SELINA Solution Class 9 Chapter 11 Inequalities Exercise 11

Question 1

From the following figure, prove that: AB > CD.

Sol:

In ΔABC,
AB = AC                 ...[ Given ]
∴ ∠ACB = ∠B ...[ angles opposite to equal sides are equal ]
∠B = 70° ...[ Given ]
⇒ ∠ACB = 70°      ...(i)

Now,
∠ACB +∠ACD = 180°...[ BCD is a straight line]
⇒ 70° + ∠ACD = 180°
⇒ ∠ACD = 110°        ...(ii)

In ΔACD,
∠CAD + ∠ACD + ∠D = 180°
⇒ ∠CAD + 110° + ∠D = 180° ...[ From (ii) ]
⇒ ∠CAD + ∠D = 70°

But ∠D = 40°        ...[ Given ]
⇒ ∠CAD + 40°= 70°
⇒ ∠CAD = 30°        …(iii)

In ΔACD,
∠ACD = 110°         ...[ From (ii) ]
∠CAD = 30°           ...[ From (iii) ]
∠D = 40°                  ...[ Given ]

∴ D > ∠CAD
⇒ AC > CD          ....[Greater angle has greater side opposite to it]

Also,
AB = AC              ...[ Given ]
Therefore, AB > CD.

Question 2

In a triangle PQR; QR = PR and ∠P = 36^o. Which is the largest side of the triangle?

Sol:

In ΔPQR,
QR = PR      ...[ Given ]
∴ ∠P = ∠Q  ...[ angles opposite to equal sides are equal ]
⇒ ∠P = 36° ..[Given]
⇒ ∠Q = 36°

In ΔPQR,
∠P + ∠Q + ∠R = 180°
⇒ 36° + 36° + ∠R = 180°
⇒ ∠R + 72° = 180°
⇒ ∠R = 108°

Now,
∠R = 108°
∠P = 36°
∠Q = 36°

Since ∠R is the greatest, therefore, PQ is the largest side.

Question 3

If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such that a is less than b.

Sol:

The sum of any two sides of the triangle is always greater than third side of the triangle.

Third side < 13 + 8 = 21 cm.

The difference between any two sides of the triangle is always less than the third side of the triangle.

Third side > 13 - 8 = 5 cm.

Therefore, the length of the third side is between 5 cm and 9 cm, respectively.

The value of a = 5 cm and  b =  21cm.

Question 4.1

In the following figure, write BC, AC, and CD in ascending order of their lengths.

Sol:

In ΔABC,
AB = AC
⇒ ∠ABC = ∠ACB  ..( angles opposite to equal sides are equal )
⇒ ∠ABC = ∠ACB = 67°
⇒ ∠BAC = 180° - ∠ABC - ∠ACB ..( angle sum property )
⇒ ∠BAC = 180° - 67° - 67° = 46°

Since ∠BAC < ∠ABC, we have
BC< AC                              ...( 1 )
Now, ∠ACD = 180° - ACB  ...( liner pair )
⇒ ∠ACD = 180° - 67° = 113°

Thus, in ΔACD,
∠CAD = 180°- ∠ACD - ∠ADC
⇒ ∠CAD = 180° - 113° - 33° = 34°
Since ∠ADC <  ∠CAD, we have
AC < CD                              ...( 2 )
 From ( 1 ) and ( 2 ), we have
BC < AC < CD.

Question 4.2

In the following figure, write BC, AC, and CD in ascending order of their lengths.

Sol:

In ΔABC,
∠BAC < ∠ABC
BC < AC                          ....( 1 )

Now, ∠ACB = 180° - ∠ABC - ∠BAC
∠ACB = 180° - 73° - 47°
∠ACB = 60°

Now, ∠ACD = 180° - ∠ACB
∠ACD = 180° - 60° = 120°

Now, in ΔACD,
∠ADC = 180° - ∠ACD - ∠CAD
∠ADC = 180° - 120° - 31°
∠ADC = 29°
Since ∠ADC < ∠CAD, we have 
AC < CD                          ....( 2 )
From ( 1 ) and ( 2 ), we have
BC < AC < CD.

Question 5

Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.

Sol:

∠BAC = 180° - ∠BAD = 180° - 137° = 43°
∠ABC = 180° - ∠ABE = 180° - 106° = 74°
Thus, in ΔABC,
∠ACB = 180° - ∠BAC - ∠ABC
⇒ ∠ACB = 180° - 43° - 74° = 63°
Now, ∠ABC = ∠OBC + ∠ABO
⇒ ∠ABC = 2∠OBC           ....( OB is biosector of ∠ABC )
⇒ 74° = 2∠OBC
⇒ ∠OBC = 37°
Similarly,
∠ACB = ∠OCB + ∠ACO
⇒ ∠ACB = 2∠OCB         ...( OC is bisector of ACB )
⇒ 63° = 2∠OCB
⇒ ∠OCB = 31.5°
Now, in ΔBOC,
∠BOC = 180° - ∠OBC - ∠OCB
⇒ ∠BOC = 180° - 37° - 31.5°
⇒ ∠BOC = 111.5°
Since, ∠BOC> ∠OBC > ∠OCB, we have
BC > OC > OB

Question 6

D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.

Sol:

AD > AC                ...( Given )
⇒ ∠C > ∠ADC        ...( 1 )
Now, ∠ADC > ∠B + ∠BAC ...( Exterior Angel Property )
⇒ ∠ADC > ∠B       ....( 2 )
From ( 1 ) and ( 2 ), we have
⇒ ∠C > ∠ADC > ∠B  
⇒ ∠C > ∠B  
⇒ AB > AC

Question 7

In the following figure, ∠BAC = 60^o and ∠ABC = 65^o.

Prove that:
(i) CF > AF
(ii) DC > DF

Sol:

In ΔBEC,
∠B + ∠BEC + ∠BCE = 180°
∠B = 65°                     ...[Given]
∠BEC = 90°                 ...[CE is perpendicular to AB]

⇒ 65° + 90° + ∠BCE = 180°
⇒ ∠BCE = 180° - 155°
⇒ ∠BCE = 25°= ∠DCF  …(i)

In ΔCDF,
∠DCF + ∠FDC + ∠CFD = 180°
∠DCF = 25°                 ....[From (i)]
∠FDC = 90°                 ...[ AD is perpendicular to BC]

⇒ 25°+ 90°+ ∠CFD = 180°
⇒ ∠CFD = 180° - 115°
⇒ ∠CFD = 65°             …(ii)

Now, ∠AFC + ∠CFD = 180°  ....[AFD is a straight line]
⇒ ∠AFC + 65° = 180°
⇒ ∠AFC = 115°                     …(iii)

In ΔACE,
∠ACE + ∠CEA + ∠BAC = 180°
∠BAC = 60°                             ....[Given]

⇒ ∠CEA = 90°        ...[CE is perpendicular to AB]
⇒ ∠ACE + 90° + 60° = 180°
⇒ ∠ACE = 180° - 150°
∠ACE = 30°                      …(iv)

In ΔAFC,
∠AFC + ∠ACF + ∠FAC = 180°
∠AFC = 115°                    ....[From (iii)]
∠ACF = 30°                   ...[From (iv)]

⇒ 115° + 30° + ∠FAC = 180°
⇒ ∠FAC = 180° - 145°
⇒ ∠FAC = 35°               …(v)

In ΔAFC,
⇒ ∠FAC = 35°               ...[ From (v) ]
⇒ ∠ACF = 30°               ...[ From (iv) ]
∴ ∠FAC > ∠ACF
⇒ CF > AF

In Δ CDF,
∠DCF = 25°                ...[From (i)]
∠CFD = 65°                 ...[From (ii)]
∴ ∠CFD > ∠DCF
⇒ DC > DF

Question 8

In the following figure ; AC = CD; ∠BAD = 110^o and ∠ACB = 74^o.

