AFD         .....( Opposite angles )

Since ED and CB are parallel lines and AC cut this line, therefore
∠ECF = ∠FAD 
From the above conditions, we have
ΔEFC = ΔAFD
Adding quadrilateral CBDF in both sides, we have
Area of // gm BDEC = Area of ΔABC.

Question 14

In the following, AC // PS // QR and PQ // DB // SR.

Prove that: Area of quadrilateral PQRS = 2 x Area of the quad. ABCD.

SoL:

In Parallelogram PQRS,
AC // PS // QR and PQ // DB // SR.

Similarly, AQRC and APSC are also parallelograms.

Since ΔABC and parallelogram AQRC are on the same base AC and between the same parallels, then
A( ΔABC ) = ${\displaystyle \frac{1}{2}}$A(AQRC)  ......(i)

Similarly,
A( ΔADC ) = ${\displaystyle \frac{1}{2}}$A( APSC ) .......(ii)

Adding (i) and (ii), we get
Area of quadrilateral PQRS = 2 x Area of the quad. ABCD.

Question 15

ABCD is a trapezium with AB // DC. A line parallel to AC intersects AB at point M and BC at point N.
Prove that: area of Δ ADM = area of Δ ACN.

Sol:

Given: ABCD is a trapezium.

AB || CD, MN || AC
Join C and M

We know that the area of triangles on the same base and between the same parallel lines are equal.
So Area of ΔAMD = Area of ΔAMC

Similarly, consider the AMNC quadrilateral where MN || AC.
ΔACM and ΔACN are on the same base and between the same parallel lines. So areas are equal.

So, Area of ΔACM = Area of ΔCAN
From the above two equations, we can say
Area of ΔADM = Area of ΔCAN

Hence Proved.

Question 16

In the given figure, AD // BE // CF.
Prove that area (ΔAEC) = area (ΔDBF)

SOl:

We know that the area of triangles on the same base and between the same parallel lines are equal.

Consider ABED quadrilateral; AD || BE.
With the common base, BE and between AD and BE parallel lines, we have
Area of ΔABE = Area of ΔBDE

Similarly, in BEFC quadrilateral, BE || CF
With common base BC and between BE and CF parallel lines, we have
Area of ΔBEC = Area of ΔBEF

Adding both equations, we have
Area of ΔABE + Area of ΔBEC = Area of ΔBEF + Area of ΔBDE
⇒ Area of AEC = Area of DBF

Hence Proved.

Question 17

In the given figure, ABCD is a parallelogram; BC is produced to point X.
Prove that: area ( Δ ABX ) = area (${\displaystyle \square }$ACXD )

Sol:

Given: ABCD is a parallelogram.
We know that
Area of ΔABC = Area of ΔACD
Consider ΔABX,
Area of ΔABX = Area of ΔABC + Area of ΔACX
We also know that the area of triangles on the same base and between the same parallel lines are equal.
Area of ΔACX = Area of ΔCXD
From the above equations, we can conclude that
Area of ΔABX = Area of ΔABC + Area of ΔACX
= Area of ΔACD+ Area of ΔCXD
= Area of ACXD Quadrilateral

Hence Proved.

Question 18

The given figure shows the parallelograms ABCD and APQR.
Show that these parallelograms are equal in the area.
[ Join B and R ]

Sol:

Join B and R and P and R.
We know that the area of the parallelogram is equal to twice the area of the triangle if the triangle and the parallelogram are on the same base and between the parallels.
Consider ABCD parallelogram:

Since the parallelogram ABCD and the triangle ABR lie on AB and between the parallels AB and DC, we have
Area(${\displaystyle \square }$ABCD ) = 2 x Area( ΔABR )        ....(1)

We know that the area of triangles with the same base and between the same parallel lines are equal.
Since the triangles ABR and APR lie on the same base AR and between the parallels AR and QP, we have,
Area ( ΔABR ) = Area ( ΔAPR )               ....(2)

From equations (1) and (2), we have,
Area(${\displaystyle \square }$ABCD) = 2 x Area( ΔAPR )        .....(3)

Also, the triangle APR and the parallelograms, AR and QR, lie on the same base AR and between the parallels, AR and QP,
Area( ΔAPR ) = ${\displaystyle \frac{1}{2}}$ x Area(${\displaystyle \square }$ARQP )    ....(4)

Using (4) in equation (3), We have,
Area(${\displaystyle \square }$ABCD ) = 2 x ${\displaystyle \frac{1}{2}\times \text{Area}(\square }$ARQP )

Area( ${\displaystyle \square \text{ABCD})=\text{Area}(\square }$ ARQP)
Hence Proved.

