Summative Assessment I

Euclid's division lemma :

Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b

The zeroes of polynomial means that value of polynomial becomes zero.

In the above graph, the curve depicts the polynomial and it gets zero at two points, therefore p(x) has two zeroes.

In ΔADE and ΔABC

∠ADE = ∠ABC [Corresponding angles as DE || BC]

∠AED = ∠ACB [Corresponding angles as DE || BC]

ΔADE ~ ΔABC [By Angle-Angle Similarity criterion]

[Corresponding sides of similar triangles are in the same ratio]

Now,

Given, AD = 3 cm

DB = 2 cm

DE = 6 cm

⇒ AB = AD + DB = 3 + 2 = 5 cm

Using this in above equation,

⇒ BC = 10 cm

Given, we know that

sin θ = cos(90° - θ)

Replacing θ by 3θ

⇒ sin(3θ) = cos(90° - 3θ)

⇒ cos(θ - 2°) = cos(90° - 3θ)

[ Given, sin 3θ = cos(θ - 2°)]

⇒ θ - 2° = 90° - 3θ

⇒ 4θ = 92°

⇒ θ = 23°

Given,

tan θ = √3

⇒ tan²θ = 3

⇒ sec²θ - 1 = 3 [As tan²θ + 1 = sec²θ]

⇒ sec²θ = 4 …[1]

Also,

⇒ as tanθ = √3

Squaring both sides,

[As cot²θ + 1 = cosec²θ]

…[2]

Putting the values from [1] and [2] into given eqn

We know that if is a rational number, such that p and q are co-prime and q has factors in the form of 2^m.5ⁿ, then, decimal expansion of will terminate after the highest power of 2 or 5 (whichever is greater).

Therefore, will terminate after 3 places of decimal.

Comparing the equation with the set of equations

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

we have,

a₁ = 6, a₂ = 2

b₁ = -3, b₂ = -1

c₁ = 10_, c₂ = 9

and we have,

and and

So, we have

and in this case, we know that equations have no solution.

As we know that, the x-coordinate of the point of intersection of the

more than ogive and less than ogive give us a median of the data.

So, the median of the data is 18.5

(7 x 5 x 3 x 2 + 3) = (210 + 3) = 213

And 213 = 71 x 3

As, this number is expressible as product of two no's other, the given number is composite.

[Composition no's are those no's which has factors other than 1 and itself]

No, because degree of remainder cannot be equal to the degree of divisor

And in this case degree of divisor, i.e. 2x + 1 = 1

And degree of remainder, i.e. x -1 = 1 is equal.

Given,

3cos²θ + 7sin²θ = 4

⇒ 3cos²θ + 3sin²θ + 4sin²θ = 4

⇒ 3(cos²θ + sin²θ) + 4sin²θ = 4

⇒ 3 + 4sin²θ = 4

[as sin²θ + cos²θ = 1]

⇒ 4sin²θ = 1

⇒ θ = 30°

[as ]

⇒ cot θ = √3

[ as cot 30° = √3]

Now, To find :

[As, (a + b)(a - b) = a² - b²]

[As sin²θ + cos²θ = 1]

DE || AC [Given]

And we know, By Basic Proportional Theorem

If a line is drawn parallel to the one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in same ratio

…[1]

And DF || AE

By Basic Proportional Theorem,

[From [1]]

Hence, Proved

We have,

BC = BD + CD

[1]

As, AD ⏊ BC

⇒ ΔADC is a right-angled triangle

By Pythagoras theorem,[i.e. hypotenuse² = perpendicular² + base²]

AD² + CD²= CA²

⇒ AD² = CA² - CD² ….[2]

Also, ΔABD is a right-angled triangle

By Pythagoras theorem,

AD² + BD² = AB²

From [2]

CA² - CD² + BD² = AB²

⇒

[From [1]]

⇒ 2CA² - BC² = 2AB²

⇒ 2CA² = 2AB² + BC²

Hence, Proved.

In the given data,

The maximum class frequency is 32. So, the modal class is 30-40.

