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[NOTE: A rhombus has 2 congruent opposite acute angles and two congruent opposite OBTUSE angles. One of the properties of a rhombus is that any two internal consecutive angles are supplementary.]

Let the acute angle be A

⇒ The obtuse angle = 3A

⇒ 4A = 180

⇒ A = 45

Now, use the formula for area of a triangle and then MULTIPLY the area by 2 

∴ Area of rhombus = $(2 \times 1/2 \times {9^2} \times \sin 45 = 81 \times \frac{1}{{\sqrt 2 }} = 57.28)$ feet² (approx) 

Total Surface AREA = 2πr (H + r) = 2 × 22/7 × 7/2 × (6 + 7/2) = 2 × 11 × 19/2

∴ Total Surface area = 11 × 19 = 209 cm²

SOLUTION:

Given: Area of square A = 24 sq.cm

To find : Area of square B = ?

Let us take the side of square A as 'a' and square B as 'b'.

So, a²= 24 sq.cm

a = √24

a = 2√6 cm.

It is given that,

Diagonal of square B = two-third of the diagonal of square A 

We KNOW that diagonal of a square = √2 of side of a square

So, √2 (b) = (⅔)√2 (a)

b = (⅔) a 

SUBSTITUTING the value of a in here we have 

b = ⅔ ( 2√6 )

b = ( 2 × 2√2 √3 ) / (√3 √3)

b = (4 √2)/√3 cm 

Area of square B = b² cm²

Area of square B =[(4 √2)/√3]² cm²

Area of square B = (16×2) / 3 cm²

Area of square B = 32/3 cm²

So,the correct option is 3). 32/3 cm²

In a parallelogram, the sum of the squares of the DIAGONALS

= 2 × (The sum of the squares of the two ADJACENT sides)

= D₁² + D₂² = 2(a² + b²)

LET the length of the second diagonal is ‘x’ cm.

∴ 16² + x² = 2(12² + 14²)

⇒ 256 + x² = 2(144 + 196)

⇒ x² = 680 – 256 = 424

⇒ x = √(424) = 20.6 cm

∴ The length of the other diagonal is 20.6 cm.

Volume of a hollow CYLINDER = πh (R² – r²)

∴ 1496 = (22/7) × 28 × (R² – r²)

⇒ (R² – r²) = 17

As R = 18 cm,

324 – r² = 17

⇒ r = √307 cm

∴ Thickness = R – r = (18 - √307) cm

According to the GIVEN INFORMATION,

AD is parallel to BC.

Slope of line BC = (4 - 3)/( - 1 - ( - 2)) = 1

⇒ Slope of line AD = 1

Line AD will pass through point A, so equation will be given by,

⇒ y - 0 = 1(X - 5)

∴ y = x - 5

Total surface area of hollow sphere = 4πr₁² + 4πr₂²

Where, r₁ = RADIUS of outer surface

r₂ = radius of inner surface

Here, r₁ = 18 cm, thickness = 10 mm = 1 cm

⇒ r₂ = 17 cm

Let r be the radius of cone, cylinder and hemisphere

And h be the height of cone, cylinder and hemisphere

As we know,

That height of a hemisphere is equal to radius of the hemisphere

∴ h = r

Now, Volume of cone = $(\FRAC{1}{3}\;\PI {r^2}\;h = \;\frac{1}{3}\;\pi {r^3})$

Volume of cylinder = π r² h = π r³

Volume of hemisphere $(= \;\frac{2}{3}\;\pi \;{r^3})$

Now, required ratio= Volume of cone: volume of hemisphere: volume of cylinder

⇒ Required ratio $(= \;\frac{1}{3}\;\pi {r^3}:\;\frac{2}{3}\;\pi \;{r^3}:\;\pi \;{r^3}\; = \;\frac{1}{3}:\frac{2}{3}:1 = 1\;:2\;:3)$

SOLUTION: 

Given :

Radius of 1st circle, R1 = 9 cm 

Radius of 2nd circle, R2 =8cm

According to the LAW of betweeness 

Distance between the CENTRES of the two circle = Sum of the radii of the two circles.