Prove that: BC > CD.

SOl:

∠ACB = 74°                             ...(i)[ Given ]
∠ACB + ∠ACD = 180°                ....[ BCD is a straight line ]
⇒ 74° + ∠ACD = 180°
⇒ ∠ACD = 106°                        …..(ii)

In ΔACD,
∠ACD + ∠ADC+ ∠CAD = 180°
Given that AC = CD
⇒ ∠ADC= ∠CAD
⇒ 106° + ∠CAD + ∠CAD = 180°     ....[From (ii)]
⇒ 2∠CAD = 74°
⇒ ∠CAD = 37° = ∠ADC                  ...(iii)

Now,
∠BAD = 110°                              ....[Given]
∠BAC + ∠CAD = 110°
∠BAC + 37° = 110°
∠ BAC = 73°                               ….(iv)

In ABC,
∠ B + ∠BAC + ∠ACB = 180°
∠B + 73° + 74° = 180°              ...[From (i) and (iv)]
∠B + 147°= 180°
∠B = 33°                                   …..(v)

∴ ∠BAC > ∠B                            ...[ From (iv) and (v)]
⇒ BC > AC
But,
AC = CD                                   ...[ Given ]
⇒ BC > CD

Question 9

From the following figure;

prove that:
(i) AB > BD
(ii) AC > CD
(iii) AB + AC > BC.

Sol;

(i) ∠ADC + ∠ADB = 180°      ...[ BDC is a straight line ]
∠ADC = 90°                          ...[ Given ]
90° + ∠ADB = 180°
∠ADB = 90°                           ....(i)

In ΔADB,
∠ADB = 90°                           ....[ From (i) ]
∴ ∠B + ∠BAD = 90°
Therefore, ∠B and ∠BAD are both acute, that is less than 90°.
∴ AB > BD                                 ….(ii)[ Side opposite 90° angle is greater than the side opposite acute angle ]

(ii) In ΔADC,
∠ADB = 90°
∴ ∠C + ∠DAC = 90°
Therefore, ∠C and ∠DAC are both acute, which is less than 90°.
∴ AC > CD                       ...(iii)[ Side opposite 90° angle is greater than side opposite acute angle ]
Adding (ii) and (iii)
AB + AC > BD + CD
⇒ AB + AC > BC

Question 10

In a quadrilateral ABCD; prove that:
(i) AB+ BC + CD > DA
(ii) AB + BC + CD + DA > 2AC
(iii) AB + BC + CD + DA > 2BD

Sol:

Const: Join AC and BD.
(i) In ΔABC,
AB + BC > AC                       ….(i)[ Sum of two sides is greater than the third side ]
In ΔACD,
AC + CD > DA                      ….(ii)[ Sum of two sides is greater than the third side ]

Adding (i) and (ii)
AB + BC + AC + CD > AC + DA
AB + BC + CD > AC + DA - AC
AB + BC + CD > DA               ….(iii)

(ii) In ΔACD,
CD + DA > AC                   ….(iv) [Sum of two sides is greater than the third side]
Adding (i) and (iv)
AB + BC + CD + DA > AC + AC
AB + BC + CD + DA > 2AC

(iii) In ΔABD,
AB + DA > BD                   ….(v)[Sum of two sides is greater than the third side]
In ΔBCD,
BC + CD > BD                   ….(vi)[Sum of two sides is greater than the third side]
Adding (v) and (vi)
AB + DA + BC + CD > BD + BD
AB + DA + BC + CD > 2BD

Question 11.1

In the following figure, ABC is an equilateral triangle and P is any point in AC;
prove that: BP > PA

Sol:

In ΔABC,
AB = BC = CA       ...[ ABC is an equilateral triangle ]
∴ ∠A = ∠B = ∠C
∴ ∠A = ∠B = ∠C = ${\displaystyle \frac{180°}{3}}$

In ΔABP,
∠A = 60°
∠ABP< 60°
∴ ∠A > ∠ABP
⇒ BP > PA         ....[ Side opposite to greater side is greater ]

Question 11.2

In the following figure, ABC is an equilateral triangle and P is any point in AC;
prove that: BP > PC

Sol:

In ΔBPC,
∠C = 60^°
∠CBP < 60^°   
∴∠C > ∠CBP
⇒ BP > PC                  ....[ Side opposite to greater side is greater ]

Question 12

P is any point inside the triangle ABC.
Prove that: ∠BPC > ∠BAC.

SOl:

Let  ∠PBC = x and ∠PCB = y
then,
∠BPC = 180° - ( x + y )              …(i)
Let ∠ABP = a and ∠ACP = b
then,
∠BAC = 180° - ( x + a ) - ( y + b )
⇒ ∠BAC = 180^° - ( x + y ) - ( a + b )
⇒ ∠BAC = ∠BPC - ( a + b )
⇒ ∠BPC = ∠BAC + ( a + b )
⇒ ∠BPC > ∠BAC.

Question 13

Prove that the straight line joining the vertex of an isosceles triangle to any point in the base is smaller than either of the equal sides of the triangle.

Sol:

We know that the exterior angle of a triangle is always greater than each of the interior opposite angles.
∴ In ΔABD,
∠ADC > ∠B                   ...(i)
In ΔABC,
AB = AC
∴∠B = ∠C                     ...(ii)

From (i) and (ii)
∠ADC > ∠C

(i) In ΔADC,
∠ADC > ∠C
∴AC > AD                 ....(iii)[ side opposite to greater angle is greater ]

(ii) In ΔABC,
AB = AC
⇒ AB > AD               ...[ From (iii) ]

Question 14

In the following diagram; AD = AB and AE bisect angle A.

Prove that:
(i) BE = DE
(ii) ∠ABD > ∠C

SOl:

Const: Join ED.

In ΔAOB and ΔAOD,
AB = AD                 ...[ Given ]
AO = AO                ....[ Common ]
∠BAO = ∠DAO      ....[ AO is bisector of A ]
∴ ΔAOB ≅ ΔAOD  ....[ SAS criterion ]

Hence,
BO = OD                                                  …(i)[ c.p.c.t. ]
∠AOB = ∠AOD                                        …(ii)[ c.p.c.t. ]
∠ABO = ∠ADO ⇒ ∠ABD = ∠ADB            …(iii)[ c.p.c.t. ]

Now,
∠AOB = ∠DOE                ...[Vertically opposite angles]
∠AOD = ∠BOE               ...[Vertically opposite angles]
∠BOE = ∠DOE               …(iv)[ From (ii) ]

(i) In ΔBOE and ΔDOE,
BO = CD                               ...[ From (i) ]
OE = OE                               ...[ Common ]
∠BOE = ∠DOE                     ...[ From (iv) ][ SAS criterion ]
Hence, BE = DE                    ...[ c.p.c.t. ]

(ii) In BCD,
∠ADB = ∠C + ∠CBD        ...[ Ext. angle = sum of opp. int. angles ]
⇒ ∠ADB > ∠C
⇒ ∠ABD > ∠C                     ...[ From (iii) ]

Question 15

Const: Join ED.