Lower limit(l) of modal class = 30

Class size(h) = 40 - 30 = 10

Frequency(f₁) of modal class = 32

Frequency(f₀) of class preceding the modal class = 12

Frequency(f₂) of class succeeding the modal class = 20

And we know,

Substituting values, we get

Let a be an positive odd integer, and let b = 4

By, using Euclid's division lemma,

a = 4q + r, where r is an integer such that, 0 ≤ r < 4

So, only four cases are possible

a = 4q or

a = 4q + 1 or

a = 4q + 2 or

a = 4q + 3

But 4q and 4q + 2 are divisible by 2, therefore these cases are not possible, as a is an odd integer.

Therefore,

a = 4q + 1 or a = 4q + 3.

Let 5 - √3 be rational,

Then, 5 - √3 can be expressed as where, p and q are co-prime integers and

q ≠ 0,

we have,

As p and q are integers, 5q - p is also an integer

is a rational number.

But √3 is an irrational number, so the equality is not possible.

This contradicts our assumption, that 5 - √3 is a rational number.

Therefore, 5 - √3 is an irrational number.

Let be rational,

Then, can be expressed as where p and q are co-prime integers and

q ≠ 0,

we have,

As p and q are integers, 5p and 3q are also integers

is a rational number.

But √3 is an irrational number, so the equality is not possible.

This contradicts our assumption, that is a rational number.

Therefore, is an irrational number.

Speed of boat in still water = 4 km/h

Let the speed of stream be 'x'

Therefore,

Speed of the boat upstream = Speed of boat in still water - Speed of stream = 4 - x

Speed of the boat downstream = Speed of boat in still water + Speed of stream = 4 + x

Time taken to go upstream

Time taken to go downstream

Given, time taken in upstream is thrice as in downstream

⇒ 4 + x = 12 - 3x

⇒ 4x = 8

⇒ x = 2

i.e. the speed of stream = x is 2 km/hour.

Let the number of correct answers = x

Let the number of wrong answers = y

Total no of questions attempted = x + y = 120

⇒ y = 120 - x ….[1]

Marks for each correct answer = 5

Marks for x correct answers = 5x

As 2 marks are deducted for each wrong question,

Marks deducted for y wrong answers = 2y

Total marks obtained by student will be 5x - 2y,

⇒ 5x - 2y = 348

⇒ 5x - 2(120 - x) = 348

⇒ 5x - 240 + 2x = 348

⇒ 7x = 588

⇒ x = 84

Hence, no of correct answers = x = 84

We know that, for a quadratic equation ax² + box + c

Sum of zeroes

Product of zeroes

Given equation = 2x² + x - 6 and zeroes are α and β

Therefore,

….[1] and

….[2]

Now, any quadratic equation having α and β as zeroes will have the form

p(x) = x² - (α + β)x + αβ

⇒ equation having α and β as zeroes will have the form

p(x) = x² - (2α + 2β)x + (2α)(2β)

⇒ p(x) = x² - 2(α + β)x + 4αβ

From [1] and [2]

Hence required equation is x² + x - 12.

Taking L.H.S

= (cosecθ - sinθ)(secθ - cosθ)

We know, sin²θ + cos²θ = 1

Therefore,

Taking R.H.S

[as sin²θ + cos²θ = 1]

LHS = RHS

Hence, Proved.

Given,

cos θ + sin θ = √2 cos θ …[1]

Squaring both side,

(cos θ + sin θ)² = 2 cos²θ

⇒ cos²θ + sin²θ + 2cosθsinθ = 2cos²θ

⇒ 2cosθsinθ = 2cos²θ - cos²θ - sin²θ

⇒ 2cosθsinθ = cos²θ - sin²θ

⇒ 2cosθsinθ = (cosθ - sinθ)(cosθ + sinθ)

⇒ 2cosθsinθ = (cosθ - sinθ)( √2 cosθ) [From [1]]

Hence, Proved.

Given: ΔABC and ΔDBC with common base BC.