So,Distance between the centres of the two circle = R1 + R2 

= 9 + 8 = 17 cm 

Distance between the centres of the two circle = 17cm

So, the CORRECT option is 1).17cm

Given,

The circumference of SPEAKER = 44 cm

⇒ 2πr = 44

⇒ r = 7

Total surface AREA of cylinder = 2πr(H + r)

⇒ 2πr(h + r) = 528

⇒ h = 5

Now,

We know that,

Volume of cylinder = πr²h = (22 × 7 × 7 × 5)/(7 × 100 × 100 × 100) = 0.00077 m³

? AREA of RHOMBUS = 1/2 × (product of its diagonals)

⇒ Area of given rhombus = 1/2 × 20.25 × 16 = 162 CM²

⇒ Area of square = 162 cm²

? Area of square = d²/2, where d is the LENGTH of its DIAGONAL

⇒ d² = 2 × 162 = 324

⇒ d = √324 = 18 cm

∴ Length of diagonal of square = 18 cm

Let side of square = 2a, then Circumradius = √2a and INRADIUS = a

Solution:

Given: ABCD is a parallelogram and P and Q are midpoints on side BC and CD.

Let us take M as the midpoint of side AD and N as the midpoint of side AB. Let us take the POINT of intersection of PM and QN as O.

It is also given that the area of ∆ABC is 24cm² then the area of WHOLE parallelogram will be,

Area of the parallelogram ABCD = 2 × 24 = 48 cm².

Now we have to find the area of ∆AQP and 

Area of ∆AQP = Area of ABCD - Area of PCQ,QDM,ABP

Area of PCQ = ⅛ × 48 = 6 cm²

Area of QDM = ¼ × 48 = 12 cm²

Area of ABP = ¼ × 48 = 12 cm²

Area of ∆AQP = Area of ABCD - (6 + 12 + 12)

Area of ∆AQP = 48 - 30

Area of ∆AQP = 18 cm²

So, the correct option is 4).18 cm²

Circumference of CIRCLE = 88 CM

⇒ 2πr = 88 cm

⇒ R = 44 × 7/22 = 14 cm

∴ Area of circle = πr²where r is the RADIUS of circle

= (22/7) × 14² = 616 sq. cm

We KNOW, area of a SQUARE = a²

⇒ a² = 30.25

⇒ a = 5.5

We know, area of an equilateral TRIANGLE = √3/4 × a²

⇒ √3/4 × a² = 49√3

Total surface AREA of the prism = Curved Surface area + 2 × Area of the BASE

Let the side of the base be x.

200 = 4 × 10x + 2x²

⇒ 100 = x² + 20x

⇒ x = - 10 + 10√2 ≈ 4.14

Volume = area of the base × height = x × x × 10

LET the side of the SQUARE be 2x

⇒ Diameter of the circle = Side of the square

⇒ Radius of the circle = x

⇒ Area wasted = Area of the square – Area of the circle= (2x)² - π x²

⇒ x² (4 – 22/7)= 6x²/7

Area of the BASE of PRISM = 6 × √3/4 × 6 × 6 = 54√3

Total surface area of the prism = 2 × Area of the base + 6 × (Side of the base) × (Height of prism)

⇒ 216√3 = 2 × 54√3 + 6 × 6 × H

⇒ 36H = 108√3

Given, vertices of a quadrilateral are (- 1, 1), (0, - 3), (5, 2) and (4, 6)

Distance between two POINTS = √((x1 – x2)² + (y1 – y2)²)

Slope of a LINE = (y1 – y2)/(x1 – x2)

First of all we will find whether the adjacent sides are EQUAL.

Let the points be A, B, C and D in order.

AB = √((- 1 – 0)² + (1 + 3)²) = √17

BC = √((0 – 5)² + (- 3 - 2)²) = 5√2

CD = √((5 – 4)² + (2 - 6)²) = √17

DA = √((4 + 1)² + (6 - 1)²) = 5√2

AB is not equal to BC, thus it can’t be a square or rhombus.