SChand Composite Mathematics Class 7 Chapter 16 Chance and Probability Exercise 16A

  Exercise 16 A

Question 1

In the following events, match correctly to indicate whether the outcomes are possible, certain, or impossible.

(1) You will throw a 7 with a normal die.

(2) Your friend will call you tonight.

(3) A cat will give birth to puppies next year.

(4) Thursday will be the day after Wednesday next week.

(5) You will have cornflakes, samosas and toasts in your breakfast today.

(6) It will be winter in Australia when it is summer in India.

(7) You will be able to see a live dinosaur in the city zoo.

(8) You will be awarded a prize for your good performance.

Ans: 

$\begin{array}{|c|c|c|c|c|c|c|c|}\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline b & a & b & c & a & c & b & a \\\hline\end{array}$

Question 2

Categorize each outcome as likely or unlikely.

(1) You will not get reservation in the train due to heavy Dusshra  rush.

(2) Your friend will go to moon next month.

(3) Someone in your class will be absent next week.

(4) It will snow in mussorie in January.

(5) There will be floods in Delhi in March next year.

(6) You will become an army officer when you grow up.

$\begin{aligned}&\text { Ans. }\\&\begin{array}{|c|c|c|c|c|c|}\hline 1 & 2 & 3 & 4 & 5 & 6 \\\hline \mathbf{a} & \mathrm{b} & \mathrm{a} & \mathrm{a} & \mathrm{b} & \mathrm{a} \\\hline\end{array}\end{aligned}$

Question 3

Choose the likelihood which matches the outcome of each event:

(1) The school football team will win the local toumament.

(2) Mr. Shah will cut the grass on his lawn when it is snowing.

(3) While going to school, you will pass by a white car.

(4) The mountaineer will be hurt if he falls off the mountain.

(5) The number on the top face of an ordinary die will be an even number.

(6) The next baby to be born will be a female.

(7) The number on the top face of an ordinary die when rolled will be less than 7 .

(8) The Titanic will float back up to the top of the ocean.

(9) You will live for 100 years.

(10) A dog will have kittens.

(11) Monday will be the day after Tuesday next week.

(12) You will have a birthday next year.

(a) Impossible

(b) Unlikely

(c) Even Chance

(d) Likely

(c) Certain

$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|}\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\\hline \mathrm{d} & \mathrm{b} & \mathrm{d} & \mathrm{d} & \mathrm{c} & \mathrm{c} &\mathrm{e} & \mathrm{a} & \mathrm{b} & \mathrm{a} & \mathrm{a} & \mathrm{e} \\\hline\end{array}$

Question 4

Multiple Choice Questions (MCQs)

 The probability of being chosen for a team is $9 \%$. This event can be described as:

(a) likely

(b) certain

(c) having an even chance

(d) unlikely

Question 5

 It will be a Thursday in one of the next 7 days.

The event is

(a) likely

(b) unlikely

(c) certain

(d) having an even chance.

Question 6

High Order Thinking Skills (HOTS)

A person with $\mathrm{O}, \mathrm{A}, \mathrm{B}$ or $\mathrm{AB}$ blood group can safely donate blood to a person with $\mathrm{AB}$ positive blood group.

Describe the event "a randomly chosen person could donate blood to a person with $\mathrm{AB}$ positive blood group" as certain or likely or unlikely or impossible.- Certain 

S Chand Class 10 CHAPTER 16 TRIGONOMETRY Exercise 16A

 Exercise 16A

Question 1

Ans: $\frac{1-\cos ^{2} \theta}{\sin ^{2} \theta}=1$

$L \cdot H \cdot S=\frac{1-\cos ^{2} \theta}{\sin ^{2} \theta}$ $\left[\because 1-\cos ^{2} \theta=\sin ^{2} \theta\right]$

$=\frac{\sin ^{2} \theta}{\sin ^{2} \theta}=1=R \cdot \mathrm{H} \cdot \mathrm{S}$