To Prove:

Construction: Draw AM ⏊ BC and DN ⏊ BC

Proof:

In ΔAMO and ΔDNO

∠AOM = ∠DON [Vertically opposite angle]

∠AMO = ∠DNO [Both 90°]

ΔAMO ~ ΔDNO [By Angle-Angle sum criterion]

[Corresponding sides of similar triangles are in the same ratio] [1]

Now, we know that

Area of a triangle

Therefore,

[From [1]]

Hence, Proved

Proof:

Given: In ΔABC, the AD is a median and E is mid-point of the AD and BE is produced to meet AC in F.

To Prove:

Construction: Draw DG || BF as shown in figure

Proof:

Now, In ΔBFC

DG || BF [By construction]

As AD is a median on BC, D is a mid-point of BC

Therefore,

G is a mid-point of CF [By mid-point theorem]

⇒ CG = FG …[1]

Now, In ΔADG

EF || DG [By Construction]

As E is a mid-point of AD [Given]

Therefore,

F is a mid-point of AG [By mid-point theorem]

⇒ FG = AF …[2]

From [1] and [2]

AF = CG = FG …[3]

And

AC = AF + FG + CG

⇒ AC = AF + AF + AF [From 3]

⇒ AC = 3AF

Hence Proved

Let us first calculate the mid-values(x_i) for each class-interval, By using the formula

Let us assume the assumed mean(a) = 75

and from that, we get the data as shown in above table.

And we know, By step-deviation method

Where, a = assumed mean

h = class size

Let us first calculate the mid-values(x_i) for each class-interval, By using the formula

By which, we get the following data

We know, that

Given, mean = 24

⇒ 1920 + 24p = 1700 + 35p

⇒ 11p = 220

⇒ p = 20

First, let us make a cumulative frequency distribution of less than type.

In this case,

Sum of all frequencies, n = 53

Now, we know the median class is whose cumulative frequency is greater than and nearest to .

As, a Cumulative frequency greater than and nearest to 26.5 is 29, the median class is 60 - 70.

Median

where l = lower limit of median class,

n = number of observations,

cf = cumulative frequency of class preceding the median class,

f = frequency of median class,

h = class size

In this case,

l = 60

n = 53

cf = 22

f = 7

h = 10

Putting values, we get,

Median

Two zeroes are √3 and -√3,

Therefore (x -√3)(x - (-√3) = (x - √3)(x + √3) is a factor of p(x).

Let us divide p(x) by (x - √3)(x + √3) = (x² - 3)

⇒ (2x⁴ - 3x³ - 5x² + 9x - 3) = (x² - 3)(2x² - 3x + 1)

= (x - √3)(x + √3)(2x² - 2x - x + 1)

= (x - √3)(x + √3)(2x(x - 1) - 1(x - 1))

= (x - √3)(x + √3)(2x - 1)(x - 1)

Hence,

2x - 1 = 0 or x - 1 = 0

or x = 1

Hence, other two zeroes are or 1.

Let ΔPQR and ΔABC be two similar triangles,

[Corresponding sides of similar triangles are in the same ratio] [1]

And as corresponding angles of similar triangles are equal

∠A = ∠P

∠B = ∠Q

∠C = ∠R

Construction: Draw PM ⏊ QR and AN ⏊ BC

In ΔPQR and ΔABC

∠PMR = ∠ANC [Both 90°]

∠R = ∠C [Shown above]

ΔPQR ~ ΔABC [By Angle-Angle Similarity]

[Corresponding sides of similar triangles are in the same ratio] [2]

Now, we know that

Area of a triangle

Therefore,

[From 2]

[From 1]

[From 1]

Hence, Proved.