In ΔAOB and ΔAOD,
AB = AD                 ...[ Given ]
AO = AO                ....[ Common ]
∠BAO = ∠DAO      ....[ AO is bisector of A ]
∴ ΔAOB ≅ ΔAOD  ....[ SAS criterion ]

Hence,
BO = OD                                                  …(i)[ c.p.c.t. ]
∠AOB = ∠AOD                                        …(ii)[ c.p.c.t. ]
∠ABO = ∠ADO ⇒ ∠ABD = ∠ADB            …(iii)[ c.p.c.t. ]

Now,
∠AOB = ∠DOE                ...[Vertically opposite angles]
∠AOD = ∠BOE               ...[Vertically opposite angles]
∠BOE = ∠DOE               …(iv)[ From (ii) ]

(i) In ΔBOE and ΔDOE,
BO = CD                               ...[ From (i) ]
OE = OE                               ...[ Common ]
∠BOE = ∠DOE                     ...[ From (iv) ][ SAS criterion ]
Hence, BE = DE                    ...[ c.p.c.t. ]

(ii) In BCD,
∠ADB = ∠C + ∠CBD        ...[ Ext. angle = sum of opp. int. angles ]
⇒ ∠ADB > ∠C
⇒ ∠ABD > ∠C                     ...[ From (iii) ]

Sol:

In ΔABC,
AB > AC,
⇒ ∠ABC < ∠ACB
∴ 180° - ∠ABC > 180° - ∠ACB
⇒ ${\displaystyle \frac{180°-\angle \text{ABC}}{2}>\frac{180°-\angle \text{ACB}}{2}}$

⇒ 90° - ${\displaystyle \frac{1}{2}\angle \text{ABC}>90°-\frac{1}{2}\angle \text{ACB}}$

⇒ ∠CBP > ∠BCP             ...[ BP is bisector of ∠CBD and CP is bisector of ∠BCE ]
⇒ PC > PB                      ...[ Side opposite to greater angle is greater ]

Question 16

In the following figure; AB is the largest side and BC is the smallest side of triangle ABC.
Write the angles x^o, y^o and z^o in ascending order of their values.

SOl:

Since AB is the largest side and BC is the smallest side of the triangle ABC.
AB > AC > BC
⇒ 180° - z° > 180° - y° > 180° - x°
⇒ - z° > -y° > - x°
⇒ z° > y° > x°.

Question 17.1

In quadrilateral ABCD, side AB is the longest and side DC is the shortest.
Prove that: C > A.

Sol:

In the quadrilateral ABCD,
Since AB is the longest side and DC is the shortest side.

∠1 > ∠2 [ AB > BC ]
∠ 7 > ∠4 [ AD > DC ]
∴ ∠1 + ∠7 > ∠2 + ∠4
⇒ ∠C > ∠A.

Question 17.2

In quadrilateral ABCD, side AB is the longest and side DC is the shortest.
Prove that: D > B.

Sol:

∠5 > ∠6 [ AB > AD ]
∠3 > ∠8 [ BC > CD ]
∴ ∠5 + ∠3 > ∠6 + ∠8
⇒ ∠D > ∠B

Question 18

In triangle ABC, side AC is greater than side AB. If the internal bisector of angle A meets the opposite side at point D,
prove that:  ∠ADC is greater than ∠ADB.

Sol:

In ΔADC,
∠ADB = ∠1 + ∠C                      ....(i)

In ADB,
∠ADC = ∠2 + ∠B                       ...(ii)
But AC > AB                               ....[ Given ]
⇒ ∠B > ∠C
Also given, ∠2 = ∠1                   ....[ AD is bisector of A ]
⇒ ∠2 + ∠B > ∠1 + ∠C               ...(iii)
From (i), (ii) and (iii)
⇒ ∠ADC > ∠ADB.

Question 19

In an isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that:
(i) AC > AD
(ii) AE > AC
(iii) AE > AD

Sol:

We know that the bisector of the angle at the vertex of an isosceles triangle bisects the base at the right angle.
Using Pythagoras theorem in AFB,
AB² = AF² + BF²                            ...(i)

In AFD,
AD² = AF² + DF²                          ...(ii)
We know ABC is isosceles triangle and AB = AC
AC² = AF² + BF²                        ..(iii)[ From (i)]

Subtracting (ii) from (iii)
AC² - AD² = AF² + BF² - AF² - DF²
AC² - AD² = BF² - DF²
Let 2DF = BF
AC² - AD² = (2DF)² - DF²
AC² - AD² = 4DF² - DF²
AC² = AD² + 3DF²
⇒ AC² > AD²
⇒ AC > AD
Similarly, AE > AC and AE > AD.

Question 20

Given: ED = EC
Prove: AB + AD > BC.

SOl:

The sum of any two sides of the triangle is always greater than the third side of the triangle.

ln ΔCEB,
CF + EB > BC
⇒ DE + EB > BC         ...[ CE = DE ]
⇒ DB > BC                 ...(i)

ln ΔADB,
AD+ AB > BD
⇒ AD + AB > BD > BC  ...[ from(i) ]
⇒ AD+ AB > BC

Question 21

In triangle ABC, AB > AC and D is a point inside BC.
Show that: AB > AD.

Sol:

Given that, AB > AC
⇒ ∠C > ∠B                   ...(i)
Also in ΔADC
∠ADB = ∠DAC + ∠C    ...[ Exterior angle ]
⇒ ∠ADB > ∠C 
⇒ ∠ADB > ∠C > ∠B     ...[ From(i) ] 
⇒ ∠ADB > ∠B  
⇒ AB > AD.

S.chand Class 8 Maths Solution Chapter 11 Time And Work Exercise 11

 Exercise 11

Question 1

1. 6 men can complete the electric fitting in a building in 7 days. How many days will it take 21 men to do the job?

2. Shashi weaves 25 baskets in 35 days. In how many days will she weave 110 baskets ?

3. Two men can do a piece of work in 3 days and 6 days respectively. If they work together, in how many days will they finish the work?

4. One man can do a piece of work in 3 hours ; another can do the same piece of work in 2 hours. How long will they take if they work together?

5. $A$ and $B$ together can do a piece of work in 10 days, but $A$ alone can do it in 15 days. How many days would $B$ alone take to do the same work ?

6. Two men can do a piece of work in 6 hours and 4 hours respectively. After the first has worked for 2 hours, he is joined by the other. By when should the work be completed ?

7. $A, B$ and $C$ can cultivate a field in 10,12 , and 15 days respectively. If they work together, in how many days will they finish the work and what fraction of the work will each of them do ?

8. $A, B$ and $C$ can do a piece of work in 10 days. $A$ alone can do it in 40 days, and $B$ alone can do it in 30 days. In how many days will $C$ alone do the same work?

9. $A$ and $B$ can do a given piece of work in 8 days; $B$ and $C$ can do the same work in 12 days and $A, B, C$ complete it in 6 days. In how many days can $A$ and $C$ finish it?

10. A cistern can be filled by one tap in 4 hours and by another in 3 hours. How long will it take to fill if both taps are opened together?

11. A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern if both taps are opened together?

12. One tap fills a bath in $12 \mathrm{~min}$. and another tap fills it in $15 \mathrm{~min}$. The waste-pipe can empty the bath in $10 \mathrm{~min}$. In what time will the bath be filled if both taps are turned on and if the waste-pipe has been left open accidentally ?

