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Side of rhombus = 30 m

Altitude of rhombus = 16 m

Rhombus is a type of parallelogram

Area of parallelogram = base × altitude

Area of parallelogram = 30 × 16 = 480 m²

Cost of levelling the garden = area × rate

Cost of levelling the garden = 480 × 2 = 960

Cost of levelling the garden is Rs 960

(i) Area of trapezium = \(\frac{1}{2}\)(sum of lengths of parallel sides) x altitude

Length of bases of trapezium = 12 dm and 20 dm

10 dm = 1 m

Therefore length of bases in m = 1.2 m and 2 m

Similarly length of altitude in m = 1 m

Area of trapezium = \(\frac{1}{2}\)(1.2 + 2.0) x 1

Area of trapezium = \(\frac{1}{2}\)x 3.2 - 1.6 m²

(ii) Area of trapezium = \(\frac{1}{2}\)(sum of lengths of parallel sides) x altitude

Length of bases of trapezium = 28 cm and 3 dm

10 dm = 1 m

Therefore length of bases in m = 0.28 m and 0.3 m

Similarly length of altitude in m = 0.25 m

Area of trapezium = \(\frac{1}{2}\)(0.28 + 0.3) x 0.25

Area of trapezium = \(\frac{1}{2}\) x 0.58 x 0.25 = 0.0725 m²

(iii) Area of trapezium = \(\frac{1}{2}\)(sum of lengths of parallel sides) x altitude

Length of bases of trapezium = 8 m and 60 dm

10 dm = 1 m

Therefore length of bases in m = 8 m and 6 m

Similarly length of altitude in m = 4 m

Area of trapezium = \(\frac{1}{2}\)(8 + 6) x 4

Area of trapezium = \(\frac{1}{2}\)x 14 x 4 = 28 m²

(iv) Area of trapezium = \(\frac{1}{2}\)(sum of lengths of parallel sides) x altitude

Length of bases of trapezium = 150 cm and 30 dm

10 dm = 1 m

Therefore length of bases in m = 1.5 m and 3 m

Similarly length of altitude in m = 0.9 m

Area of trapezium = \(\frac{1}{2}\)(1.5 + 3) x 0.9

Area of trapezium = \(\frac{1}{2}\)x 4.5 x 0.9 = 2.025 m²

Side of rhombus = 6 cm

Altitude of rhombus = 4 cm

Since rhombus is a parallelogram, therefore area of parallelogram = base× altitude

Area of parallelogram = 6 × 4 = 24 cm²

Area of parallelogram = area of rhombus

Area of rhombus = \(\frac{1}{2}\)x d₁ x d₂

24 = \(\frac{1}{2}\)x 8 x d₂

d₂ = \(\frac{24}{4}\) = 6

Hence, length of other diagonal of rhombus is 6 cm

Length of base of triangle = 24.8 cm

Length of altitude of triangle= 16.5 cm

Area of triangle = \(\frac{1}{2}\)x base x altitude

Area of triangle = \(\frac{1}{2}\)x 24.8 x 16.5 = 204.6

Area of rhombus = \(\frac{1}{2}\)x d₁ x d₂

204.6 = \(\frac{1}{2}\)x 22 x d₂

d₂ = \(\frac{204.6}{11}\) = 18.6

Hence, the length of other diagonal is 18.6 cm

Perimeter of square field = \(\frac{cost\,of\,fencing}{rate\,of\,fencing}\)

Perimeter of square field = \(\frac{1200}{0.6}\) = 2000 m

Perimeter of square = 4 × side

Side of square = \(\frac{perimeter}{4}\) = \(\frac{2000}{4}\) = 500 m

Area of square = side²

Area of square = 500 × 500 = 250000 m²

Cost of reaping = \(\frac{250000\times 0.5}{100}\) = 1250

Cost of reaping is Rs 1250

Cost of fencing per metre = Rs 30 

Total cost of fencing = Rs 12000 

So, the length of fencing (perimeter) = Total cost/ Cost per metre 

= 1200/30

= 400 m 

Now, length of fencing = Perimeter of the square field 

 = 4 × side of the field 

Therefore, 4 × side of the field = 400 m 

or, side of the field = 400/4 m = 100 m 

So, area of the field = 100 m × 100 m = 10000 sq m. 

Let the length of one parallel side of trapezium = x meter

Length of other parallel side of trapezium = (x + 8) m

Area of trapezium = 91 cm²

Area of trapezium = \(\frac{1}{2}\)(sum of sides) x distance between parallel sides

91 = \(\frac{1}{2}\)(x + x + 8) x 7

x + 4 = \(\frac{91}{7}\) 

x + 4 = 13

x = 13 - 4 = 9

Therefore length of one parallel side of trapezium = 9 cm

Length of other parallel side of trapezium = 9 + 8 = 17 cm

Length of parallel sides of trapezium = 10 m and 6 m

Distance between parallel sides of trapezium = x meter

Area of trapezium = \(\frac{1}{2}\)(sum of sides) x distance between parallel sides

72 = \(\frac{1}{2}\)(10 + 6) x x

x = \(\frac{72}{8}\) = 9

Therefore depth of river is 9 m

Radius of given sphere = (r) = 14 cm 

Formula for surface area of sphere = 4πr² 

∴ Surface area of given sphere = 4 × \(\frac{22}{7}\) × 14 × 14 = 88 × 28 

= 2464 cm²

Diagonals

Explanation:

We know that the area of a rhombus = pq/2

Where p and q are diagonals.