Question 2

Ans: $\frac{1-\sin ^{2} \theta}{\cos ^{2} \theta}$=1

L.H.S $=\frac{1-\sin ^{2} \theta}{\cos ^{2} \theta} \quad\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$

$=\frac{\cos ^{2} \theta}{\cos ^{2} \theta}=1=R \cdot H \cdot S$

Question 3

Ans: $\sin A \cdot \cot A=\cos A$

L.H.S=$\sin A \cdot \cot A$

$=\sin A \cdot \frac{\cos A}{\sin A}$  $\left[\cot A=\frac{\cos A}{\sin A}\right]$

$=\cos A=R . H .S$

Question 4

Ans: $\frac{1}{\cos ^{2} \theta}-\tan ^{2} \theta=1$

$L . H . S=\frac{1}{\cos ^{2} \theta}-\tan ^{2} \theta \quad\left[\frac{1}{\cos ^{2} \theta}=\sec ^{2} \theta\right]$

$=\sec ^{2} \theta-\tan ^{2} \theta$

$=1=R \cdot H \cdot S$

Question 5

Ans: $\tan ^{2} \cos ^{2} A=1-\cos ^{2} A$

$L \cdot H \cdot S=\tan ^{2} A \cos ^{2} A$ $\left[\tan ^{2} A=\frac{\sin ^{2} A}{\cos ^{2} A}\right]$

$=\frac{\sin ^{2} A}{\cos ^{2} A} \cos ^{2} A$

$=\sin ^{2} A=1-\cos ^{2} A=R \cdot H \cdot S$

Question 6

Ans: $\tan \theta=\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$

 $\Rightarrow R . H . S=\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$ $\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$

$=\frac{\sin \theta}{\sqrt{\cos ^{2} \theta}}$

$=\frac{\sin \theta}{\cos \theta}=\tan \theta=$ L.H.S

Question 7

Ans: $\frac{1+\cos \theta}{\sin ^{2} \theta}=\frac{1}{1-\cos \theta}$

L.H.S $=\frac{1+\cos \theta}{\sin ^{2} \theta}=\frac{1+\cos \theta}{1-\cos ^{2} \theta}\left[\sin ^{2} \theta=1-\cos ^{2} \theta\right]$

$=\frac{1+\cos \theta}{(1-\cos \theta)(1-\cos \theta)}\left[a^{2}-b^{2}=(a-b)\left(a+
 b\right)\right.$

$=\frac{1}{1-\cos \theta}=$ R.H.S

Question 8

Ans: $\cot ^{2} \theta\left(1-\cos ^{2} \theta\right)=\cos ^{2} \theta \quad\left[\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$

$L \cdot H \cdot S=\cot ^{2} \theta\left(1-\cos ^{2} \theta\right)$ $\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right]$

$=\frac{\cos ^{2} \theta}{\sin ^{2} \theta} \times \sin ^{2} \theta$

$=\cos ^{2} \theta$ = R.H.S

Question 9

Ans: $\tan ^{2} \theta\left(1-\sin ^{2} \theta\right)=\sin ^{2} \theta$

L.H.S $=\tan ^{2} \theta\left(1-\sin ^{2} \theta\right)$

$\begin{array}{ll}=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \cos ^{2} \theta & {\left[\tan \theta=\frac{\sin \theta}{\cos \theta}\right]} \\ & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]}\end{array}$

$=\sin ^{2} \theta=R \cdot H \cdot S$

Question 10

Ans: $\left(1-\sin ^{2} \theta\right) \operatorname{sic}^{2} \theta=1$

L.H.S $=\left(1-\sin ^{2} \theta\right) \sec ^{2} \theta \quad\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]$

$=\cos ^{2} \theta \cdot \sec ^{2} \theta$

$=1=$ R.H.S

Question 11

Ans: 

$\begin{aligned}\left(1-\cos ^{2} \theta\right) \operatorname{cosec}^{2} \theta &=1 \\ L \cdot H \cdot S=\left(1-\cos ^{2} \theta\right) & \operatorname{cosec}^{2} \theta \\ 1-\cos ^{2} \theta &=\sin ^{2} \theta \\ \operatorname{cosec} \theta &=\frac{1}{\sin \theta} \\ &=\sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta} \\ 1^{-} &=R \cdot H \cdot S \end{aligned}$

