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 cm^{2}. 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

n_{21} = 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 sin^{2}θ + cos^{2}θ = 1

And tan^{2}θ × cot^{2}θ = 1

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