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Correct option is  B) 2

ABCD is a rectangle, where CD = 6 cm and AD = 8 cm (Given) 

From figure: Parallelogram CDEF and rectangle ABCD on the same base and between the same parallels, which means both have equal areas. 

Area of parallelogram CDEF = Area of rectangle ABCD ….(1) 

Area of rectangle ABCD = CD x AD 

= 6 x 8 cm² 

= 48 cm² 

Equation (1) => Area of parallelogram CDEF = 48 cm².

We know that

Area of parallelogram PQRS = ½ (Area of parallelogram ABCD)

Construct diagonals AC, BD and SQ

From the figure we know that S and R are the midpoints of AD and CD

Consider △ ADC

Using the midpoint theorem

SR || AC

From the figure we know that P and Q are the midpoints of AB and BC

Consider △ ABC

Using the midpoint theorem

PQ || AC

It can be written as

PQ || AC || SR

So we get

PQ || SR

In the same way we get SP || RQ

Consider △ ABD

We know that O is the midpoint of AC and S is the midpoint of AD

Using the midpoint theorem

OS || AB

Consider △ ABC

Using the midpoint theorem

OQ || AB

It can be written as

SQ || AB

We know that ABQS is a parallelogram

We get

Area of △ SPQ = ½ (Area of parallelogram ABQS) ……. (1)

In the same way we get

Area of △ SRQ = ½ (Area of parallelogram SQCD) …….. (2)

By adding both the equations

Area of △ SPQ + Area of △ SRQ = ½ (Area of parallelogram ABQS + Area of parallelogram SQCD)

So we get

Area of parallelogram PQRS = ½ (Area of parallelogram ABCD)

Therefore, it is proved that ar (||gm PQRS) = ½ × ar (||gm ABCD).

Correct option is  A) 3

\(\frac{xyz^2}{-9xz}\) = \((\frac{1}{-9}x^{1-1}yz^{2-1})\) = \(-\frac{1yz}{9}\) = \(-\frac{yx}{9}\) [Using aⁿ÷a^m = a^n-m] and [a⁰ = 1]

We have,

(acx² + (bc + ad) x + bd) / (ax + b)

(acx² + (bc + ad) x + bd) / (ax + b) 

= cx + d + 0/ (ax + b)

= cx + d

∴ The answer is (cx + d).

We have,

[(a² + 2ab + b²) – (a² + 2ac + c²)] / (2a + b + c)

[(a² + 2ab + b²) – (a² + 2ac + c²)] / (2a + b + c) 

= b – c + 0/(2a + b + c)

= b – c

∴ The answer is (b – c)

\(-\frac{4a^3}{2a}+\frac{4a^2}{2a}+\frac{a}{2a}\) = -2a² + 2a + \(\frac{1}{2}\) [Using aⁿ÷a^m = a^n-m]

We have,

(-4a³ + 4a² + a) / 2a

-4a³/2a + 4a²/2a + a/2a

By using the formula aⁿ / a^m = a^n-m

-2a^3-1 + 2a^2-1 + 1/2 a^1-1

-2a² + 2a + 1/2

We have,

(ax² – ay²)/ (ax+ay)

(ax² – ay²)/ (ax+ay) 

= (x – y) + 0/(ax+ay)

= (x – y)

∴ The answer is (x – y).

We have,

(–x⁶ + 2x⁴ + 4x³ + 2x²) / √2x²

-x⁶/√2x² + 2x⁴/√2x² + 4x³/√2x² + 2x²/√2x²

By using the formula aⁿ / a^m = a^n-m

-1/√2 x^6-2 + 2/√2 x^4-2 + 4/√2 x^3-2 + 2/√2 x^2-2

-1/√2 x⁴ + √2x² + 2√2x + √2

We have,

72xyz² / -9xz

By using the formula aⁿ / a^m = a^n-m

72/-9 x^1-1 y z^2-1

-8yz

We have,

(9x²y – 6xy + 12xy²) / -3/2xy

9x²y/(-3/2xy) – 6xy/(-3/2xy) + 12xy²/(-3/2xy)

By using the formula aⁿ / a^m = a^n-m

(-9×2)/3 x^2-1y^1-1 – (-6×2)/3 x^1-1y^1-1 + (-12×2)/3 x^1-1y^2-1

-6x + 4 – 8y

(C) Perimeter of ABCD > Perimeter of ABEM

Explanation:

In rectangle ABEM,

AB = EM …(eq.1) [sides of rectangle]

In parallelogram ABCD,

CD = AB …(eq.2)

Adding, equations (1) and (2),

We get

AB + CD = EM + AB …(i)

We know that,

Perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.

BE < BC and AM < AD

[because, in a right angled triangle, the hypotenuse is greater than the other side]

On adding both above inequalities, we get

SE + AM <BC + AD or BC + AD> BE + AM

On adding AB + CD both sides, we get

AB + CD + BC + AD> AB + CD + BE + AM

⇒ AB+BC + CD + AD> AB+BE + EM+ AM    [∴ CD = AB = EM]

Hence,

We get,

Perimeter of parallelogram ABCD > perimeter of rectangle ABEM

Hence, option (C) is the correct answer.

We have,

(3x³y² + 2x²y + 15xy) / 3xy

3x³y²/3xy + 2x²y/3xy + 15xy/3xy

By using the formula aⁿ / a^m = a^n-m

3/3 x^3-1y^2-1 + 2/3 x^2-1y^1-1 + 15/3 x^1-1y^1-1

x²y + 2/3x + 5

\(\frac{5z^3}{2z}-\frac{6z^2}{2z}+\frac{7z}{2z}\) = \(\frac{5z^2}{2}-3z+\frac{7}{2}\)[Using aⁿ÷a^m = a^n-m]

We have,

6x³y²z² / 3x²yz

By using the formula aⁿ / a^m = a^n-m

6/3 x^3-2 y^2-1 z^2-1

2xyz

\(\frac{2\times 9x^2y}{-3xy}-\frac{2\times 6xy}{-3xy}-\frac{2\times 12xy^2}{-3xy}\) = - 6x²y + 4y - 8y [Using aⁿ÷a^m = a^n-m]

We have,

(4z³ + 6z² – z) / -1/2z

4z³/(-1/2z) + 6z²/(-1/2z) – z/(-1/2z)

By using the formula aⁿ / a^m = a^n-m

-8 z^3-1 – 12z^2-1 + 2 z^1-1

-8z² – 12z + 2

\(\frac{2\times 4z^3}{-1z}+\frac{2\times 6z^2}{-1z}+\frac{2\times z}{-1z}\) = - 8z² - 12z + 2 [Using aⁿ÷a^m = a^n-m]

We have,

(5x³ – 15x² + 25x) / 5x

5x³/5x – 15x²/5x + 25x/5x

By using the formula aⁿ / a^m = a^n-m

5/5 x^3-1 – 15/5 x^2-1 + 25/5 x^1-1

x² – 3x + 5

\(\frac{5x^3}{5x}-\frac{15x^2}{5x}+\frac{25x}{5x}\) = 5x² - 3x + 5 [Using aⁿ÷a^m = a^n-m]

2x + 3y = 8  ......(1)

4x + 6y = 7   .....(2)

Step 1 : Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal. Then we get the equations as :

4x + 6y = 16   ......(3)

4x + 6y = 7   .........(4)

Step 2 : Subtracting Equation (4) from Equation (3),

(4x – 4x) + (6y – 6y) = 16 – 7

i.e., 0 = 9, which is a false statement.

Therefore, the pair of equations has no solution.

Given pair of linear equations is

0.4x – 1.5y = 6.5 …(i)

And 0.3x + 0.2y = 0.9 …(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of x equal, we get the equation as

1.2x – 4.5y = 19.5 …(iii)

1.2x + 0.8y = 3.6 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

1.2x + 0.8y – 1.2x + 4.5y = 3.6 – 19.5

⇒ 5.3y = – 15.9

y = -15.9/5.3

⇒ y = – 3

On putting y = – 3 in Eq. (ii), we get

0.3x + 0.2y = 0.9

⇒ 0.3x + 0.2( – 3) = 0.9

⇒ 0.3x – 0.6 = 0.9

⇒ 0.3x = 1.5

⇒ x = 1.5/0.3

⇒ x = 5

Hence, x = 5 and y = – 3 , which is the required solution.

Given pair of linear equations is

8x + 5y = 9 …(i)

And 3x + 2y = 4 …(ii)

On multiplying Eq. (i) by 2 and Eq. (ii) by 5 to make the coefficients of y equal, we get the equation as

16x + 10y = 18 …(iii)

15x + 10y = 20 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

15x + 10y – 16x – 10y = 20 – 18

⇒ – x = 2

⇒ x = – 2

On putting x = – 2 in Eq. (ii), we get

3x + 2y = 4

⇒ 3( – 2) + 2y = 4

⇒ – 6 + 2y = 4

⇒ 2y = 4 + 6

⇒ 2y = 10

y = 10/2 = 5

Hence, x = 2 and y = 5 , which is the required solution.

Given pair of linear equations is

2x + 3y = 8 …(i)

And 4x + 6y = 7 …(ii)

On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as

4x + 6y = 16 …(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

4x + 6y – 4x – 6y = 16 – 7

⇒ 0 = 9

Which is a false equation involving no variable.

So, the given pair of linear equations has no solution i.e. this pair of linear equations is inconsistent.

Given pair of linear equations is

2x + 5y = 1 …(i)

And 2x + 3y = 3 …(ii)

On subtracting Eq. (ii) from Eq. (i), we get

2x + 3y – 2x – 5y = 3 – 1

⇒ – 2y = 2

⇒ y = – 1

On putting y = – 1 in Eq. (ii), we get

2x + 3( – 1) = 3

⇒ 2x – 3 = 3

⇒ 2x = 6

⇒ x = 6/2

⇒ x = 3

Hence, x = 3 and y = – 1 , which is the required solution.

Let the length and the breadth of the rectangle be x m and y m, respectively. 

∴ Area of the rectangle = (xy) sq.m 

Case 1: 

When the length is reduced by 5m and the breadth is increased by 3 m: 

New length = (x – 5) m 

New breadth = (y + 3) m 

∴ New area = (x – 5) (y + 3) sq.m 

∴ xy – (x – 5) (y + 3) = 8 

⇒ xy – [xy – 5y + 3x – 15] = 8 

⇒ xy – xy + 5y – 3x + 15 = 8 

⇒ 3x – 5y = 7 ………(i) 

Case 2: 

When the length is increased by 3 m and the breadth is increased by 2 m: 

New length = (x + 3) m 

New breadth = (y + 2) m 

∴ New area = (x + 3) (y + 2) sq.m 

⇒ (x + 3) (y + 2) – xy = 74 

⇒ [xy + 3y + 2x + 6] – xy = 74 

⇒ 2x + 3y = 68 ………(ii)

On multiplying (i) by 3 and (ii) by 5, we get: 

9x – 15y = 21 ……….(iii) 

10x + 15y = 340 ………(iv) 

On adding (iii) and (iv), we get: 

19x = 361 

⇒ x = 19 

On substituting x = 19 in (iii), we get: 

9 × 19 – 15y = 21 

⇒171 – 15y = 21 

⇒15y = (171 – 21) = 150 

⇒y = 10 

Hence, the length is 19m and the breadth is 10m.

(1) The poet wishes to build a small cabin at Innisfree to be made with sticks and clay.

(2) Innisfree symbolises a place of peace and tranquillity.

(3) The poet wishes to stay at Innisfree :

• To live in peaceful environment and
• To escape from hectic schedule – of the city life.

Answer is (A) π/4

Answer is (b) √3

The correct option is (C) IsinΦ.

The apparent power between v and i is given as, 

v_i = vicosФ + jvisinФ 

Where vicosФ is known as the active power which is responsible for heat produced and visinФ is known as the reactive power or wattless component of v and i which is required to maintain the reactive power consumption in the circuit.

Answer is (a) [(1,2n),(0,1)] 

Answer is (d) π/12

Answer is (C) √2

Answer is (A) 0.06

Answer is (a) P(A).P(B/A)

Answer is (d) 5/8

Answer is (d) 4/5

∫(√tan x + √cot x) dx, for x ∈ [0,π/2]

∫(√(sin x/cos x) + √(cos x/sin x)) dx, for x ∈ [0,π/2]

∫(sin x + cos x)/√(sin x.cos x) dx, for x ∈ [0,π/2]

√2∫(sin x + cos x)/√(2sin x.cos x) dx, for x ∈ [0,π/2]

√2∫(sin x + cos x)/√(1 - (1 - 2sin x.cos x)) dx, for x ∈ [0,π/2]

√2∫(sin x + cos x)/√(1 - (sin x - cos x)²) dx, for x ∈ [0,π/2]

Put sin x - cos x = t

(cos x + sin x) dx = dt

When x = 0, t = 1 and when x = π/2, t = 1

So, ∫(√tan x + √cot x) dx, for x ∈ [0,π/2]

√2∫(dt/√(1 - t²), for t ∈ [-1,1]

= √2.[sin^-1 t], for t ∈ [-1,1]

= √2[sin^-1 1 - sin^-1 (-1)] = √2((π/2) - (-π/2))

= √2.π

Answer is (b) (1/2a) log|(x + a)/(x - a)| 

Answer is (A) π/4

Answer is (b) 0.06

Answer is (a) (3/2)x^2/3 + k

Answer is (a) π/√2

Answer is (a) (3/2)x^2/3 + c

Answer is (b) 0

Answer is (c) c 

Answer is (a) 108√3/35

Answer is (a) 5/7