Let us consider a triangle ABC, in which

AC² = BC² + AB² …[1]

To Prove: Angle opposite to the first side i.e. AC is right angle or

∠ ABC = 90°

Construction:

Let us draw another right-angled triangle PQR right-angled at Q, with

AB = PQ

BC = QR

Now, By Pythagoras theorem, In ΔPQR

PR² = QR² + PQ²

But QR = BC and PQ = AB

⇒ PR² = BC² + AB²

But From [1] we have,

AC² = PR²

⇒ AC = PR

In ΔABC and ΔPQR

AB = PQ [Assumed]

BC = QR [Assumed]

AC = PR [Proved above]

⇒ ΔABC ≅ ΔPQR [By Side-Side-Side Criterion]

⇒ ∠ABC = ∠PQR [Corresponding parts of congruent triangles are equal]

But, ∠PQR = 90°

⇒ ∠ABC = 90°

Hence, Proved !

Taking LHS

Dividing by cosθ in numerator and denominator

Using and

Putting 1 = sec²θ - tan²θ in numerator

Using a² - b² = (a + b)(a - b)

= tan θ + sec θ

Now, taking RHS

Multiplying and dividing by secθ + tanθ = 1

= tanθ + secθ [As sec²θ - tan²θ = 1]

LHS = RHS

Hence Proved.

Using cosec(90° - θ) = secθ

and cot(90° - θ) = tanθ

we have,

Now, sin(90 - θ) = cos θ and

tan(90 - θ) = cot θ we have

[ Since,

tan²θ - sec²θ = 1

sin²θ + cos²θ = 1

tan 60° = √3]

Taking RHS

Now, sec²θ - tan²θ = 1 and (a + b)² = a² + b² + 2ab

Now, and , using these we have

= LHS

Hence, Proved !

Equation 1:

2x - y = 1

Plot the line with equation 1 on graph.

Equation 2:

x - y = -1

Plot the line with equation 2 on graph.

From the graph We observe point of intersection of two lines is (2, 3)

Region bound by these lines and y-axis is shaded in the graph.

Let us draw cumulative frequency with table for the above data

Taking Yield as x-axis and Cumulative frequencies as y-axis, we draw its more than 'ogive'

Eqn1 : ax + by - a + b = 0

⇒ ax + by = a - b

Multiplying both side by b

⇒ abx + b²y = ab - b² …[1]

Eqn2 : bx - ay - a - b = 0

⇒ bx - ay = a + b

Multiplying both side by a

⇒ abx - a²y = a² + ab …[2]

Subtracting [2] from [1]

abx - a²y - (abx + b²y) = a² + ab - (ab - b²)

⇒ abx - a²y - abx - b²y = a² + ab - ab + b²

⇒ -y(a² + b²) = a² + b²

⇒ -y = 1

⇒ y = 1

Putting value of y in eqn1, we get

ax + b(-1) - a + b = 0

⇒ ax - b - a + b = 0

⇒ ax = a

⇒ x = 1

So, x = 1 and y = -1

Taking LHS

Multiplying and dividing by (1 - cos θ)

As sin²θ + cos²θ = 1

Hence Proved.

Given: A ΔABC right-angled at B, and D is the mid-point of BC, i.e. BD = CD

To Prove: AC² = (4AD² - 3AB²)

Proof:

In ΔABD,

By Pythagoras theorem, [i.e. Hypotenuse² = Base²+ Perpendicular²]

AD² = AB² + BD²

[ as D is mid-point of BC, therefore,

⇒ 4AD² = 4AB² + BC²

⇒ BC² = 4AD² - 4AB² [1]

Now, In ΔABC, again By Pythagoras theorem

AC² = AB² + BC²

AC² = AB² + 4AD² - 4AB² [From 1]

AC² = 4AD² - 3AB²

Hence Proved !

Let us make the table for above data and containing cumulative frequency and mid-values for each data

MEAN

We know, that

MODE

In the given data,

The maximum class frequency is 30. So, the modal class is 30-40.

Lower limit(l) of modal class = 30

Class size(h) = 40 - 30 = 10

Frequency(f₁) of modal class = 30

Frequency(f₀) of class preceding the modal class = 18

Frequency(f₂) of class succeeding the modal class = 20

And we know,

Substituting values, we get

MEDIAN

In this case,

Sum of all frequencies, n = 100

Now, we know the median class is whose cumulative frequency is greater than and nearest to .

