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)`
Answer: Adding initial two equations, we get:
`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