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If a POINT lies on the bisector of an ANGLE, then it is equidistant from the sides of the angle.

Side of rhombus = ½ √(sum of SQUARES of diagonal)

Given, diagonals of a rhombus are 16 CM and 12 cm respectively.

⇒ Side $(= \FRAC{1}{2} \times \SQRT {{{16}^2} + {{12}^2}})$

⇒ Side = 10 cm

Let the BREADTH of room be 

Length of room be 2X

Area of FLOOR = x × 2x = 2x²

Area of floor = 5000/100 = 50 m²

⇒ 2x² = 50

⇒ x = 5 m

∴ Length of room = 2x = 10 m

Width of rectangle = 28 m

Diagonal of rectangle = 53 m

Length of rectangle = √ (53² - 28²) = 45 m [by USING PYTHAGORAS theorem]

Perimeter of rectangle = 2[length + width] = 2[45 + 28] = 146 m

Fencing cost = 146 ×15 = Rs. 2190

Area of rectangle = length × width = 45 × 28 = 1260 m²

Area of wheat PLANTATION = [3/(3 + 5)] × 1260 = 472.5 m²

Area of rice plantation = 1260 - 472.5 = 787.5 m²

Cost for wheat plantation = 472.5 × 25 = Rs. 11,812.5

Cost for rice plantation = 787.5 × 30 = Rs. 23,625

∴ Total cost = 2190 + 11812.5 + 23625 = Rs. 37,627.5

AREA of rectangle = LENGTH × BREADTH

Let Length of rectangle be a and breadth be b

∴ ORIGINAL area = a × b = AB

New area = (a – 25% of a) × (b + 30% of b)

⇒ New area = 0.75a × 1.3b

⇒ New area = 0.975ab

? 0.975ab < ab, area has decreased

Percent decrease in area = (Original area - New area) × 100/ Original area

⇒ Percent decrease = (ab - 0.975ab) × 100/ab

SOLUTION:

Given: Length of the playground = 125 m 

Width of the playground = 75m

TOTAL area of the rectangular playground = Length × Breadth

= 125 × 75

 = 9375 m²

It is given that 5m wide walking strip is made in the middle of the playground along the shorter side that is along the width of the playground.

So, width of the walking strip = 5m

Length of the walking strip = 75m

Area of the rectangular walking strip = Length × Breadth

= 75 × 5

= 375 m²

Area of the playground WITHOUT walking strip

= Area of the total playground - Area of the walking strip 

= 9375 - 375

= 9000 m²

So, the CORRECT option is 2). 9000 sq.m

Volume of cylinder = πr²H where r is the RADIUS of cylinder.

Volume of cone = πr²H/3, where r is the radius of cone and H is the HEIGHT of cone

Given, Volume of cylinder = Volume of cone

⇒ πr²h = πr²H/3

⇒ H = 8 × 3 = 24 cm

∴ Height of the cone = 24 cm

As we know, perimeter of a circle of radius r = 2πr

And perimeter of a rectangle of length L and breadth B = 2 (L + B)

ACCORDING to the given information:

 A circle and a rectangle have the same perimeter

∴ 2π r = 2 (L + B)

Whereas, Length of the rectangle = 12 cm

And breadth of the rectangle = 10 cm

∴ 2 × (22/7) × r = 2 × (12 + 10) = 44

⇒ r = 7

Now, Area of the circle of radius r = π r²

∴ (22/7) × 7 × 7 = 154

Let the breadth and length of the room be ‘X’ and ‘2X’;

Height of the room is given 4 m;

Area of 4 walls = 2(l + b) × h

⇒ 120 = 2(3x) × 4

⇒ 3x = 15

⇒ x = 5 m

Breadth = 5 m and Length = 10 m

Solution:

Given: Length of the side of the square PARK = 10m

Let us TAKE the length of the side of the park as 'a'.

Area of the total park = a²sq.units

Area of the total park = 10² 

Area of the total park =100 m²

Length of the side of the park without the pavement = 10 - (2×2)

= 10 - 4 

= 6 m 

Area of the park without including the pavement = 6² = 36m²

So,Area of the pavement

= Area of the total park - Area of the park without pavement

= 100 - 36

= 64 m²

So,the correct option is 1). 64 m²

Given the length of the WIRE is 88 cm

When the wire is bent in the shape of the square, then the total length of the wire is EQUAL to the PERIMETER of the square

Let us consider the SIDE of the square = ‘X’

⇒ Perimeter of the square = 4x

⇒ 4x = 88

⇒ x = 88/4

⇒ 22 cms

⇒ Area of the square = side × side

⇒ 22 × 22 = 484 cm²

∴ Area = 484 cm²

SURFACE area of cuboid = 2(lb + BH + hl)