Question 12

Ans: $\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}=1$ 

$L \cdot H \cdot S=\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}$

$=\sin ^{2} \theta+\frac{1}{\sec ^{2} \theta} \quad\left[1+\tan ^{2} \theta=\sec ^{2} \theta\right]$

$\doteq \sin ^{2} \theta+\cos ^{2} \theta$ $\left[\frac{1}{\sec ^{2} \theta}=\cos ^{2} \theta\right]$

$R \cdot H .S=1$

Question 13

Ans:  $\cos ^{2} \theta+\frac{1}{1+\cot ^{2} \theta}=1$

$L \cdot H S=\cos ^{2} \theta+\frac{1}{1+\cot ^{2} \theta}$

$\left[1+\cot ^{2} \theta=\cos \operatorname{sc}^{2} \theta\right]$

$=\cos ^{2} \theta+\frac{1}{\operatorname{cosec}^{2} \theta}$

$\left\{\frac{1}{\operatorname{cosec}}=\sin \theta\right.$

$=\cos ^{2} \theta+\sin ^{2} \theta$

$R \cdot H \cdot S=1$

Question 14

Ans: $\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sec ^{2} \theta-\tan ^{2} \theta}=1$

$L \cdot H \cdot S=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sec ^{2} \theta-\tan ^{2} \theta}=1$

$\sin ^{2} \theta+\cos ^{2} \theta=1$

$\sec ^{2} \theta-\tan ^{2} \theta=1$

$R \cdot H \cdot S=1$

Question 15

Ans: $\left[\frac{\cos ^{2} A}{\sin ^{2} A}+1\right] \tan ^{2} A=\frac{1}{\cos ^{2} A}$

$L \cdot H \cdot S=\left[\frac{\cos ^{2} \cdot A}{\sin ^{2} A}+1\right] \tan ^{2} \theta$

$\frac{\cos A}{\sin A}=\cot A$

$=\left(\cot ^{2} A+1\right) \tan ^{2} \theta$

$\quad \cos ^{2} A+1=\operatorname{cosec}^{2} A$

$=\operatorname{cosec}^{2} A \times \tan ^{2} A$

$\quad \operatorname{cosec} A=\frac{1}{\sin A}$

$=\frac{1}{\sin ^{2} A} \times \frac{\sin ^{2} A}{\cos ^{2} A}$

$R \cdot H \cdot S=\frac{1}{\cos ^{2} A}$

Question 16

Ans: $\sin ^{2} \theta+\sin ^{2} \theta \cos ^{2} \theta=\sin ^{2} \theta .$

$\begin{aligned} L \cdot H \cdot S=& \sin ^{4} \theta+\sin ^{2} \theta \cos ^{2} \theta \\=& \sin ^{2} \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right.\\ & \sin ^{2} \theta+\cos ^{2} \theta=1 \\=& \sin ^{2} \theta \times 1 \\ R \cdot H \cdot S=& \sin ^{2} \theta \end{aligned}$

Question 17

Ans: $\sin ^{4} \theta+2 \sin ^{2} \theta \cos ^{2} \theta+\cos 4 \theta=1$

$\begin{aligned} \text { L.H.S }=& \sin ^{2} \theta+2 \sin ^{2} \theta(a)^{2} \theta+\cos ^{4} \theta \\ & a^{2}+2 a b+b^{2}=(a+b)^{2} \\=&\left(\sin ^{2} \theta\right)^{2}+2 \sin ^{2} \theta \cos ^{2} \theta+\left(\cos ^{2} \theta\right)^{2} \\=&\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2} \\ & \sin ^{2} \theta+\cos ^{2} \theta=1 \\=&(1)^{2} \end{aligned}$

R.H.S= 1

Question 18

Ans: $\sin 4 A \operatorname{cosec}^{2} A+\cos 4 A \sec ^{2} A=1$

L.H.S $=\sin 4 A \operatorname{cosec}^{2} A+\cos ^{4} A \sec ^{2} A$

$=\sin 4 A \times \frac{1}{\sin ^{2} A}+\cos ^{4} A \times \frac{1}{\cos ^{2} A}$  $\left[\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$

$\therefore \sin ^{2} \theta+\cos ^{2} \theta=1$ $\left[\sec \theta=\frac{1}{\cos \theta}\right]$

