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Correct answer is

(A) Rs. 165, Rs. 132, Rs. 110

Correct answer is

(A) same units

(a) \(\frac{n+1}{2}\)

Reqd. nth term = \(\frac{1+2+3+....+n}{n}\)

= \(\frac{n(n+1)}{2}\) x \(\frac{1}{n}\) = \(\frac{n+1}{2}\).

(d) \(\frac{n}{6}(4n^2+15n+17)\)

Let 

S = 2·3 + 3·5 + 4·7 + ..... + (n + 1) (2n + 1)

\(\displaystyle\sum_{k=1}^{n}(k+1)\)(2k + 1) = \(\displaystyle\sum_{k=1}^{n}(2k^2+3k+1)\)

= \(2\displaystyle\sum_{k=1}^{n}k^2\) + \(2\displaystyle\sum_{k=1}^{n}k+n\)

= 2 x \(\frac{n(n+1)(2n+1)}{6}\) + 3 x \(\frac{n(n+1)}{2}\) + n

= \(\frac{n}{6}\)[2(n + 1) (2n + 1) + 9 (n + 1) + 6]

= \(\frac{n}{6}\) [4n² + 6n + 2 + 9 + 6]

= \(\frac{n}{6}(4n^2+15n+17)\)

(b) 66.

Let S = 1² – 2² + 3² – 4² + ..... + 11² 

= (1² + 2² + 3² + 4² + ..... + 11²) – 2(2² + 4² + 6² + 8² + 11²) 

= (1² + 2² + 3² + 4² + ..... + 11²) – 2³(1² + 2² + 3² + 4² + 5²)

= \(\frac{11\times(11+1)\times(2\times11\times1)}{6}\) - \(\frac{8\times5\times(5+1)\times(2\times5+1)}{6}\)

= \(\frac{11\times12\times23}{6}\) - \(\frac{8\times5\times6\times11}{6}\) = 22 x 23 - 40 x 11

= 506 – 440 = 66.

(a) n² (2n² – 1).

Let Sn = 1³ + 3³ + 5³ + 7³ + ..... upto n terms 

nth term of the series = (2n – 1)³

∴ S_n = \(\displaystyle\sum_{k=1}^{n}(2k-1)^3\) = \(\displaystyle\sum_{k=1}^{n}[8k^3-1-12k^2+6k]\)

= \(8\displaystyle\sum_{k=1}^{n}k^3\) - \(12\displaystyle\sum_{k=1}^{n}k^2\) + \(6\displaystyle\sum_{k=1}^{n}k-n\)

= 8 \(\frac{n^2(n+1)^2}{4}\)- 12 x \(\frac{n(n+1)(2n+1)}{6}\) + 6 x \(\frac{n(n+1)}{2}\) - n

= 2n² (n² + 2n + 1) – 2n (2n² + 3n + 1) + 3n(n + 1) – n 

= 2n⁴ + 4n³ + 2n² – 4n³ – 6n² – 2n + 3n² + 3n – n 

= 2n⁴ + n² = n² (2n² – 1).

(a) S_n = \(T_n^2\)

Sn = ∑n³

= \(\frac{n^2(n+1)^2}{4}\) = \(\big(\frac{n(n+1)}{2}\big)^2\)                ....(i)

T_n = \(\frac{n(n+1)}{2}\)                   ......(ii)

∴ From (i) and (ii) 

S_n = \(T_n^2\)

1 hectare = 10000 m²

(b) 200 cm²

From the question it is given that, radius of circle 10 cm.

Square has diagonal equal to its diameter.

Diameter = 2 × radius

= 2 × 10

= 20 cm

Then, let us assume the side of the square be P.

So, area of square be P².

By the rule of Pythagoras theorem,

Diagonal² = height² + base²

20² = P² + P²

20² = 2P²

400 = 2P²

P² = 400/2

P² = 200 cm²

Therefore, area the largest square that can be cut out of a circle of radius 10 cm is 200 cm².

(c) 3 times

We know that, area of the circle = πr²

Where, r = radius of original circle

Then, radius is doubled = 2r

Then, area of new circle = π(2r)²

= π4r²

Hence, the area will increase by 3 times the area of the original circle.

(c) circumference/diameter

We know that, circumference of circle = 2πr

Then, π = circumference/2r

Therefore, π = circumference/diameter … [2r = diameter]

(a) more than three times of its diameter

As we know that, 1 hectare = 10000 m²

So, 5 hectare = 5 x 10000 m² = 50000 m²

True

∵ Circumference : Radius = 2πr: r = 2π : 1

Given, area of square MNOP = 144 cm²

Since, there are 8 identical triangles in the given square MNOP.

Hence, area of each triangle =1/8 x Area of square MNOP= 1/8 x 144= 18 cm²

True

It is clear from the figure that all triangles may not have the same perimeter.

False

∵ Area of triangle = 1/2 x Base x Height

So, area of triangle does not only depend on base, it also depends on height. Hence, if triangles have equal base and equal height, then only their areas are equal.

False

It is not necessary that all parallelograms having equal areas have same perimeters as their base and height may be different.

When we change the shape, then the perimeter remains same as the length of wire is fixed, but area changes as shape changes.

Given, ABCD is a square and AB = 15 cm

Diagonal of square ABCD = √2a = √2 x 15 = 15√2 cm

From the figure, diagonal of square ABCD is the side of square BDEF

Area of the square BDFE = (side)² 

= (15√2)² 

= 15 x 15 x √2 x √2

= 225 x 2

= 450 cm²

True.

From the given figure, all triangles have the same base and the same altitude.

True

Congruent triangles have equal shape and size. Hence, their areas are also equal.