Where, l = length of cuboid = 30 cm

b = BREADTH of the cuboid = 20 cm

h = height of the cuboid = 15 cm

We know that,

Volume of CUBE = a³ where, a = side of the cube

Volume of sphere = (4/3) π R³ where, r = radius of the sphere

Let,

Side of the cube = a

Since, the sphere exactly fit INSIDE the cube

∴ Radius of the sphere = r = a/2

∴ Required ratio

$(= \;\frac{{{a^3}}}{{\frac{4}{3}{\RM{\;}} \times {\rm{\;\PI }} \times {{(\frac{a}{2})}^3}}} = \;\frac{{3 \times 8 \times \;{a^3}}}{{4 \times \;{\rm{\pi }} \times {\rm{\;}}{a^3}}} = \;\frac{6}{{\rm{\pi }}} = 6\;:{\rm{\pi }})$

Let ‘a’ be the side of an equilateral triangle.

⇒ HEIGHT of the triangle = $(\frac{{\SQRT 3 a}}{2})$

⇒ Area of triangle = $(\frac{1}{2} \TIMES \frac{{\sqrt 3 a}}{2} \times a = \frac{{\sqrt 3 {a^2}}}{4})$

⇒ √(3 / 4)a² = 60√3

⇒ a² / 4 = 60

⇒ a² = 240

⇒ a = √240 = 4√15

∴ Height of equilateral triangle = √(3/2) × 4√15 = 6√5 m

Longest Diagonal of a cube = a√3 = 6√3

⇒ a = 6 cm

⇒ SURFACE AREA of cube = 6 × a² = 6 × 6²

AREA of SQUARE field = 45 × 45 = 2025 m²

Surface AREA of a CYLINDER = 2πrh + 2πr² = 2π(5)(11) + 2π(5)² = 160π SQ. Units

Perimeter of semi-circle = R(π + 2)

⇒ 144 = r(22/7 + 2)

⇒ 144 = 36/7 × r

LET the old Radius be ‘R’ then the new Radius will be ‘4r’.

New CIRCUMFERENCE = 2 × π × (4r) = 8πr

Square of the new circumference = 64π²r²

New area = π × (4r)² = 16πr²

The perimeter of a square = 4(SIDE)

The CIRCUMFERENCE of a circle =2 π r

According to the GIVEN information:

⇒ 4(side)= 2 π r

$(\begin{array}{l} \Rightarrow \FRAC{{{\rm{Side\;}}}}{{\rm{r}}} = \frac{{2\pi }}{4}\\ \Rightarrow \frac{{{\rm{Side\;}}}}{{\rm{r}}} = \frac{\pi }{2}\end{array})$

CIRCUMFERENCE of the CIRCLE = PERIMETER of the square

2πr = 4s

∴ s = πr/2

⇒ 22/7 × 98/2 = 154 CM

LET the BREADTH be m.

Then length will be (m + 10)

We have the FORMULA for

Area of rectangle = L × B

Where L and B are the length and breadth respectively.

Area of the rectangular plot = Length × Breadth

⇒Area = m × (m + 10)

We have area = 15m.

⇒15m = m² + 10m

⇒m² – 5M = 0

⇒m = 5

Hence, breadth of rectangular plot = 5 metres.

GIVEN: DIAMETER of a HEMISPHERE = 21 cm

⇒ VOLUME of a hemisphere = (2/3)πr³

⇒ Volume of a hemisphere = (2/3)π(21/2)³

⇒ Volume of a hemisphere = 771.75π

⇒ Volume of a hemisphere = 2425.5

We know that,

Area of a rhombus with DIAGONALS d₁ and d₂ = d₁d₂/2

⇒ Area of rhombus = (8× 6)/2 = 48/2 = 24 sq. m.

Half of each of the two diagonal forms the two SIDES of a right - angled triangle whose hypotenuse is the side of the rhombus.

Length of Diagonal 1 = 24

Length of Diagonal 2 = 10

⇒ Sides of the triangle = 12 and 5

Since it’s a right - angled triangle, it should follow Pythagoras theorem.

⇒ (Side)² = (12)² + (5)²

⇒ Side = 13

∴ Side of the rhombus = 13.

CIRCUMFERENCE of CIRCLE = 2πr, where r is the RADIUS of circle.