R.H.S =1

Question 19

Ans: $\sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A=1$

$\begin{aligned} L \cdot H \cdot S=& \sin ^{2} A \cot ^{2} A+\cos ^{2} A \tan ^{2} A \\=& \sin ^{2} A \times \frac{\cos ^{2} A}{\sin ^{2} A}+\cos ^{2} A \times \frac{\sin ^{2} A}{\cos ^{2} A} \\ & \frac{\cos A}{\sin A}=\cot \theta, \frac{\sin \theta}{\cos \theta}=\tan \theta \\=& \cos ^{2} A+\sin ^{2} A \cdot=1 \\ \text { R.H.S }=& 1 \end{aligned}$

Question 20

Ans: $\tan \theta+\cot \theta=\sec \theta \cdot \operatorname{cosec} \theta \cdot .$

$L \cdot H \cdot S=\tan \theta+\cot \theta$

${\left[\tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta=\frac{\cos \theta}{\sin \theta}\right] }$

$=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$

$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos \theta \sin \theta}$

$=\left[\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta}\right]$

$=\frac{1}{\cos \theta \sin \theta}$

${\left[\frac{1}{\cos \theta}=\sec \theta, \frac{1}{\sin \theta}=\operatorname{cosec} \theta\right] }$

$R \cdot H \cdot S=\operatorname{Sec} \theta \cdot \operatorname{cosec} \theta$

Question 21

Ans:  $(\tan A+\cot A) \sin A \cos A=1$

$L \cdot H \cdot S=(\tan A+\cot A)(\sin A \cos A)$

$\begin{aligned} & {\left[\tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta \doteq \frac{\cos \theta}{\sin \theta}\right] } \\=& {\left[\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right] \sin A \cos A . } \\=& \frac{\sin ^{2} A+\cos ^{2} A}{\cos A \sin A} \times \sin A \cos A \\ & {\left[\sin ^{2} \theta+\operatorname{Cos}^{2} \theta=1\right] }\end{aligned}$

$=\frac{1}{\cos A \sin A} \times \sin A \cos A$

R.H.S = 1

Question 22

Ans:   $\begin{aligned} \frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\cot \theta & \\ L \cdot H \cdot S=& \frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)} \\=& \frac{1-\sin ^{2} \theta+\cos \theta}{\sin \theta(1+\cos \theta)} \\ & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right] } \\=& \frac{\cos \theta+\cos ^{2} \theta}{\sin \theta(1+\cos \theta} \end{aligned}$

$=\frac{\cos \theta(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$

$=\frac{\cos \theta}{\sin \theta}$

${\left[\frac{\cos \theta}{\sin \theta}=\cot \theta\right] }$

$R \cdot H \cdot S=\cot \theta$

Question 23

Ans: 

 $\frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}=2 \operatorname{cosec}^{2} \theta$

$\begin{aligned} \text { L.H.S } &=\frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta} \\ &=(a+b)(a-b)=a^{2}-b^{2} \\ &=\frac{1+\cos \theta+1-\cos \theta}{(1-\cos \theta)(1+\cos \theta)}=\frac{2}{1-\cos ^{2} \theta} \end{aligned}$

$\begin{aligned} & {\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right] } \\=& \frac{2}{\sin ^{2} \theta} . \\ & {\left[\frac{1}{\sin \theta}=\operatorname{cosec} \theta\right] }\end{aligned}$

$R \cdot H \cdot S=2 \operatorname{cosec}^{2} \theta$

Question 24

Ans:  $\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}=\tan ^{2} \theta .$

$L \cdot H \cdot S=\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}=\frac{1-\tan ^{2} \theta}{\frac{1}{\tan ^{2} \theta}-1}$ 

$=\frac{1-\tan ^{2} θ}{\frac{1-\tan ^{2} θ }{\tan ^{2} \theta}}$

$=\tan ^{2}\theta=R.H.S$

Question 25

Ans:  $\frac{1}{\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta}$

$L \cdot H \cdot S=\frac{1}{\operatorname{Sec} \theta+\tan \theta}$

$=\frac{\sec \theta-\tan \theta}{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}$

$=\frac{\sec \theta-\tan \theta}{\sec ^{2} \theta-\tan ^{2} \theta} \quad\left[\sec ^{2} \theta-\tan ^{2} \theta=1\right]$