13. A pipe can fill a cistern in 6 hours. Due to a leak in the bottom it is filled in 7 hours. When the cistern is full, in how much time will it be emptied by the leak ?

14. A cistern has two inlets $A$ and $B$ which can fill it in 15 hours and 20 hours respectively. An outlet car empty the full cistern in 12 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely?

Multiple Choice Questions (MCQs)

15. $A$ and $B$ together take 5 days to do a work, $B$ and $C$ take 7 days to do the same and $A$ and $C$ take 4 days to do it. Who among these will take the least time, if put to do it alone?

(a) $A$

(b) $B$

(c) $C$

(d) data is not adequate

16. A tap can empty a tank in one hour. A second tap can empty it in 30 minutes. If both the taps operate simultaneously, how much time is needed to empty the tank?

(a) 20 min.

(b) 30 min.

(c) 40 min.

(d) 45 min.

High Order Thinking Skills (HOTS)

17. $A$ and $B$ can finish a work in 30 days. They worked for it for 20 days and then $B$ left. The remaining work was done by $A$ alone in 20 more days. $A$ alone can finish the work in

(a) 54 days

(b) 60 days

(c) 48 days

(d) 50 days

18. Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. The tank is half full. All the three pipes are in operation simultaneously. After how much time will the tank will be full.

RS AGGARWAL CLASS 9 CHAPTER 11 AREAS OF PARALLELOGRAMS AND TRIANGLES EXERCISE 11

 EXERCISE 11 

PAGE NO-387

Question 1:

Which of the following figures lie on the same base and between the same parallels. In such a case, write the comon base and the two parallels.

Answer 1:

(i) No, it doesnt lie on the same base and between the same parallels.
(ii) No, it doesnt lie on the same base and between the same parallels.
(iii) Yes, it lies on the same base and between the same parallels. The same base is AB and the parallels are AB and DE.
(iv) No, it doesnt lie on the same base and between the same parallels.
(v) Yes, it lies on the same base and between the same parallels. The same base is BC and the parallels are BC and AD.
(vi) Yes, it lies on the same base and between the same parallels. The same base is CD and the parallels are CD and BP.

Question 2:

In the adjoining figure, show that ABCD is a parallelogram.
Calculate the area of || gm ABCD.

Answer 2:

Given: A quadrilateral ABCD and BD is a diagonal.
To prove: ABCD is a parallelogram.
Construction: Draw AM ⊥ DC and CL ⊥ AB   (extend DC and AB). Join AC, the other diagonal of ABCD.

Proof: ar(quad. ABCD) = ar(∆ABD) + ar(​∆DCB)
                                      = 2 ar(​∆ABD)                    [∵ ar​(∆ABD) = ar(​∆DCB)]
∴ ar(​∆ABD) = $\frac{1}{2}$ar(quad. ABCD)                 ...(i)

Again, ar(quad. ABCD) = ar(∆ABC) + ar(​∆CDA)
                                    = 2 ar(​∆ ABC)                    [∵ ar​(∆ABC) = ar(​∆CDA)]
∴ ar(​∆ABC) = $\frac{1}{2}$ar(quad. ABCD)                ...(ii)
From (i) and (ii), we have:
 ar(​∆ABD) = ar(​∆ABC) = $\frac{1}{2}$ AB ⨯ BD = $\frac{1}{2}$ AB ⨯ CL
 ⇒ CL = BD
 ⇒ DC |​​| AB
Similarly, AD |​​| BC.
Hence, ABCD is a paralleogram.
∴ ar(​|​| gm ABCD) = base ​⨯ height = 5 ​⨯ 7 = 35 cm²

Question 3:

In a parallelogram ABCD, it is being given that AB = 10 cm and the altitudes corresponding to the sides AB and AD are DL = 6 cm and BM = 8 cm, respectively. Find AD.

Answer 3:

ar(parallelogram ABCD) = base ​⨯ height
⇒ AB ​⨯DL = AD ​⨯ BM
⇒ 10 ​​⨯ 6 = AD ​⨯ BM
⇒ AD ​⨯ 8 = 60 cm²
⇒ AD =  7.5 cm
∴ AD = 7.5 cm

PAGE NO-388

Question 4:

Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm.

Answer 4:

Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. 
Join the diagonals, AC and BD.
In ∆ ABC, we have:
PQ ∣∣ AC and PQ = $\frac{1}{2}$AC                    [By midpoint theorem]
$\mathrm{PQ}=\frac{1}{2}\times 16=8 \mathrm{cm}$
Again, in ∆DAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR = $\frac{1}{2}$AC                    [By midpoint theorem]
$\mathrm{SR}=\frac{1}{2}\times 12=6 \mathrm{cm}$
$\mathrm{Area} \mathrm{of} \mathrm{PQRS}=\mathrm{length}\times \mathrm{breadth}=6\times 8=48 {\mathrm{cm}}^2$

Question 5:

Find the area of a trapezium whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm.

Answer 5:

ar(trapezium) = $\frac{1}{2}$ ⨯ (sum of parallel sides) ⨯ (distance between them)
 = $\frac{1}{2}$ ⨯ (9 + 6) ⨯ 8
= 60 cm²
Hence, the area of the trapezium is 60 cm².

Question 6:

(i) Calculate the area of quad. ABCD, given in Fig. (i).
(ii) Calculate the area of trap. PQRS, given in Fig. (ii).

Answer 6:

(i) In $△$BCD, 
$$\begin{gathered} {\mathrm{DB}}^2+{\mathrm{BC}}^2={\mathrm{DC}}^2 \\ \Rightarrow {\mathrm{DB}}^2={17}^2-8^2=225 \\ \Rightarrow \mathrm{DB}=15 \mathrm{cm} \end{gathered}$$
Ar($△$BCD) = $\frac{1}{2}\times b\times h=\frac{1}{2}\times 8\times 15=60 {\mathrm{cm}}^2$
In $△$BAD,
$$\begin{gathered} {\mathrm{DA}}^2+{\mathrm{AB}}^2={\mathrm{DB}}^2 \\ \Rightarrow {\mathrm{AB}}^2={15}^2-9^2=144 \\ \Rightarrow \mathrm{AB}=12 \mathrm{cm} \end{gathered}$$
Ar($△$DAB) = $\frac{1}{2}\times b\times h=\frac{1}{2}\times 9\times 12=54 {\mathrm{cm}}^2$

Area of quad. ABCD = Ar($△$DAB) + Ar($△$BCD) = 54 + 60 = 114 cm² .

(ii) Area of trap(PQRS) = $\frac{1}{2}\left(8+16\right)\times 8=96 {\mathrm{cm}}^2$

Question 7:

In the adjoining figure, ABCD is a trapezium in which AB || DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the length of DC and hence, find the area of trap. ABCD.

Answer 7:

∆ADL is a right angle triangle.
So, DL = $\sqrt{\left(5^2 - 4^2 \right)} = \sqrt{9} = 3 \mathrm{cm}$
Similarly, in ∆BMC, we have:
MC = $\sqrt{\left(5^2 - 4^2 \right)} = \sqrt{9} = 3 \mathrm{cm}$
∴ DC =  DL + LM + MC =  3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = $\frac{1}{2}$⨯ (sum of parallel sides) ⨯ (distance between them)
=$\frac{1}{2}$ ⨯ (7 + 13) ⨯ 4
= 40 cm²
​Hence, DC = 13 cm and area of trapezium = 40 cm²

Question 8:

BD is one of the diagonals of a quad. ABCD. If AL ⊥ BD and CM ⊥ BD, show that $\mathrm{ar}(\mathrm{quad}. ABCD)=\frac{1}{2}\times BD\times (AL + CM)$.