True.

We know that congruent triangle have equal size and shapes.

False

It is clear from the figure that all triangles have only base line is equal and no such other lines are equal to each other.

BD = 42 cm, AC = 28 cm, OD= 12 cm

Area of Triangle ABC = 1/2 (AC x OB)

= 1/2 (AC x (BD – OD))

= 1/2 (28 cm x (42 cm – 12 cm))

= 1/2 (28 cm x 30 cm)

= 14 cm x 30 cm

= 420 cm²

Area of Triangle ADC = 1/2 (AC x OD) = 1/2 (28 cm x 12 cm)

= 14 cm x 12 cm

= 168 cm²

Hence, Area of the quadrilateral ABCD = Area of ABC + Area of ADC

= (420 + 168) cm²

= 588 cm²

(c)  \( \frac{11x^2}{12}\) m²

Each side of the square = x m 

⇒ Area of the square = x² m² 

Given, one side of the rectangle = \(\frac{x}2\) m and other side of the square = \(\frac13\) x \(\frac{x}2\) m = \(\frac{x}2\) m

∴ Area of the rectangle = \(\frac{x}2\) m x \(\frac{x}2\) m = \(\frac{x^2}{12}\) m²

∴ Area of the remaining portion = \(x^2-\frac{x^2}{12} = \frac{11x^2}{12}\) m²

(d)   \(\frac{x^2}{18}\) m²

Let the breadth of the rectangular field be a m. Then, its length = 2a m

Given, 2(2a + a) = x ⇒ 6a = x ⇒ a = \(\frac{x}{6}\) m

∴ Length = 2a = 2 x \(\frac{x}{6}\) = \(\frac{x}{3}\) m

∴ Area of the rectangular field = \(\frac{x}{3}\) x \(\frac{x}{6}\) =   \(\frac{x^2}{18}\) m²

(c) 52 cm²

Area of trapezium = 1/2 x Sum of parallel sides x Distance between them

= 1/2 x (9.7 + 6.3) x 6.5

= 8 x 6.5

= 52. 0 cm²

The area of 10 cm side square will be 100 sq. cm.

When we cut the triangles and keep on each other then they do not cover each other. If BA and CA” would have same length, then ∆ABC and A”BC will cover each other.

The formula of the area of right angle triangle will be 1/2 x Base x Height.

The correct option is (A) πr².

Formula of area of rectangle is l x b.

Area of the trapezium = 1/2 x  (sum of the parallel sides) x distance between the parallel sides

= 1/2 x (12+9) x 8

= 21 x4

= 84 sq m

So, the area of the trapezium is 84 cm².

(i) Perimeter

(ii) Area

(iii) Area

(iv) Perimeter

For obtained rectangle

Length of rectangle = 1/2 x circumference

⇒ Area of rectangle = length x breadth

= 1/2 x circumference x breadth

If radius of circle is r, then

Area = 1/2 x 2πr x r = πr²

1. 14.4 cm

2. 5/11

3. 145kg 875g

1. 96

2.Thousandths

3. 8

Longer side = 32 cm

Shorter side = 24 cm

Let the distance between the shorter sides be x cm.

Area of a parallelogram = Longer side x Distance between the longer sides

= Shorter side x Distance between the shorter sides

or, 32 x 17.4 = \(24\times x\)

or, \(x=\frac{32\times 17.4}{24}=23.2\) cm

∴ Distance between the shorter sides = 23.2 cm

Area of the rhombus = 1/2 (Product of diagonal)

= 1/2 (48 x 20)

= 480 cm²

Let × be the CP of TV Set 

CP = x 

SP = (x) × 6/5 

= 6x/5 

Gain = SP – CP 

= 6x/5 – x 

= x/5

\(Gain\%=\frac{Gain\times100}{CP}\)

= (x/5 × 100) / x 

= 20% 

So, If TV set is sold at 6/5 price of its CP. Then Gain percent will be 20%.

Let × be the CP of Pen 

SP of 1 pen = x/16 

CP of 1 Pen = x/12 

Loss = CP – SP = x/12 – x/16 = x/48

\(Loss\%=\frac{Loss\,\times\,100}{CP}\)

\(=\frac{\frac{x}{48}\times100}{\frac{x}{12}}\)

= 25%

Correct answer is (a) 3 5/13

Correct answer is 3.27 minutes

Correct answer is  (d)3/2

Correct option is  B) 1

Since A = {1, 2, 3, 4} and B = {5, 7, 9}. Therefore,

(i) A × B = {(1, 5), (1, 7), (1, 9), (2, 5), (2, 7), (2, 9), (3, 5), (3, 7), (3, 9), (4, 5), (4, 7), (4, 9)}

(ii) B × A = {(5, 1), (5, 2), (5, 3), (5, 4), (7, 1), (7, 2), (7, 3), (7, 4), (9, 1), (9, 2), (9, 3), (9, 4)}

(iii) No, A x B ≠ B x A. Since A x B and B x A do not have exactly the same ordered pairs.

(iv) n (A x B) = n (A) x n (B) = 4 x 3 = 12

n(B x A) = n(B) x n(A) = 4 x 3 = 12

Hence, n(A x B) = n(B x A)

(c) aluminium

(d) Fluorine

We know that an element having low ΔiH is less stable & its ion is more stable. It is experimentally observed that sum of first & second ionization enthalpies of Ni is less than that of Pt so, Ni(II) compounds are more stable than Pt(II) compounds. It is further observed that sum of the first four ionization enthalpies of Ni is much higher than that of Pt. Hence Ni (IV) compounds are less stable than Pt (IV) compounds.