Volume of the cylinder = πr²h = 448π 

Here h = 7cm ,r = r 

then πr² × 7 = 448π

r² = \(\frac{448}{7}\) = 64 

∴ Radius (r) = 8 cm

πb²

Explanation:

When the cylinder exactly fits in the cube of side “b”, the height equals to the edges of the cube and the radius equal to half the edges of a cube.

It means that,

h = b, and r = b/2

Then the CSA of a cylinder be = 2πrh

= 2π (b/2)(b)

= πb²

πh²r

Given, radius of cylinder = h and height of cylinder = r.

Now, volume of a cylinder

= π x (Radius)² x Height 

= π x h² x r 

= πh²r

Surface area of a football (sphere) = 4πr² . 

4πr² = 616 

πr² = 154 

r² = 154 × \(\frac{7}{22}\)

r² = 7 × 7 = 49 

∴ r = 7 cm

 πa³ /4

Explanation:

When the cylinder exactly fits in the cube of side “a”, the height equals to the edges of the cube and the radius equal to half the edges of a cube.

It means that,

h = a,and r = a/2

Then the volume of a cylinder be = πr²h

= π(a/2)²(a)

= πa³/4

Radius of a cylinder = r = 7 cm 

Height of a cylinder = h – 10 cm 

CSA of a cylinder = A = 2πrh 

= 2 × \(\frac{22}{7}\) × 7 × 10 

= 44 × 10 = 440 sq.cm

2πh(r + h)

Explanation:

Given radius = r and height = h

TSA of cylinder = CSA of cylinder + Area of top surface + Base area

TSA = 2πrh + πr² + πr²

2πrh

Explanation:

The CSA of a cylinder with radius “r” and height “h” is

CSA = 2π(r)(h)

50%

Explanation:

The CSA of cylinder with radius “r” and height “h” is 2πrh

When the height is halved, then new CSA is 2πr (h/2) = πrh

Hence, the percentage reduction in CSA = [(2πrh – πrh) (100)]/ 2πrh = 50%

Equal

Explanation:

A cuboid is made up of 6 rectangular faces, but the opposite sides have equal length and breadth. Hence, the opposite areas are equal.

The correct answer is option (a) lw /2

Explanation:

We know that, the area of a triangle is (1/2) x base x height

Let ABCD be a triangle with length “l” and width “w”.

Here, we have to construct a triangle of maximum area inside the rectangle in all possible ways.

Now, the maximum base length is “l”

Maximum height is “w”.

Therefore, the area of a largest triangle is (1/2) x l x w.

¼ times

Explanation:

Volume of cylinder = πr²h (when radius is r and height is “h”)

When the radius is halved, then it becomes

V = π (r/2)²h

V = ¼ (πr²h)

Rectangular shape, different

Explanation:

It is known that, a cuboid is made up of 6 rectangular face which different lengths and breadths. Hence, it has different area.

The correct answer is option (b) Volume of the cylinder will remain unchanged.

Explanation:

We know that, the volume of a cylinder is π × r² × h

We know that, base radius and height of the cylinder is “r” and “h” respectively.

Now, height “h” becomes (1/4)h and “r” becomes “2r”, then the volume of the cylinder is:

(V) = π × 4r² × (1/4) h = πr²h = v

Therefore, the volume of new cylinder = the volume of original cylinder.

The correct answer is option (c) Curved surface area of the cylinder will be halved.

Explanation:

We know that the curved surface area of a cylinder with radius “r” and height “h” is given as

The curved surface area of a cylinder = 2πrh … (1)

Now, the new curved surface area of cylinder with radius 2r and height (1/4)h, then the new curved surface area is

= 2π(2r)(1/4)h

= πrh

Now, multiply an divide the new curved surface area by 2, we will get

= (1/2) (2) πrh …. (2)

Now, by comparing (1) and (2), we get:

The new curved surface area of a cylinder is (1/2) times of the original curved surface area of a cylinder.

3, equal areas

Explanation:

By using SSS congruency rule of triangle, we can show that a trapezium can be divided into three equilateral triangle with equal areas.

4 times

Explanation:

Let p and q be the two diagonals of the rhombus

We know that area of a rhombus = pq/2

If the diagonals are doubled, we will get

A= (4p)(4q)/2

Take 4 outside, we will get

A = 4(pq/2)