Answer 8:

ar(quad. ABCD) = ar(∆​ABD) + ar (∆DBC)
ar(∆ABD) = $\frac{1}{2}$⨯ base ⨯ height = $\frac{1}{2}$⨯ BD ⨯ AL             ...(i) 
ar(∆DBC) = $\frac{1}{2}$ ⨯ BD ⨯ CL                  ...(ii)
From (i) and (ii), we get:
​ar(quad ABCD) = $\frac{1}{2}$ ⨯BD ⨯​ AL + $\frac{1}{2}$ ⨯ BD ⨯​ CL
$\Rightarrow$​ar(quad ABCD) = $\frac{1}{2}$ ⨯ BD ⨯ ​(AL + CL)
Hence, proved.

Question 9:

M is the midpoint of the side AB of a parallelogram ABCD. If ar(AMCD) = 24 cm², find ar(∆ABC).

Answer 9:

Join AC. 
AC divides parallelogram ABCD into two congruent triangles of equal areas. 
$\mathrm{ar}\left(△\mathrm{ABC}\right)=\mathrm{ar}\left(△\mathrm{ACD}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)$
M is the midpoint of AB. So, CM is the median. 
CM divides $△$ABC in two triangles with equal area. 
$\mathrm{ar}\left(△\mathrm{AMC}\right)=\mathrm{ar}\left(△\mathrm{BMC}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$
ar(AMCD) = ar($△$ACD) + ar($△$AMC) = ar($△$ABC) + ar($△$AMC) = ​ar($△$ABC) + $\frac{1}{2}$​ar($△$ABC)
$$\begin{gathered} \Rightarrow 24=\frac{3}{2}\mathrm{ar}\left(△\mathrm{ABC}\right) \\ \Rightarrow \mathrm{ar}\left(△\mathrm{ABC}\right)=16 {\mathrm{cm}}^2 \end{gathered}$$

Question 10:

In the adjoining figure, ABCD is a quadrilateral in which diag. BD = 14 cm. If AL ⊥ BD and CM ⊥ BD such that AL = 8 cm and CM = 6 cm, find the area of quad. ABCD.

Answer 10:

​ar(quad ABCD) = ar($△$ABD) + ar($△$BDC)
= $\frac{1}{2}$ ⨯BD ⨯​ AL  +$\frac{1}{2}$ ⨯BD ⨯​ CM
= $\frac{1}{2}$ ⨯BD ⨯​ ( AL + CM)
By substituting the values, we have;
ar(quad ABCD) = ​$\frac{1}{2}$ ⨯ 14 ⨯ ( 8 + 6)
= 7 ​⨯14
= 98 cm²

Question 11:

If P and Q are any two points lying respectively on the sides DC and AD of a parallelogram ABCD then show that ar(∆APB) = ar(∆BQC).

Answer 11:

We know
ar(∆APB) = $\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)$           .....(1)                     [If a triangle and a parallelogram are on the same base and between the same parallels then the area of the triangle is equal to half the area of the parallelogram]
Similarly, 
ar(∆BQC) = $\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)$           .....(2)
From (1) and (2)
ar(∆APB) = ar(∆BQC)
Hence Proved

PAGE NO-389

Question 12:

In the adjoining figure, MNPQ and ABPQ are parallelograms and T is any point on the side BP. Prove that
(i) ar(MNPQ) = are(ABPQ)
(ii) ar(∆ATQ) = $\frac{1}{2}$ar(MNPQ).

Answer 12:

(i) We know that parallelograms on the same base and between the same parallels are equal in area.
So, ar(MNPQ) = are(ABPQ)                   (Same base PQ and MB || PQ)                      .....(1)

(ii) If a parallelogram and a triangle are on the same base and between the same parallels then the area of the triangle is equal to half the area of the parallelogram. 
So, ar(∆ATQ) = $\frac{1}{2}$ar(ABPQ)                 (Same base AQ and AQ || BP)                       .....(2)
From (1) and (2)
ar(∆ATQ) = $\frac{1}{2}$ar(MNPQ)


 

Question 13:

In the adjoining figure, ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect at O. Prove that ar(∆AOD) = ar(∆BOC).

Answer 13:

∆CDA and ​∆CBD lies on the same base and between the same parallel lines.
So, ar(​∆CDA) = ar(CDB)            ...(i)
Subtracting ar(​∆OCD) from both sides of equation (i), we get:
ar(​∆CDA) $-$ ar(​∆OCD) = ar(​​∆CDB) $-$ ar (​​∆OCD)
⇒ ar(​​∆AOD) = ar(​​∆BOC)

Question 14:

In the adjoining figure, DE || BC. Prove that
(i) ar(∆ACD) = ar(∆ABE),
(ii) ar(∆OCE) = ar(∆OBD),

Answer 14:

∆DEC and ​∆DEB lies on the same base and between the same parallel lines.
So, ar(​∆DEC) = ar(∆DEB)                      ...(1)

(i) On adding​ ar(∆ADE)​ in both sides of equation (1), we get:
  ar(​∆DEC) + ar(∆ADE)​ = ar(∆DEB) + ar(∆ADE)​ ​                 
⇒ ar(​​∆ACD) = ar(​​∆ABE) 

  (ii) On subtracting​ ar(ODE)​ from both sides of equation (1), we get:​
   ar(​∆DEC) $-$ ar(∆ODE)​ = ar(∆DEB) $-$ ar(∆ODE)​ ​      ​
⇒ ar(​​∆OCE) = ar(​∆OBD)

Question 15:

Prove that a median divides a triangle into two triangles of equal area.

Answer 15:

Let AD is a median of ∆ABC and D is the midpoint of BC. AD divides ∆ABC in two triangles: ∆ABD and ∆ADC.
To prove: ar(∆ABD) = ar(∆ADC)
Construction: Draw AL ⊥ BC.
Proof:
Since D is the midpoint of BC, we have:
BD = DC
Multiplying with $\frac{1}{2}$AL on both sides, we get:
$\frac{1}{2}$ × BD × AL = $\frac{1}{2}$ × DC × AL 
⇒ ar(∆ABD) = ar(∆ADC)

Question 16:

Show that a diagonal divides a parallelogram into two triangles of equal area.

Answer 16:

Let ABCD be a parallelogram and BD be its diagonal.
To prove: ar(∆ABD) = ar(∆CDB)

Proof: 
In ∆ABD and ∆CDB, we have:
AB = CD                    [Opposite sides of a parallelogram]
AD = CB                   [Opposite sides of a parallelogram]​

 BD  = DB                  [Common]

i.e., ∆ABD ≅ ∆CDB           [ SSS criteria]
∴ ar(∆ABD) = ar(∆CDB)

Question 17:

In the adjoining figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ΔABC) = ar(ΔABD)

Answer 17:

Line segment CD is bisected by AB at O                   (Given)
CO = OD                                .....(1)
In ΔCAO, 
AO is the median.                (From (1))
So, arΔCAO = arΔDAO          .....(2)
Similarly, 
In ΔCBD, 
BO is the median                 (From (1))
So, arΔCBO = arΔDBO          .....(3)
From (2) and (3) we have
arΔCAO + arΔCBO = arΔDBO + arΔDAO
$\Rightarrow$ar(ΔABC) = ar(ΔABD)



 

Question 18:

D and E are points on sides AB and AC respectively of ∆ABC such that ar(∆BCD) = ar(∆BCE). Prove that DE || BC.

Answer 18:

ar(∆BCD) = ar(∆BCE)                     (Given)
We know, triangles on the same base and having equal areas lie between the same parallels.
Thus, DE || BC. 

Question 19:

P is any point on the diagonal AC of a parallelogram ABCD. Prove that ar(∆ADP) = ar(∆ABP).

Answer 19:

Join BD. 
Let BD and AC intersect at point O. 
O is thus the midpoint of DB and AC. 
PO is the median of $△$DPB, 
So, 
$$\begin{gathered} \mathrm{ar}\left(△\mathrm{DPO}\right)=\mathrm{ar}\left(△\mathrm{BPO}\right) .....\left(1\right) \\ \mathrm{ar}\left(△\mathrm{ADO}\right)=\mathrm{ar}\left(△\mathrm{ABO}\right) .....\left(2\right) \\ \mathrm{Case} 1: \\ \left(2\right)-\left(1\right) \\ \Rightarrow \mathrm{ar}\left(△\mathrm{ADO}\right)-\mathrm{ar}\left(△\mathrm{DPO}\right)=\mathrm{ar}\left(△\mathrm{ABO}\right)-\mathrm{ar}\left(△\mathrm{BPO}\right) \end{gathered}$$
Thus, ar(∆ADP) = ar(∆ABP)

Case II: 

$\mathrm{ar}\left(△\mathrm{ADO}\right)+\mathrm{ar}\left(△\mathrm{DPO}\right)=\mathrm{ar}\left(△\mathrm{ABO}\right)+\mathrm{ar}\left(△\mathrm{BPO}\right)$
Thus, ar(∆ADP) = ar(∆ABP)

Question 20:

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.
If BO = OD, prove that
ar(∆ABC) = ar(∆ADC),

Answer 20:

Given:  BO = OD
To prove: ar(∆ABC) = ar(∆ADC)
Proof: 
Since BO = OD, O is the mid point of BD.
We know that a median of a triangle divides it into two triangles of equal areas.
CO is a median of ∆BCD.
i.e., ar(∆COD) = ar (∆COB)            ...(i)

AO is a median of ∆ABD.
i.e., ar(∆AOD) = ar(∆AOB)              ...(ii)

From (i) and (ii), we have:
ar(∆COD) + ar(∆AOD) = ar(∆COB) + ar(∆AOB)
∴ ar(∆ADC )​ = ar(∆ABC)

Question 21:

The vertex A of ∆ABC is joined to a point D on the side BC. The midpoint of AD is E.
Prove that $\mathrm{ar}(∆BEC)=\frac{1}{2}\mathrm{ar}(∆ABC)$.

Answer 21:

Given:  D is the midpoint of BC and E is the midpoint of AD.
To prove: $\mathrm{ar}(∆BEC)=\frac{1}{2}\mathrm{ar}(∆ABC)$
Proof: 
Since E is the midpoint of AD, BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED ) = $\frac{1}{2}$ar(∆ABD)                 ...(i)
Also, ar(∆CDE ) =$\frac{1}{2}$ ar(∆ADC)             ...(ii)

From (i) and (ii), we have:
ar(∆BED) + ar(∆CDE)​ = $\frac{1}{2}$ ⨯​ ar(∆ABD)​ + $\frac{1}{2}$ ⨯​ ar(∆ADC)   
⇒ ar(∆BEC )​ = $\frac{1}{2}$⨯ [ar(∆ABD) + ar(∆ADC)] 
⇒ ​ar(∆BEC )​ =​ $\frac{1}{2}$ ⨯​ ar(∆ABC)

PAGE NO-390

Question 22:

D is the midpoint of side BC of ∆ABC and E is the midpoint of BD. If O is the midpoint of AE, prove that ar(∆BOE) = $\frac{1}{8}$ar(∆ABC).

Answer 22:

D is the midpoint of side BC of ∆ABC. 
$\Rightarrow$AD is the median of ∆ABC. 
$\Rightarrow$$\mathrm{ar}\left(△\mathrm{ABD}\right)=\mathrm{ar}\left(△\mathrm{ACD}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$
E is the midpoint of BD of ∆ABD, 
$\Rightarrow$AE is the median of ∆ABD
$\Rightarrow$$\mathrm{ar}\left(△\mathrm{ABE}\right)=\mathrm{ar}\left(△\mathrm{AED}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABD}\right)=\frac{1}{4}\mathrm{ar}\left(△\mathrm{ABC}\right)$
Also, O is the midpoint of AE, 
$\Rightarrow$BO is the median of ∆ABE, 
$\Rightarrow$$\mathrm{ar}\left(△\mathrm{ABO}\right)=\mathrm{ar}\left(△\mathrm{BOE}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABE}\right)=\frac{1}{4}\mathrm{ar}\left(△\mathrm{ABD}\right)=\frac{1}{8}\mathrm{ar}\left(△\mathrm{ABC}\right)$
Thus, ar(∆BOE) = $\frac{1}{8}$ar(∆ABC)

 

Question 23:

In a trapezium ABCD, AB || DC and M is the midpoint of BC. Through M, a line PQ || AD has been drawn which meets AB in P and DC produced in Q, as shown in the adjoining figure. Prove that ar(ABCD) = ar(APQD).

Answer 23:

In $△$MQC and $△$MPB, 
MC = MB                            (M is the midpoint of BC)
$\angle$CMQ = $\angle$BMP                (Vertically opposite angles)
$\angle$MCQ = $\angle$MBP                (Alternate interior angles on the parallel lines AB and DQ)
Thus, $△$MQC $\cong$ $△$MPB   (ASA congruency)
$\Rightarrow$ar($△$MQC) = ar($△$MPB)
$\Rightarrow$ar($△$MQC) + ar(APMCD) = ar($△$MPB) + ar(APMCD)
$\Rightarrow$ar(APQD) = ar(ABCD)

Question 24:

In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Prove that ar(∆ABP) = ar(quad. ABCD).

Answer 24:

We have:
​ar(quad. ABCD) = ar(∆ACD) + ar(∆ABC)
ar(∆ABP) = ar(∆ACP)​​ + ar(∆ABC) 

∆ACD and ∆ACP are on the same base and between the same parallels AC and DP.
∴ ar(∆ACD) = ar(∆ ACP)​
By adding ar(∆ABC) on both sides, we get:
ar(∆ACD) + ar(∆ABC) = ar(∆ACP)​​ + ar(∆ABC)                
⇒​ ar (quad. ABCD) = ar(∆ABP)
Hence, proved.

Question 25:

In the adjoining figure, ∆ABC and ∆DBC are on the same base BC with A and D on opposite sides of BC such that ar(∆ABC) = ar(∆DBC). Show that BC bisects AD.

Answer 25:

Given: ∆ABC and ∆DBC are on the same base BC.
ar(∆ABC) = ar(∆DBC)​
To prove: BC bisects AD
Construction: Draw AL ⊥ BC and DM ⊥ BC.
Proof: 
Since ∆ABC and ∆DBC are on the same base BC and they have equal areas, their altitudes must be equal.
i.e., AL = DM
Let AD and BC intersect at O.
Now, in ∆ALO and ∆DMO, we have:
AL = DM
∠ALO = ∠DMO =  90^o
∠​AOL = ∠DO​M                  (Vertically opposite angles)
i.e., ∆ ALO ≅ ∆ DMO
​​∴​ OA = OD
Hence, BC bisects AD.

Question 26:

ABCD is a parallelogram in which BC is produced to P such that CP = BC, as shown in the adjoining figure. AP intersects CD at M. If ar(DMB) = 7 cm², find the area of parallelogram ABCD.

Answer 26:

In $△$MDA and $△$MCP, 
$\angle$DMA = $\angle$CMP                  (Vertically opposite angles)
$\angle$MDA = $\angle$MCP                  (Alternate interior angles)
AD = CP                                 (Since AD = CB and CB = CP)
So, $△$MDA $\cong$ $△$MCP         (ASA congruency)
$\Rightarrow$DM = MC                         (CPCT)
$\Rightarrow$M is the midpoint of DC
$\Rightarrow$BM is the median of $△$BDC
$\Rightarrow$ar($△$BMC) = ar($△$DMB) = 7 cm² 
ar($△$BMC) + ar($△$DMB) = ar($△$DBC) = 7 + 7 = 14 ${\mathrm{cm}}^2$
Area of parallelogram ABCD = 2 $\times$ ar($△$DBC) = 2 $\times$ 14 = 28 ${\mathrm{cm}}^2$ 
 

Question 27:

In a parallelogram ABCD, any point E is taken on the side BC. AE and DC when produced meet at a point M. Prove that
ar(∆ADM) = ar(ABMC)

Answer 27:

Join BM and AC. 
ar(∆ADC) = $\frac{1}{2}bh$ = $\frac{1}{2}\times \mathrm{DC}\times h$
ar(∆ABM) = $\frac{1}{2}\times \mathrm{AB}\times h$
AB = DC                               (Since ABCD is a parallelogram)
So, ar(∆ADC) = ar(∆ABM)
$\Rightarrow$ar(∆ADC) + ar(∆AMC) = ar(∆ABM) + ar(∆AMC)
$\Rightarrow$ar(∆ADM) = ar(ABMC)
Hence Proved

Question 28:

P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of || gm ABCD. Show that PQRS is a parallelogram and also show that
$\mathrm{ar}\left(\left|\right| \mathrm{gm} PQRS\right)=\frac{1}{2}\times \mathrm{ar}\left(\right||\mathrm{gm} ABCD)$.

Answer 28:

Given:  ABCD is a parallelogram and P, Q, R and S are the midpoints of sides AB, BC, CD and DA, respectively.
To prove: ar(parallelogram PQRS ) = $\frac{1}{2}$ × ar(parallelogram ABCD )
Proof: 
In ∆ABC, PQ || AC and PQ = $\frac{1}{2}$ × AC              [ By midpoint theorem] 
Again, in ∆DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR || AC and SR = $\frac{1}{2}$ × AC                                   [ By midpoint theorem] 
Now, PQ || AC and SR || AC  
⇒ ​PQ || SR
Also, PQ = SR = $\frac{1}{2}$ × AC
∴ PQ || SR and PQ = SR
Hence, PQRS is a parallelogram.

Now, ar(parallelogram​ PQRS) = ar(∆PSQ) + ar(∆SRQ)                       ...(i)
also, ar(parallelogram ABCD) = ar(parallelogram ABQS) + ar(parallelogram SQCD)            ...(ii)

∆PSQ and parallelogram ABQS are on the same base and between the same parallel lines.
So, ar(∆PSQ ) =$\frac{1}{2}$ × ar(parallelogram ABQS)                         ...(iii)
Similarly, ∆SRQ and parallelogram SQCD are on the same base and between the same parallel lines.
So, ar(∆SRQ ) = $\frac{1}{2}$ × ar(parallelogram SQCD)                        ...(iv)
Putting the values from (iii) and (iv) in (i), we get:
  ar(parallelogram​ PQRS) = $\frac{1}{2}$ × ar(parallelogram ABQS)​ + $\frac{1}{2}$ × ar(parallelogram SQCD)
From (ii), we get:
ar(parallelogram​ PQRS) = $\frac{1}{2}$ × ar(parallelogram ABCD)

Question 29:

In a triangle ABC, the medians BE and CF intersect at G. Prove that ar(∆BCG) = ar(AFGE).

Answer 29:

Figure
CF is median of $△$ABC.
$\Rightarrow$ar($△$BCF) = $\frac{1}{2}$($△$ABC)                     .....(1)
Similarly, BE is the median of $△$ABC,
$\Rightarrow$ar($△$ABE) = $\frac{1}{2}$($△$ABC)                     .....(2)
From (1) and (2) we have 
ar($△$BCF) = ar($△$ABE)
$\Rightarrow$ar($△$BCF) $-$ ar($△$BFG) = ar($△$ABE) $-$ ar($△$BFG)
$\Rightarrow$ar(∆BCG) = ar(AFGE)
 

PAGE NO-391

Question 30:

The base BC of ∆ABC is divided  at D such that $BD=\frac{1}{2}DC$. Prove that $\mathrm{ar}(∆ABD)=\frac{1}{3}\times \mathrm{ar}(∆ABC)$.

Answer 30:

Given: D is a point on BC of ∆ABC, such that BD = $\frac{1}{2}$DC
To prove:  ar(∆ABD) = $\frac{1}{3}$ar(∆ABC)
Construction: Draw AL ⊥ BC.
Proof: 
In ∆ABC, we have:
BC = BD + DC
 ⇒​ BD​ + 2 BD = 3 × BD
Now, we have:
ar(∆ABD)​ = $\frac{1}{2}$​ ×​ BD ×​ AL
ar(∆ABC)​ = $\frac{1}{2}$​ ×​ BC ×​ AL
⇒  ar(∆ABC) = $\frac{1}{2}$ ×​ 3BD ×​ AL = 3 ×​ $\left(\frac{1}{2}\times BD\times AL\right)$
⇒ ar(∆ABC)​ = 3 × ar(∆ABD)
∴ ​ar(∆ABD) = ​$\frac{1}{3}$​ar(∆ABC)

Question 31:

In the adjoining figure, BD || CA, E is the midpoint of CA and BD = $\frac{1}{2}$ CA. Prove that ar(∆ABC) = 2ar(∆DBC).

Answer 31:

E is the midpoint of CA. 
So, AE = EC                            .....(1)
Also, BD = $\frac{1}{2}$ CA                    (Given)
So, BD = AE                            .....(2)
From (1) and (2) we have
BD = EC
BD || CA and BD = EC so, BDEC is a parallelogram
BE acts as the median of $△$ABC
so, ar($△$BCE) = ar($△$ABE) = $\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$                  .....(1)
ar($△$DBC) = ar($△$BCE)                  .....(2)                  (Triangles on the same base and between the same parallels are equal in area) 
From (1) and (2) 
ar(∆ABC) = 2ar(∆DBC)
 

Question 32:

The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, meets BA produced at G, and CF, drawn parallel to DB, meets AB produced at F. Show that ar(pentagon ABCDE) = ar(∆DGF).

Answer 32:

Given:  ABCDE is a pentagon.  EG || DA and CF || DB.
To prove: ar(pentagon ABCDE ) =  ar( ∆DGF) 
Proof: 
ar(pentagon ABCDE )​ = ar(∆DBC) + ar(∆ADE ) + ar(∆ABD)               ...(i)
Also, ar(∆DGF) = ar(∆DBF) + ar(∆ADG) + ar(∆ABD )                ...(ii)

Now, ∆DBC and ∆DBF lie on the same base and between the same parallel lines. 
∴ ar(∆DBC) = ar(∆DBF)                         ...(iii)               
Similarly, ∆ADE and ∆ADG lie on same base and between the same parallel lines.  
 ∴ ar(∆ADE) = ar(∆ADG)                       ...(iv)

From (iii) and (iv), we have:
ar(∆DBC) + ar(∆ADE) = ar(∆DBF) + ar(∆ADG)
Adding ar(∆ABD) on both sides, we get:
ar(∆DBC) + ar(∆ADE) + ar(∆ABD) = ar (∆DBF) + ar(∆ADG) + ar(∆ABD) 
By substituting the values from (i) and (ii), we get:
ar(pentagon ABCDE) =  ar(∆DGF) 

Question 33:

In the adjoining figure, CE || AD and CF || BA. Prove that ar(∆CBG) = ar(∆AFG).

Answer 33:

$\mathrm{ar}\left(△\mathrm{CFA}\right)=\mathrm{ar}\left(△\mathrm{CFB}\right)$                            (Triangles on the same base CF and between the same parallels CF || BA will be equal in area)
$$\begin{gathered} \Rightarrow \mathrm{ar}\left(△\mathrm{CFA}\right)-\mathrm{ar}\left(△\mathrm{CFG}\right)=\mathrm{ar}\left(△\mathrm{CFB}\right)-\mathrm{ar}\left(△\mathrm{CFG}\right) \\ \Rightarrow \mathrm{ar}\left(△\mathrm{AFG}\right)=\mathrm{ar}\left(△\mathrm{CBG}\right) \end{gathered}$$
Hence Proved
 

Question 34:

In the adjoining figure, the point D divides the side BC of ∆ABC in the ratio m : n. Prove that ar(∆ABD) : ar(∆ADC) = m : n.

Answer 34:

Given: D is a point on BC of ∆ ABC, such that BD : DC =  m : n
To prove:  ar(∆ABD) : ar(∆ADC) = m : n
Construction: Draw AL ⊥ BC.
Proof: 
ar(∆ABD)​ = $\frac{1}{2}$ ×​ BD ×​ AL                     ...(i)
ar(∆ADC)​ = $\frac{1}{2}$​ ×​ DC ×​ AL                   ...(ii)
Dividing (i) by (ii), we get:
$$\begin{gathered} \frac{\mathrm{ar}(△ABD)}{\mathrm{ar}(∆ADC}=\frac{{\displaystyle \frac{1}{2}}\times BD\times AL}{{\displaystyle \frac{1}{2}}\times DC\times AL} \\ =\frac{BD}{DC} \\ =\frac{m}{n} \end{gathered}$$

∴ ar(∆ABD) : ar(∆ADC) = m : n

Question 35:

In a trapezium ABCD, AB || DC, AB = a cm, and DC = b cm. If M and N are the midpoints of the nonparallel sides, AD and BC respectively then find the ratio of ar(DCNM) and ar(MNBA).

Answer 35:

Construction: Draw a perpendicular from point D to the opposite side AB, meeting AB at Q and MN at P.
Let length DQ = h 
Given, M and N are the midpoints of AD and BC respectively. 
So, MN || AB || DC and MN = $\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)=\left(\frac{a+b}{2}\right)$
M is the mid point of AD and MP || AB so by converse of mid point theorem,
MP || AQ and P will be the mid point of DQ. 
$$\begin{gathered} \mathrm{ar}\left(\mathrm{DCNM}\right)=\frac{1}{2}\times \mathrm{DP}\left(\mathrm{DC}+\mathrm{MN}\right)=\frac{1}{2}h\left(b+\frac{a+b}{2}\right)=\frac{h}{4}\left(a+3b\right) \\ \mathrm{ar}\left(\mathrm{MNBA}\right)=\frac{1}{2}\times \mathrm{PQ}\left(\mathrm{AB}+\mathrm{MN}\right)=\frac{1}{2}h\left(a+\frac{a+b}{2}\right)=\frac{h}{4}\left(b+3a\right) \end{gathered}$$
ar(DCNM) : ar(MNBA) = (a + 3b) : (3a + b)
 

Question 36:

ABCD is a trapezium in which AB || CD, AB = 16 cm and DC = 24 cm. If E and F are respectively the midpoints of AD and BC, prove that ar(ABFE) = $\frac{9}{11}$ ar(EFCD).

Answer 36:

Construction: Draw a perpendicular from point D to the opposite side CD, meeting CD at Q and EF at P.
Let length AQ = h 
Given, E and F are the midpoints of AD and BC respectively. 
So, EF || AB || DC and EF = $\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)=\left(\frac{a+b}{2}\right)$
E is the mid point of AD and EP || AB so by converse of mid point theorem,
EP || DQ and P will be the mid point of AQ. 
$$\begin{gathered} \mathrm{ar}\left(\mathrm{ABFE}\right)=\frac{1}{2}\times \mathrm{AP}\left(\mathrm{AB}+\mathrm{EF}\right)=\frac{1}{2}h\left(b+\frac{a+b}{2}\right)=\frac{h}{4}\left(a+3b\right) \\ \mathrm{ar}\left(\mathrm{EFCD}\right)=\frac{1}{2}\times \mathrm{PQ}\left(\mathrm{CD}+\mathrm{EF}\right)=\frac{1}{2}h\left(a+\frac{a+b}{2}\right)=\frac{h}{4}\left(b+3a\right) \end{gathered}$$
ar(ABEF) : ar(EFCD) = (a + 3b) : (3a + b)
Here a = 24 cm and b = 16 cm
So, $\frac{\mathrm{ar}\left(\mathrm{ABEF}\right)}{\mathrm{ar}\left(\mathrm{EFCD}\right)}=\frac{24+3\times 16}{16+3\times 24}=\frac{9}{11}$

Question 37:

In the adjoining figure, D and E are respectively the midpoints of sides AB and AC of ∆ABC. If PQ || BC and CDP and BEQ are straight lines then prove that ar(∆ABQ) = ar(∆ACP).

Answer 37:

In $△$PAC, 
PA || DE and E is the midpoint of AC
So, D is the midpoint of PC by converse of midpoint theorem.
Also, $\mathrm{DE}=\frac{1}{2}\mathrm{PA}$                         .....(1)
Similarly, $\mathrm{DE}=\frac{1}{2}\mathrm{AQ}$                  .....(2)
From (1) and (2) we have 
PA = AQ
∆ABQ and ∆ACP are on same base PQ and between same parallels PQ and BC
ar(∆ABQ) = ar(∆ACP) 

PAGE NO-392

Question 38:

In the adjoining figure, ABCD and BQSC are two parallelograms. Prove that ar(∆RSC) = ar(∆PQB).

Answer 38:

In ∆RSC and ∆PQB
$\angle$CRS = $\angle$BPQ                       (CD || AB) so, corresponding angles are equal)
$\angle$CSR = $\angle$BQP                        ( SC || QB so, corresponding angles are equal)
SC = QB                                    (BQSC is a parallelogram)
So, ∆RSC $\cong$ ∆PQB                   (AAS congruency)
Thus, ar(∆RSC) = ar(∆PQB)