$=\frac{\sec \theta-\tan \theta}{1}$

$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$

R.H.S=$ \frac{1-\sin \theta}{\cos \theta}$

Question 26

Ans: $(\operatorname{cosec} A-\sin A) \cdot(\sec A-\cos A)(\tan A+\cot A)=1$

L.H.S $=(\operatorname{cosec} A-\sin A) \cdot(\sec A-\cos A)(\tan A+\cot A)$

$\left[\operatorname{cosec} A=\frac{1}{\sin A}, \sec A=\frac{1}{\cos A}, \tan A=\frac{\sin A}{\cos A}\right.$

$\left.\cot A=\frac{\cos A}{\sin A}\right]$

$=\left[\frac{1}{\sin A}-\sin A\right]\left[\frac{1}{\cos A}-\cos A\right]\left[\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right]$

$=\frac{1-\sin ^{2} A}{\sin A} \times \frac{1-\cos ^{2} A}{\cos A} \times \frac{\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}$

$=\frac{\left[\sin ^{2} A+\cos ^{2} A=1\right]}{\sin A} \times \frac{\sin ^{2} A}{\cos A} \times \frac{1}{\sin A \cos A}$

$=\frac{\cos ^{2} A \sin ^{2} A}{\sin ^{2} A \cos ^{2} A}$

R.H.S $=1$

Question 27

Ans: $\frac{1+\sin \theta}{1-\sin \theta}=(\sec \theta+\tan \theta)^{2}$

$\begin{aligned} \text { L.H.S. } &=\frac{1+\sin \theta}{1-\sin \theta} \\ &=\frac{(1+\sin \theta)(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)} \\ &=\frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta} \end{aligned}$

$\begin{aligned} & {\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right] } \\=& \frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta} \\=& {\left[\frac{1+\sin \theta}{\cos \theta}\right]^{2} } \\=& {\left[\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right]^{2} }\end{aligned}$

$\frac{1}{\cos \theta}=\sec \theta, \frac{\sin \theta}{\operatorname{con} \theta}=\tan \theta$

$R \cdot H \cdot S=(\sec \theta+\tan \theta)^{2}$

Question 28

Ans: $\frac{1+\cos \theta}{1-\cos \theta}=(\operatorname{cosec} \theta+\cot \theta)^{2}$

$\begin{aligned} L \cdot H \cdot S &=\frac{1+\cos \theta}{1-\cos \theta} \\ &=\frac{(1+\cos \theta)(1+\cos \theta)}{(1-\cos \theta)(1+\cos \theta)} \cdot\left[1-\cos ^{2} \theta=\sin ^{2} \theta\right] \end{aligned}$

$=\frac{\left(1+\cos (\theta)^{2}\right.}{1-\cos ^{2} \theta}$

$=\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}$

$=\left[\frac{1+\cos \theta}{\sin \theta}\right]^{2}$

$=\left[\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\right]^{2}$

$\left(\frac{1}{\sin \theta}=\operatorname{cosec} \theta, \frac{\cos \theta}{\sin \theta}=\cot \theta\right)$

$R \cdot H \cdot S=(\cos \theta+\theta+\cot \theta)^{2}$

Question 29

Ans: $\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}=\operatorname{cosec} \theta+\cot \theta-\frac{1+\cos \theta}{\sin \theta}$

$L \cdot H \cdot S=\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{\cot A+\operatorname{cosec} \theta-\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{\cot \theta+\operatorname{cosec} \theta+\left(\cot ^{2} \theta-\operatorname{cosec} 2 \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{(\cot \theta+\operatorname{cosec} \theta)+\cot \theta+\operatorname{cosec} \theta)(\cot \theta-\operatorname{cosec} \theta)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{(\cot \theta+\operatorname{cosec} \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{\cot \theta-\operatorname{cosec} \theta+1}$

$=\frac{(\operatorname{cotc}+\operatorname{cosec} \theta)(\cot \theta-\operatorname{cosec} \theta+1)}{(\cot \theta-\operatorname{cosec} \theta+1}$

$=\cot \theta+\operatorname{cosec} \theta$

$=\operatorname{cosec} \theta+\cot \theta$

$=\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta} .$

$R \cdot H \cdot S=\frac{1+\cos \theta}{\sin \theta}$

Question 30

Ans: $\tan \theta+\cot \theta=2$

$(\tan \theta+\cot \theta)^{2}=4 \quad$ [Square both side]