As, Cumulative frequency greater than and nearest to 50 is 63, the median class is 30 - 40.

Median

where l = lower limit of median class,

n = number of observations,

cf = cumulative frequency of class preceding the median class,

f = frequency of median class,

h = class size

In this case,

l = 30

n = 100

cf = 33

f = 30

h = 10

Putting values, we get,

Median

We know that Dividend = Divisor × Quotient + Remainder

According to the problem :

Dividend 1 = 245

Dividend 2 = 1029

Dividend - Remainder = Divisor × Quotient

So Dividend 1- Remainder = 240 = Divisor × Quotient 1

Prime Factor of 240 = 2⁴ × 3 × 5

Dividend 2 – Remainder = 1024 = Divisor × Quotient 2

Prime Factor of 1024 = 2⁴ × 2⁶

Since, the Divisor is common for both the numbers we need to find the Highest Common Factor between both the numbers. From the Prime factors, we find the

Highest Common Factor between the two numbers is 2⁴ = 16

Given Equation :ax² – 6x – 6 = 0

which is of the form ax² + bx + c = 0 (General Form)

The product of the roots of the general form of equation

So according to the given Equation Product of the roots

The Value Of a for which the equation has product of root 4 =

Given :

Area of ΔABC = 25 cm²

Area of ΔPQR = 49 cm²

Length of QR = 9.8 cm.

Since both the triangles are similar so according to the Area –Length relations of similar triangle we can write

The length Of The side BC is 7 cm.

Given sin (θ + 34⁰) = cos θ …Equation 1

Since sin θ & cos θ are complementary to each other

so sin θ = cos (90⁰ – θ)

Using the above relations in Equation 1 we get

cos (90⁰ – θ – 34⁰) = cos θ

Since both L.H.S. and R.H.S. are functions of cosine and θ + 34⁰ is acute so we can write

90⁰ – θ – 34⁰ = θ

⇒ 2θ = 56⁰

⇒ θ = 28⁰

Given cos θ = 0.6

⇒ sin θ = 0.8

According to the question, the required problem needs us to find

5 sin θ- 3 tan θ

The value of the expression is 0

Prime factorization of 1095 = 5×3×73

Prime factorization of 1168 = 2⁴ ×73

So

Since 73 is a common factor for both numerator and denominator so it cancels out

The Simplest form is

Equation 1: 4x - 5y = 20

Equation 2: 3x + 5y = 15

Both the equations are in the form of :

a₁x + b₁y= c₁ & a₂x + b₂y = c₂ where

According to the problem:

a₁ = 4

a₂ = 3

b₁ = -5

b₂ = 5

c₁ = 20

c₂ = 15

We compare the ratios

Since , So

Ithas a Unique solution

Given: mode = x(median) - y(mean)

According to an empirical relation, the relation between Mean, Median & Mode is given by

Mode = 3 Median – 2 Mean …Eq(1)

This empirical relation is very much close to the actual value of mode which is calculated. So this relation is valid.

Comparing the Relation given with equation 1 we find

x = 3 & y = 2

When a number ends with 0 it has to be divisible by the factors of 10 which are 5 and 2

Now 6ⁿ = (3 × 2)ⁿ …Equation 1

From Equation 1 We can see the factors of 6 are only 3 & 2.

There are no factors as powers of 5 in the factorization of 6

Hence 6ⁿ cannot end with 0

Given Equation : 9x² - 5 = 0

which is of the form ax² + bx + c = 0 (General Form)

For finding the zeroes of the polynomial we use the method of Factorization

9x² - 5 = 0

⇒ 9x² = 5

⇒ x² =

⇒ x =

The zeroes of the polynomial expression are

Given 2 sin 2θ = √3

⇒ sin 2θ = sin 60⁰

⇒ 2θ = 60⁰

⇒ θ = 30⁰

Given: 7 sin² θ + 3 cos² θ = 4

Since sin² θ + cos² θ = 1 …Equation 1

So the equation becomes

4 sin² θ = 1

⇒ sin² θ =

From Equation 1 we get

cos² θ =

Since

⇒

Hence Proved

Given :

AD = 5 cm

DB = 8 cm

AC = 6.5 cm

DE ||BC

In ΔABC & ΔADE

∠ADE = ∠ABC (Corresponding Angles)

∠AED = ∠ACB (Corresponding Angles)

So ΔABC & ΔADE are similar by the A.A. (Angle-Angle) axiom of Similarity

AB = AD + BD = 13 cm.

