 CBSE Board 2020 Solution for Math for Class 10 Section C

# CBSE Board 2020

## Mathematics

### Question Paper Solution

#### Section C

##### 3 Marks Questions

Question 27: If a circle touches the side BC of a triangle ABC at P and extended sides AB and Ac at Q and R, respectively, prove that

AQ=1/2(BC+CA+AB) Answer: BP = BQ, CP = CR and AQ = AR (because tangents from same point are equal)

BP + CP = BC

Or, BC = BQ + CR

AQ = AB + BQ = AB + BP

AR = AC + CR = AC + CP

So, AQ + AR = AB + AC + BP + CP

Or, 2AQ = AB + AC + BC

Or, AQ = 1/2(BC+CA+AB) proved

Question 28: The area of a circular playground is 22176 cm2. Find the cost of fencing this ground at the rate of Rs. 50 per meter.

Answer: Area of circle = πr^2

Or, πr^2=22176

Or, r^2=22176×7/(22)=7056

Or, r=sqrt(7056)=84 cm

Circumference =2πr

=2×(22)/7×84=528 cm

= 5.28 m

Cost =5.28×50=264 Rupees

Question 29: If the mid-point of the line segment joining the points A (3, 4) and B (k, 6) is P(x, y) and x+y-10=0, find the value of k.

Answer: x+y-10=0

Or, x+y=10

Or, x=10-y

From midpoint formula:

x=(3+k)/2

y=(4+6)/2=5

Using this value, we get:

x=10-5=5

Substituting the value of x in midpoint formula;

5=(3+k)/2

Or, 3+k=10

Or, k=10-3=7

OR

Find the area of triangle ABC with A(1, -4) and the mid-points of sides through A being (2, -1) and (0, -1).

Answer: Let us assume midpoints to be M and N

Area of triangle AMN can be calculated as follows:

1/2(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))

=1/2(1(-1+1)+2(-1+4)+0(-4+1))

=1/2(1×0+2×3+0)

=3 sq units

Area of triangle ABC will be 4 times the area of triangle AMN, because ratio of areas is square of ratios of sides.

So, area of triangle ABC = 12 sq units

Question 30: In following figure, if Δ ABC ∼ Δ DEF and their sides of lengths (in cm) are marked along them, then find the lengths of sides of each triangle. Answer: (3x)/(6x)=1/2

So, (2x-1)/(18)=1/2

Or, 2x-1=9

Or, 2x=10

Or, x=5

Using the value of x, we can calculate different sides as follows:

AB =2x-1=2×5-1=10-1=9 cm

BC =2x+2=2×5+2=10+2=12

AC =3x=3×5=15

So, DF =6×5=30

EF =12×2=24

Question 31: If 2x+y=23 and 4x-y = 19, find the value of (5y-2x) and (y/x-2)

2x+y+4x-y=23+19

=6x=42

Or, x=7

Substituting the value of x in second equation, we get:

4x-y=19

Or, 4×7-y=19

Or, 28-y=19

Or, y=28-19=9

Now, 5y-2x=5×9-2×7

=45-14=31

Similarly, (y/x-2)=9/7-2

=(9-14)/7=-5/7

OR

Solve for x:1/(x+4)-1/(x+7)=(11)/(30), x≠-4,7

Answer: (1)/(x+4)-(1)/(x-7)=(11)/(30)

Or, (x-7-x-4)/((x+4)(x-7))=(11)/(30)

Or, (x^2-7x+4x-28)/(30)=-1

Or, x^2-3x-28=-30

Or, x^2-3x-28+30=0

Or, x^2-3x+2=0

Or, x^2-2x-x+2=0

Or, x(x-2)-1(x-2)=0

Or, (x-1)(x-2)=0

Hence,roots are; 1 and 2

Question 32: Which term of the AP, 20, 191/4, 181/2, 173/4, ……….is the first negative term.

Answer: Value of d can be calculated as follows:

20-191/4=20-(77)/4

=(80-77)/4=3/4

20 is not divisible by 3/4, so let us take 21 and divide it by 3/4 to guess the possible number of terms

21÷3/4=28

Let us find 28th term:

n_(28)=a-(28-1)×3/4

=20-27×3/4

=20-(81)/4=(80-81)/4=-1/4

28th term is the first negative term

OR

Find the middle term of the AP 7, 13, 19, ……. 247

Answer: Here, a = 7, d = 6 and last term = 247

247=7+(n-1)6

Or, 6n-6=240

Or, 6n=246

Or, n=246÷6=41

So, Middle term = 21

n21 = 7 + 20 × 6 = 127

Question 33: Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm standing water is required?

Answer: Length in 1 hour = 10000 m so it will be 5000 m in 30 minutes

Volume in 30 minutes = Length × Width × Depth

5000×6×1.5

So, area for 8 cm = 0.08 m

=5000×6×1.5÷0.08=562500 sq m

Question 34: Show that

(text(cos)^2(45°+θ)+text(cos)^2(45°-θ))/(text(tan)(60°+θ)text(tan)(30°-θ))=1

Answer: LHS =(text(cos)^2(45°+θ)+text(sin)^2(90°-45°-θ))/(text(tan)(60°+θ)text(cot)(90°-60°-θ))

=1/1=1

Because sin2θ + cos2θ = 1

And tan2θ × cot2θ = 1