CBSE Board 2020


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


Circle and triangle

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:



Using this value, we get:


Substituting the value of x in midpoint formula;


Or, `3+k=10`

Or, `k=10-3=7`


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:




`=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.

Similar triangles

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)`

Answer: Adding initial two equations, we get:



Or, `x=7`

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


Or, `4×7-y=19`

Or, `28-y=19`

Or, `y=28-19=9`

Now, `5y-2x=5×9-2×7`


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



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


Let us find 28th term:




28th term is the first negative term


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

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


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


So, area for 8 cm = 0.08 m

`=5000×6×1.5÷0.08=562500` sq m

Question 34: Show that


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


Because sin2θ + cos2θ = 1

And tan2θ × cot2θ = 1

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