Since the two triangles are similar so their lengths of sides must be in proportion.

⇒

⇒

AE = 2.5cm.

Given:

ADC = BAC

D is a point on the side BC

∠ACB = ∠ACD (Common Angle)

So ΔABC & ΔADC are similar by the A.A. (Angle-Angle) axiom of Similarity

Since the two triangles are similar so their lengths of sides must be in proportion

Cross Multiplying We Get

CA² = DC x CB

Which is the required expression

Hence Proved

Class corresponding to maximum frequency = (4-8)

f₁ (Frequency of the modal class) = 8

f₀ (Frequency of the class preceding the modal class) = 4

f₂ (Frequency of the succeeding modal class) = 5

l(lower limit) = 4

h(width of class) = 4

Mode =

⇒ Mode =

Mode = 6.29

According to Euclid’s algorithm p = 6q + r

where r is any whole number 0< = r<6 and p is a positive integer

Since 6q is divisible by 2 so the value of r will decide whether it is odd or even.

Also since r<6 so only 6 cases are possible

For r = 1 , 3, 5 we get three odd numbers and for r = 0 , 2 , 4 we get three even numbers

So (6q + 1) , (6q + 3) & (6q + 5) represents positive odd integers .

Hence Proved

Let us assume (3 -√15 ) is rational

(Assume)

where a & b are integers (b≠0)

⇒

⇒

Now let’s solve the R.H.S. Of the above equation

Let

Squaring we get

In The above equation since 15 divides p² so it must also divide p

so p is a multiple of 15

let p = 15k where k is an integer

Putting in Equation 1 the value of p we get

15q² = 225k²

⇒ q² = 15k²

Since 15 divides q² so it must also divide q

so q is a multiple of 15

But this contradicts our previously assumed data since we had considered p & q has been resolved in their simplest form and they shouldn't have any common factors.

So √15 is irrational and hence

(3 -√15) is also irrational

Hence Proved

Let us consider to be rational

where a & b are integers (b≠0)

Rearranging we get

The R.H.S of the above expression is a rational number since it can be expressed as a numerator by a denominator

Let L.H.S = where p and q are integers (q≠0)

⇒

⇒

Squaring both sides we get

2q² = p²…Equation 1

Since 2 divides p² so it must also divide p

so p is a multiple of 2

let p = 2k where k is an integer

Putting in Equation 1 the value of p we get

2q² = 4k²

⇒ q² = 2k²

Since 2 divides q² so it must also divide q

so q is a multiple of 2

But this contradicts our previously assumed data since we had considered p & q has been resolved in their simplest form and they shouldn't have any common factors.

So √2 is irrational and hence

is also irrational

Hence Proved

Let the number added to each of the numbers to make them in proportion be x

When any four numbers (a, b, c, d)are in proportion then

Applying the above equation for our problem we get

⇒ (5 + x)(27 + x) = (17 + x)(9 + x)

⇒ 135 + 32x + x² = 153 + 26x + x²

⇒ 6x = 18

The number added should be 3

Let the two numbers be x & Y

x + y = 18 (Given) …Equation 1

(Given) …Equation 2

Solving Equation 2 We get

Putting the value from Equation 1 we get

⇒ xy = 72

…Equation 3

Putting the value of Equation 3 in Equation 1 We get

⇒ x² + 72 = 18x

⇒ x² - 18x + 72 = 0

⇒ (x-6)² = 0

⇒ x = 6

Putting the value of x in Equation 1 we get y = 12

The two numbers are 6 & 12

Given Equation : x² - x - 12 = 0

which is of the form ax² + bx + c = 0 (General Form)

The product of the roots of the general form of equation

Sum of Roots of the general equation =

So

⇒ α + β = 1

⇒ 2(α + β) = 2 ….Equation 1

Similarly

α×β = -12

⇒ 2α×2β = -48 …Equation 2

The new equation will be formed by combining the results of Equation 1 & 2

The New Polynomial Formed from the new roots is x² -2x-48

Given L.H.S. = (sin θ + cosec θ)² + (cos θ + sec θ)²

We know

⇒ sin² θ + cosec² θ + 2 + cos² θ + sec² θ + 2

Also From the Trigonometrical identities

sin²θ + cos²θ = 1

cosec² θ = 1 + cot² θ

sec² θ = 1 + tan² θ

⇒ 1 + 1 + cot² θ + 2 + 1 + tan² θ + 2

⇒ 7 + cot² θ + tan² θ

So, L.H.S = R.H.S

Hence Proved

Given sec θ + tan θ = m

So, we can write

Squaring both sides we get

Since cos² θ = 1- sin² θ

⇒

⇒

Applying Componendo & Dividendo i.e.

is equivalent to

we get

Hence Proved

Given :

AB|| CD

AB = 2 x CD

⇒

∠AOB = ∠COD (Vertically Opposite angles)

∠DCO = ∠OAB (Alternate Angles)

So ΔAOB & ΔDOC are similar by the A.A. (Angle Angle) axiom of Similarity

Since both the triangles are similar so according to the Area –Length relations of similar triangle we can write

⇒

Area of ∆DOC = 21cm²

Given:

AB BC

GFBC

DE AC

Since AB BC so ∠DAE & ∠GCF are complementary angles i.e.

∠DAE + ∠GCF = 90⁰ ….Equation 1

Similarly since GFBC so ∠CFG & ∠GCF are complementary angles i.e.

∠CGF + ∠GCF = 90⁰ ….Equation 2

Combining Equation 1 & 2 We can say that

∠CGF = ∠DAE

Also ∠CFG = ∠DEA (Perpendicular Angles)

So ΔCGF is similar to ΔADE By A.A. (Angle Angle)axiom of similarity

Hence Proved

h (Represents the class width) = 10

a (Assumed mean) = 25

So Mean according to Step Deviation method:

Mean =

Mean = 25

Mean = 78 (Given)

According to the direct method

Mean =

⇒ 2886 + 78p = 2665 + 95p

⇒ 17p = 221

Value of p is 13

Total frequency(n) = 30

15 lies in the interval 55-60

so l (lower limit) = 55

c_f(Cumulative frequency of the preceding class of median class) = 13

f (frequency of median class) = 6

h (class size) = 5

Median =

Median =

Median = 56.67Kg

Given: p(x) = 2x⁴ + 7x³ - 19x²- 14x + 30

Since x = √2 & - √2 is a solution so

x- √2 & x + √2 are two factors of p(x)

Multiplying the two factors we get x²-2 …Equation 1

which is also a factor of p(x)

To get the other two factors we need to perform long division

On performing long division we will get

2x² + 7x -15 …Equation 2

Equation 2 is also a factor of p(x)

To find the other two zeroes of the polynomial we need to solve Equation 2

We use the method of factorization for solving Equation 2

2x² + 7x -15 = 0

⇒ 2x² + 10x -3x -15 = 0

⇒ 2x(x + 5) -3(x + 5) = 0

⇒ (2x-3)(x + 5) = 0

The two roots areand -5

Let us assume BFEC is a square , ΔABF is an equilateral triangle described on the side of the square & Δ CFD is an equilateral triangle describes on diagonal of the square

Now since ΔABF & Δ CFD are equilateral so they are similar

Let side CE = a,

So EF = a

CF² = a² + a²

CF² = 2a²

Since both the triangles are similar so according to the Area –Length relations of similar triangle we can write