⇒ 2πr = 440

∴ r = 440/2π = 70 cm

⇒ Diameter of the circle = 2r = 140 cm

Diameter of circle = DIAGONAL of the square

⇒ √2a = 140

∴ a = 140/√2

Radius = diameter/2 = 28/2 = 14cm

TOTAL surface area of the walls OFA room = 2 × ( LENGTH × height + breadth × height)

Given,a room is 8 m long, 4 m broad and 3 m high

⇒ Total surface area of the given room = 2 × (24 + 12) = 72 m²

Let the length of paper be ‘x’

Given, breadth of paper is 80 cm.

Area of paper = 0.8 × x sq. m.

∴ 0.8x = 72

Volume of cylinder = πr²h

Volume of CONE = (1/3)πr'²h'

⇒ πr²h = (1/3) πr'²h'

⇒ π × 10² × h = (1/3) × π × 100² × 21

Length of rectangle = y cm

Width of rectangle = x cm

Area of rectangle = Length × width

⇒ 216 = y × x

⇒ x = 216/y----(I)

PERIMETER of rectangle = 2(Length + width)

⇒ 60 = 2(y + x)

⇒ 30 = y + x

Substitute the value of Eq. (I) in above Equation,

⇒ 30 = y + 216/y

⇒ 30y = y² + 216

⇒ y² – 30y + 216 = 0

⇒ y² – 18y – 12y + 216 = 0

⇒ y (y – 18) – 12 (y – 18) = 0

⇒ (y – 18) (y – 12) = 0

⇒ y = 18 [? Length is larger than width]

⇒ x = 216/18 = 12

VOLUME of a hemisphere $(= \frac{2}{3}\PI {R^3})$

Volume of a cone $(= \frac{1}{3}\pi {r^2}h)$

For a hemisphere, HEIGHT will be same as radius.

Given, height of the hemispherical part is 7 cm.

∴ radius = 7cm

Radius of the base of the cone is also same as radius of hemisphere = 7cm

Given, height of the cone is same as radius of the hemisphere.

Volume of ice-cream $(= \frac{2}{3}\pi {r^3} + \frac{1}{3}\pi {r^2} \times r = \pi {r^3})$ 

SOLUTION:

GIVEN: In ∆ABC,

Let us take the ratio as 'k'.

That is,

So, 3k + 3k + 4k = 180⁰ (sum of all the angles of a triangle is 180⁰)

10k = 180⁰

k = 18⁰

Therefore,

Since LINE PQ is parallel to BC,

So,

= 72⁰ - 54⁰

=18⁰

So, the CORRECT option is 2). 18⁰

LET the radius of the circle be X cm.

So, area = πx² cm².

The radius of a circle is increased by 25%.

So, the increased radius = x × (125/100) = 5x/4 cm

Now, area = π × (5x/4)² cm² = 25πx²/16 cm²

So, INCREASE in area = (25πx²/16) – πx² = 9πx²/16 cm²

PERIMETER of a square = 22 cm

⇒ 4a = 22

⇒ a = 5.5 cm

∴ Area of square = a² = (5.5)² = 30.25 cm²

Given Volume of the CYLINDER = 847 m³

Formula for Volume of cylinder = πr²h

⇒ πr²h = 847 (I)

Curved surface AREA of the cylinder = 242 m²

Formula for Curved surface of cylinder = 2πrh

⇒ 2πrh = 242 (II)

⇒ Dividing Equation (I) with Equation (II), we get

⇒ πr²h/2πrh = 847/242

⇒ R/2 = 7/2

⇒ r = 7 m

Substituting the value of r in Equation 2, we get

⇒ 2 × (22/7) × 7 × h = 242

⇒ h = (11/2) m

∴ Required ratio = h/r = (11/2)/7 = 11/14

As we KNOW, the diagonal of a parallelogram divides it into two EQUAL triangles.

Hence,

Area of parallelogram = 2 × area of triangle FORMED by the two adjacent sides and the diagonal

⇒ Area of parallelogram = 2 × √s(s – a)(s – b)(s – c)

Where s = (a + b + c)/2

Putting a = 10 cm, b = 8 cm, c = 6 cm, we get,

s = (10 + 8 + 6)/2 = 24/2 = 12

LET RADIUS of cylinder be r and radius of sphere be R

Volume of a right CIRCULAR cylinder = πr²h

Volume of a sphere = (4/3)πR³

Given,

R = 2r

Ratio of the volumes of a right circular cylinder and a sphere is 3 : 2

$(\Rightarrow \;\FRAC{{\pi {r^2}h}}{{\frac{4}{3}\pi {R^3}}}\; = \;\frac{3}{2})$

⇒ h = 16r

Curved Surface area of a right circular cylinder = 2πrh

Surface area of the sphere = 4πR²

∴ Ratio of the surface areas of the cylinder and the sphere $(= \;\frac{{2\pi rh}}{{4\pi {R^2}}}\; = \;\frac{{32{r^2}}}{{4 \times 4{R^2}}}\; = \;2\;:\;1)$