$\tan ^{2} \theta+\cot ^{2} \theta+2 \tan \theta \cot \theta=4$

${[\tan \theta \cdot \cot \theta=1] }$

$\begin{aligned} \tan ^{2} \theta+\cot ^{2} \theta+2 \times 1 &=4 \\ \tan ^{2} \theta+\cot ^{2} \theta+2 &=4 \\ \tan ^{2} \theta+\cot ^{2} \theta &=4-2 \\ &=2 \\ \tan ^{2} \theta+\cot ^{2} \theta &=2 \end{aligned}$

Question 31

Ans:  $a^{2}+b^{2}=(\sin \theta+\cos \theta)^{2}+(\sin \theta-\cos \theta)^{2}$

$=-\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta+\sin ^{2} \theta+\cos ^{2} \theta-2 \sin \epsilon \cos \theta$

$=2 \sin ^{2} \theta+2 \cos ^{2} \theta$

$=2\left(\sin ^{2} \theta+\left(\cos ^{2} \theta\right)\right.$

$=2 \times 1$

=2

RS Aggarwal solution class 8 chapter 16 Parallelograms Exercise 16A

Exercise 16A

Page-193

Q1 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 1:

ABCD is a parallelogram in which ∠A = 110°. Find the measure of each of the angles ∠B, ∠C and ∠D.

Answer 1:

Q2 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 2:

Two adjacent angles of a parallelogram are equal. What is the measure of each of these angles?

Answer 2:

Q3 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 3:

Two adjacent angles of a parallelogram are in the ratio 4 : 5. Find the measure of each of its angles.

Answer 3:

                                 

Q4 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 4:

Two adjacent angles of a parallelogram are (3x − 4)° and (3x + 16)°. Find the value of x and hence find the measure of each of its angles.

Answer 4:

Q5 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 5:

The sum of two opposite angles of a parallelogram is 130°. Find the measure of each of its angles.

Answer 5:

Q6 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 6:

Two sides of a parallelogram are in the ratio 5 : 3. If its perimeter is 64 cm, find the lengths of its sides.

Answer 6:

                             
 

Q7 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 7:

The perimeter of a parallelogram is 140 cm. If one of the sides is longer than the other by 10 cm, find the length of each of its sides.

Answer 7:

                           
                        

 

Q8 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 8:

In the adjacent figure, ABCD is a rectangle. If BM and DN are perpendiculars from B and D on AC, prove that ∆BMC ≅ ∆DNA. Is it true that BM = DN?

Answer 8:

Refer to the figure given in the book.

Q9 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 9:

In the adjacent figure, ABCD is a parallelogram and line segments AE and CF bisect the angles A and C respectively. Show that AE||CF.

Answer 9:

Refer to the figure of the book.

Q10 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 10:

The lengths of the diagonals of a rhombus are 16 cm and 12 cm respectively. Find the length of each of its sides.

Answer 10:

Q11 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 11:

In the given figure ABCD is a square. Find the measure of ∠CAD.

Answer 11:

Refer to the figure given in the book.

Q12 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 12:

The sides of a rectangle are in the ratio 5 : 4 and its perimeter is 90 cm. Find its length and breadth.

Answer 12:

                            

Q13 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 13:

Name each of the following parallelograms.
(i) The diagonals are equal and the adjacent sides are unequal.
(ii) The diagonals are equal and the adjacent sides are equal.
(iii) The diagonals are unequal and the adjacent sides are equal.
(iv) All the sides are equal and one angle is 60°.
(v) All the sides are equal and one angle is 90°.
(vi) All the angles are equal and the adjacent sides are unequal.

Answer 13:

Page-194

Q14 | Ex-16A | Class 8 | RS AGGARWAL | chapter 16 | Parallelograms 

OPEN IN YOUTUBE

Question 14:

Which of the following statements are true and which are false?
(i) The diagonals of a parallelogram are equal.
(ii) The diagonals of a rectangle are perpendicular to each other.
(iii) The diagonals of a rhombus are equal.
(iv) Every rhombus is a kite.
(v) Every rectangle is a square.
(vi) Every square is a a parallelogram.
(vii) Every square is a rhombus.
(viii) Every rectangle is a parallelogram.
(ix) Every parallelogram is a rectangle.
(x) Every rhombus is a parallelogram.

Answer 14: