Here are the answer keys (solutions ) of CBSE 10 Mathematics question paper -2019 Section C

Section C

13.

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the  mean number of days a student was absent.

Solution:

Frequency distribution :Mean number of days a student was absent =14.1

14.

In Figure 2. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at point T. Find the length TP

Solution:

Join OQ

Let TP=x and TR =y

 

 

 

In OPT and OQT,

OP=OQ……………………………..Radii of the same circle

PT=QT………………………………Tangents from the same

OT = OT…………………………..Common side

OPT≅  OQT ……………..SSS postulate

So  angle ∠PTR = ∠QTR …………………….CPCT……………1

Consider triangles, PRT and QRT

PT= QT……………………….Tangents from same point to a circle are equal

∠PTR = ∠QTR…………………….Proved in 1

TR=TR………………………Common side

So  PRT ≅ QRT……………SAS

So ∠PRT=∠QRT………………..CPCT

But ∠PRT+∠QRT=180°…………Linear pair

So∠PRT=∠QRT=90°

OR⊥ PQ

Perpendicular from the centre bisects the chord.

∴ PR=QR= 4cm

Triangle OPR is also a right angle triangle

∴ PO² = PR² +OR²

OR² = PO² – PR²

= 5² – 4²

=9

OR=3 cm

So, triangle PRT is a right angles triangle.

PT²= TR² + PR²………Pythagoras theorem

x²=y² + PR²

x²=y² + 16……………………………3

OP⊥PT………Tangent is perpendicular to radius

∴ In triangle OPT

PT²= PO² +OT²……………………….Pythagoras theorem

(y+3)² = x² + 5²

∴ x²=(y+3)²   –  5²…………………………..4

Comparing equation 3 and 4

y² + 16= (y+3)²   –  5²

y² + 16= y²+6y+9-25

6y= 32

y=32/6= 16/3

x²=(16/3)² + 4²

=(256/9) + 16

= 400/9

x=√400/9

=20/3 cm

 

15.

A, B and C are interior angles of a triangle ABC. Show that

(i) sin[(B+C)/2]= cos A/2

(ii) If ∠A=90° , then find the value of tan[ (B+C)/2]

Solution:

(i)

In triangle ABC

∠A+∠B+∠C=180°

∠B+∠C=180°-∠A

Divide by 2

(∠B+∠C)/2= (180°-∠A)/2

(∠B+∠C)/2 = 90- (∠A)/2)

sin[(∠B+∠C)/2 ]= sin [ 90- (∠A)/2)]

sin[(∠B+∠C)/2 ]= cos (∠A)/2)…………….(sin(90-θ)= cosθ

(ii)

∠A=90°

∠A+∠B+∠C=180°………………. Angle sum property of triangle

∴ ∠B+∠C=90°

Divide by 2

(∠B+∠C)/2= 45°

tan [(∠B+∠C)/2]= tan 45°

=1

 

                                                                            OR

If tan (A+B) = 1 and tan (A-b)= 1/√3 , 0<A+B<90° , A> B, then find the values of A and B

Solution:

tan(A+B)= 1

tan 45° =1

∴ (A+B) =45°…………………….1

tan (A-B)= 1/√3

tan 30°= 1/√3

(A-B) = 30°……………………….2

Solving equation 1 and 2

(A+B) =45°

(A-B) = 30°

———————-

2A=75°

A= 37.5°

Substitute A= 37.5° in A+B= 45°

37.5°+ B= 45°

B= 45°  – 37.5°

=7.5°

A= 37.5°  B=7.5°

 

16.

Prove that √3 is an irrational number.

Solution :

Let us assume √3 is a rational number.

So √3 can be expresses in the form of a/b where a  and b are integers  having no common factor and b≠ 0

∴ √3 = a/b

a= √3b

Squaring bot the sides

a² = 3b²…………………………..(1)

∴3 divides a²

⇒3 divides a

Let a = 3c, and c is an integer

Putting a = 3c in (1)

3b² = 9c²  ⇒b² = 3c²

⇒ 3 divides b²

⇒ 3 divides b

Thus, 3 is common factor of  and b.

But, this contradicts the fact that a and b have no common factor other than 1.

The contradiction arises by assuming that √3 is rational.

Hence ,√3 is rational.

                                                      OR

Find the largest number which on dividing 1251,9377 and 15628 leaves remainder 1,2 and 3.

Solution:

On dividing 1251, 9377 and 15628 the number leaves remainder,  1,2 and 3 respectively.

∴ the number divides 1251- 1= 1250

the number divides 9377-2=9375

the number divides 15628-3 =15625

The number is HCF(1250, 9375, 15625)

1250= 5⁴ X 2

9375=5⁵ X 3

15625=5⁶

^{HCF((1250, 9375, 15625)=} 5⁴

∴ the number is 625

17.

Draw the graph of equations, x-y+1=0 and 3x+2y-12=0. Using the graph, find the values of x and y which satisfy both the equations.

Solution:

Equation (1)

x-y+1=0

∴ y= x+1

 

 

Equation (2)

3x+2y-12=0

3x+2y=12

2y=12-3x

y=(12-3x)/2

 

 

 

The graph:

The x and y values satisfying both the equations are :

x=2 and y =3

18.

Water in a canal , 6m wide and 1.5 m deep, is flowing with a speed of 10km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed.

Solution:

Width of the canal b =6m

Depth of the canal h =1.5 m

Speed of water in the canal l =10 km/h

i.e. length covered by water in 1hr= 10 km =10000m

∴ the volume of water flows in 1 hour = l X b X h

= 6 X 1.5 X 10000

So , volume of water flows in 30 minutes= (6 X 1.5 X 10000)  /2  m³

=45000 m³

Same volume of water fills the field.

Height of the standing water in the field (h)   = 8cm =0.08 m

Volume of water filled in the field = height X area = 45000 m³

Height X area = 45000

0.08 X Area = 45000

Area= 45000/0.08

= 562500 m²

Area irrigated in 30 minutes = 562500 m²

=56.25 hectare ( 10000 m² = 1 hectare )

19.

The perpendicular  from A on side BC of triangle ABC meets BC at D such that DB= 3CD. Prove that

2 AB² = 2 AC² + BC².

Solution:

It is given,

DB = 3CD

BC= DB+ CD

= 3CD+ CD

= 4 CD

∴CD= BC /4…………………………(1)

DB= 3CD

DB   = 3 BC/4…………………………(2)

Consider triangle ADC

∠ ADC = 90º

∴ AC² = AD² + CD²…………………… Pythagoras theorem

∴ AD²=AC² –  CD²………………….(3)

Consider triangle ABD

∠ADB=90º

∴ AB² = AD² +DB²

AD² = AB² – BD²……………..(4)

From  eqn(2) and eqn(3)

(AC²- CD²) = (AB² – BD²)

AC² – (BC/4)² = AB² -( 3BC/4)²

AC² – BC²/16 = AB²- 9BC²/16

(16 AC²- BC²)/16= (16 AB²-9BC²)/16

16 AC²- BC² = 16 AB² – 9BC²

16AC²= 16AB²  – 8BC²

Dividing by 8

2AC² = 2AB² – BC²

2AB² = 2AC² + BC²

                                                                   OR

AD and PM are medians of triangles ABC and PQR respectively where ABC∼ PQR. Prove that AB/PQ= AD/PM.

Solution:

AD is median of Δ ABC

∴BD=DC

PM is median of Δ PQR

∴QM=MR

Δ ABC∼ΔPQR

So,

AB/PQ= BC/QR

AB/PQ= (BC/2)/(QR/2)

AB/PQ= BD/QM………..(1)………Since 1/2 BC= BD, 1/2 QR= QM

In ΔABD and ΔPQM

∠B= ∠Q……………Given

AB/PQ= BD/QM………….Proved in (1)

∴ΔABD ∼ ΔPQM

20.

A chord of a circle of radius 14cm subtends an angle of 60°, at the centre. Find the area of the corresponding minor segment of the circle.(Use π= 22/7 and√3=1.73)

Solution:

 

 

 

 

 

 

 

 

Radius of the circle = 14 cm.

i.e. OA=OB=14 cm

Let the sector formed by OA and Ob is OAXB

Area of segment AXB= Area of sector OAXB- Area of ΔOAB………..(1)

Area of sector = (θ/360) X π r²

Area of sector OAXB= (60/360)x( 22/7)(14²)

=102.66cm²……………………………..(2)

In ΔAOB

AO= BO…………….Radii of same circle

∠OAB=∠ABO

∠OAB+∠ABO+∠AOB= 180º

∠OAB+∠ABO+60º= 180º………………………..∠AOB= 60º

∠OAB+∠ABO=120º

∴∠OAB=∠ABO=60º

So, ΔAOB is an equilateral triangle.

Area of equilateral triangle= (√3/4)side²

∴Area of ΔAOB = (√3/4)  X 14²

=(1.73/4)X 14²

Area of ΔAOB  =84.77 cm²…………………………………..(3)

From (1),(2) and (3)

Area of segment AXB=102.66cm²-84.77 cm²

=17.89 cm²

21)

Find the value of k such that the area of triangle ABC with A(k+1,1), B(4,-3) and C(7,-k) is 6 square units.

Solution:

Vertices of the triangle are A(k+1,1), B(4,-3) and c(7,-k)

Area of the triangle ABC= 6 units

We know,

Formula to Area of triangle = 1/2[x1(y2-y1) +x2(y3-y1)+x3(y1-y2)]

Area of triangle ABC= 1/2[(k+1)(-3+k)+4(-k-1)+7(1+3)]

=1/2[-3k-3+k² +k-4k-4+28]

= 1/2(k²-6k +21)

1/2(k²-6k +21)= 6…………………….Given

k²-6k +21=12

k² – 6k +9=0

k² -3k – 3k +9=0

k(k-3)-3(k-3)=0

(k-3)(k-3)=0

k-3=0

∴k=3

22.

If 2/3 and -3 are the zeroes of the polynomial ax² +7x+b, then find the values of a and b.

Solution :

Let p(x)=ax² +7x+b

2/3 is zero of p(x)

∴p(2/3)=0

i.e. a(2/3)² + 7X (2/3) +b=0

a(4/9) + 14/3 +b=0

(4a+42+9b)/9=0

4a+9b=-42…………………………..(1)

-3 is zero pf p(x)

∴p(-3)=0

a(-3)² +7 X -3 +b=0

9a+b=21………………………….(2)

Solving eqn(1) and (2)

4a+9b=-42

9a+b=21

Multiplying eqn (2) by 9

4a      +9b=-42

81a    +9b=189

(-)      (-)    (-)

—————-

-77a=  -231

a=-231/-77

a=3

Substituting a=3, in eqn(1)

4a +9 b=-42

4X3 +9b= -42

9b= -42-12

=-54