Volume of HEMISPHERE = 2/3 × πr³

⇒ 19404 = 2/3 × 22/7 × r³

⇒ r³ = 2261

⇒ r³ = 21 × 21 × 21

⇒ r = 21

Radius of hemisphere (r) = 21 cm

3 cos²α + 4 sin² α = 3 cos²α + 4 (1 – cos² α) = 4 – cos² α

The minimum value occurs when cos α = 1

⇒ Let the side of the square be 100 cm

⇒ Area of the square = 100 × 100 = 10000 cm²

⇒ As side of the square is INCREASED by 6%, the new vaue of side = 100 + (6% of 100) = 100 + 6 = 106 cm

⇒ New area of the square = 106 × 106 = 11236 cm²

⇒ Percentage increase in area = [(Final Area - ACTUAL Area)/Actual Area] × 100%

⇒ [(11236 - 10000)/10000] × 100% = (1236/100)% = 12.36%

Solution:

Given: AREA of the rectangular hall = 744 m² and 

LENTH of the rectangular hall is 7M more than its breadth.

Let us TAKE 'l' as the length and 'b' as the breadth of the hall.

So, we have l = 7 + b 

Area of the rectangular hall = 744 m²

Length × Breadth = 744 

l × b = 744

Substituting 'l' here we get 

(7+b)b = 744

b² + 7b - 744 = 0

(b + 31)(b -24) = 0

So, b = 24m

Substituting the value of 'b' in l = 7 + b 

We have l = 7 + 24

So, l = 31m

The CORRECT option is 4). 31m

 As the shown in the figure, PAINTING needs to be done in the space between two circles

Radius of the inner circle (r₁) = 16/2 = 8 cm

Radius of the OUTER circle (r₂) = 24/2 = 12 cm

Area of the PAINTED part = π(r₂)² - π(r₁)² = π[(r₂)² - (r₁)²]

⇒ Area = $(\frac{{22}}{7}\; \times \;\left( {{{12}^2} - {8^2}} \right))$

⇒ Area = 22/7 (144 – 64)

⇒ Area = 22/7 × 80

⇒ Area = 251.4286 cm²

⇒ For 100 cm² Price is Rs. 7

∴ For 251.4286 cm², price = (251.4286 × 7)/100 = Rs. 17.6

As we know, the DIAGONAL of a PARALLELOGRAM divides it into TWO equal triangles.

Hence,

Area of parallelogram = 2 × area of triangle FORMED by the two adjacent sides and the diagonal

⇒ Area of parallelogram = 2 × √s(s – a)(s – b)(s – c)

Where s = (a + b + c)/2

Putting a = 8 CM, b = 4 cm, c = 6 cm, we get,

s = (8 + 4 + 6)/2 = 18/2 = 9

Let the length of smaller parallel side be 6X.

∴ Length of bigger parallel side = 8x.

Now, SINCE we know AREA of trapezium = (Length of parallel sides) × Height/2

⇒ Area = (6x + 8x) × 32 / 2

⇒ 1120 = 14x × 16

⇒ 14x = 70

⇒ x = 5

SUM of the three ANGLE of triangle is 180°

∠P + ∠Q + ∠R = 180°

∠R + ∠R = 180°

2∠R = 180°

∠R = 90°

So, PQ is hypotenuse

As PER Pythagoras THEOREM,

PQ² = PR^{2 +} QR²

= 36 + 64

PQ² = 100

PQ = 10

Solution :

GIVEN 

The SIDES of triangle ABC are $25$ cm, $24$cm and $7$ cm . Triangle $ABC$ is a right angle triangle as $25$ cm, $24$cm and $7$ cm are Pythagoras triplets.

A pyramid is made up of four triangles namely BASE triangle $ABC$, triangle $BCD$, triangle $ACD$ and triangle $ABD$. 

Triangles $BCD$ and $ACD$ are ALSO right angle triangles.

Area of triangle $ABC = (1/2 )× 24 × 7 = 84$ sq.cm

Area of triangle $BCD = (1/2 )× 24 × 34 = 408$sq.cm

Area of triangle $ACD = (1/2 )× 7 × 34 = 119$ sq.cm

Given : Area of triangle $ABD = 40$ sq.cm

Total surface area of the pyramid $= ( 84 + 408 + 119 + 40) $sq.cm

= 651 sq.cm

The correct option is 3). 651 sq.cm