⇒

So Area Of Δ CFD = 2 ΔABF

Hence Proved

Let us assume ΔABC & ΔPQR are similar

Area of ΔABC = 0.5 ×AD ×BC

Area of ΔPQR = 0.5 ×PS ×QR

Now since the two triangles are similar so the length of sides and perpendiculars will also be in proportion

…Equation 1

…Equation 2

From Equation 1 We get

Putting in Equation 2 we get

⇒

So we can see ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides

Hence Proved

Given: L.H.S. =

Since we know

So L.H.S.
=
=

Multiplying Numerator & Denominator with sin θ - (1 - cos θ) we get

L.H.S.

=

=

Since sin² θ + cos² θ = 1

=

=

=

L.H.S. = R.H.S.

Hence Proved

Given: …Equation 1

We know

sec θ = cosec (90⁰ – θ)

tan θ = cot (90⁰ – θ)

sin θ = cos (90⁰ – θ)

Using the above three relations in Equation 1 we get

We also know

sin² θ + cos² θ = 1

⇒ sin² 55 + cos² 55 = 1

And,

∴

⇒

⇒

⇒

Given sec θ + tan θ = m

So we can write

Squaring both sides we get

Since cos² θ = 1- sin² θ

⇒

⇒

Applying Componendo & Dividendo i.e.

is equivalent to

we get

Hence Proved

Given: The equations 3x + y- 11 = 0 and x-y- 1 = 0.
To find: the region bounded by these lines and the y-axis.
Solution:
For 3x + y - 11 = 0
y = 11 - 3x
Now for x = 0
y = 11 - 3(0)
y = 11
For x = 3
y = 11 - 3(3)
y = 11 - 9
y = 2
Table for equation 3x + y -11 = 0 is

For x-y- 1 = 0
y = x - 1

Now for x = 0
y = 0 - 1
y = -1
For x = 3
y = 3 - 1
y = 2
Table for equation x-y- 1 = 0 is

Plot the points (0,-1),(3,2)
The graph is shown below:

Given:

Equation 1: 2x - 3y = 7

Equation 2: (k + 1)x + (1 - 2k)y = (5k - 4)

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

For the system of linear equations to have infinitely many solutions, we must have

………(i)

According to the problem:

a₁ = 2

a₂ = k + 1

b₁ = -3

b₂ = 1-2k

c₁ = 7

c₂ = 5k-4

Putting the above values in equation (i) we get:

⇒ 2(1-2k) = -3(k + 1)

⇒ 2-4k = -3k-3

⇒ k = 5

Thevalue of k for which the system of equations has infinitely many solutions is k = 5

To Prove: (sin - cosec)(cos- sec) =

L.H.S. = (sin θ –cosec θ)(cos θ- sec θ)

⇒

⇒

Since sin²θ + cos²θ = 1 , So

⇒

After Cancellation we get

L.H.S. = sin θ cos θ

Dividing the numerator and denominator with cos θ we get

⇒

We know

⇒

Since sec²θ = 1 + tan²θ

⇒

Dividing The Numerator and denominator by tan θ we get

⇒

Since

= R.H.S

Since L.H.S. = R.H.S

Hence Proved

Given:

AC = BC

AB² = 2AC² …(Equation 1)

Equation 1 can be rewritten as

AB² = AC² + AC²

Since AC = BC we can write

AB² = AC² + BC² …Equation 2

Equation 2 represents the Pythagoras theorem which states that

Hypotenuse² = Base² + Perpendicular²

Since Pythagoras theorem is valid only for right-angled triangle so

So ABC is a right angled triangle right angled at C

HenceProved

According to the direct method

Mean =

⇒ Mean =

⇒ Mean = 205

Total frequency(n) = 30

15 lies in the interval 200-250

so l (lower limit) = 200

c_f(Cumulative frequency of the preceding class 200-250) = 13

f (frequency of median class) = 12

h (class size) = 50

Median =

Median =

Median = 208.33