Slope of line AB = (- 3 – 1)/(0 + 1) = - 4

Slope of line BC = (- 2 + 3)/(5 – 0) = 1/5

AB and BC are not perpendicular as Slope of AB × slope of BC is not equal to - 1

AB and CD are parallel.

DIMENSIONS = 80 cm × 60 cm × 40 cm

Required SQ. cm to cover the box = Surface AREA of the box

= 2(lb + bh + hl)

= 2(80 × 60 + 60 × 40 + 40 × 80)

Let l and b be the length and BREADTH of rectangle respectively.

Then perimeter of rectangle, 2(l + b) = 42 cm

⇒ l + b = 21cm

⇒ (l + b)² = 441 cm.......(1)

Length of DIAGONAL = 2√41

⇒ l² + b² = (2√41)² (Using Pythagoras theorem)

⇒ l² + b² = 164.......(2)

? (l + b)² = l² + b² + 2l × b (Using IDENTITY, (a + b)² = a² + b² + 2ab)

⇒ 441 = 164 + 2l × b

⇒ l × b = 138.5

∴ Area of rectangle = 138.5 cm² (? area of rectangle is length × breadth)

Let be the NUMBER of spherical BALLS that can be made

So, we have

⇒ 2/3(π)(6)³ = n × 4/3(π)(1)³

Given that PATH has width 2.5 m

LENGTH of the GARDEN INCLUSIVE of the path = 10 + 2(2.5) = 15

Breadth of the garden inclusive of the path = 8 + 2 (2.5) = 13

∴ AREA of the garden, inclusive of the path = length × breadth = 15 × 13 = 195 sq.m

Given length of the tank = 1 m = 100 CM

 BREADTH of the tank = ½ m = 50 cm

Height of the tank =1/2 m = 50 cm

Volume of the tank = 100 × 50 × 50

= 250000 cm³

Given, L = 2B ⇒ B = L/2

Let the SIDE of the SQUARE be A.

We have

2 (L + B) = 4A

⇒ 3L/2 = 2A

As we know, total surface area of cuboid = 2(lb + BH + hl)

And, maximum length of a ROD that can be put inside the BOX = √(L² + b² + h²)

Given, l + b + h = 24 cm and 2(lb + bh + hl) = 376 cm²

As we know, (l + b + h)² = l² + b² + h² + 2(lb + bh + hl)

⇒ 24² = l² + b² + h² + 376

⇒ l² + b² + h² = 576 – 376

⇒ l² + b² + h² = 200

⇒ √(l² + b² + h²) = √200 = 10√2 cm

As per the given data

Let deep l = 27 CM, height H = 6CM and wide b = 9 cm

Volume of the cuboid = length × breadth × height = 27 × 6 × 9 cm³

Also given that each cube is 3 cm as its edge

Volume of the each cube = l³ = 3³ = 27 cm³

∴ no of cubes will it hold = volume of the cuboid/volume of the cube = (27 × 6 × 9)/27 = 54

Given that r = 6 cm h = 12 cm

Volume of CONE = $(\frac{1}{3}\PI {r^2}h)$

$(= \frac{1}{3}\pi {6^2}12 = 144\pi)$ cm³

As per the given DATA,

Area of the circle = π × r²

Where, r = radius of circle

Area of the circle = 616 sq.m

⇒ π × r² = 616

⇒ r² = 616/π = 196

⇒ r = √196 = 14

∴ Diameter = 2r = 2 × 14 = 28 m

We know, the CURVED surface area of cylinder = 2πrh

Given, the curved surface area of cylinder = 352 cm² & diameter of cylinder = 14 cm

So, radius of cylinder = 14/2 = 7 cm

The cured surface area of given cylinder = 352

⇒ 2πrh = 352

⇒ 2 × 22/7 × 7 × h = 352

⇒ h = 8 cm

Length of the rectangle = 8 cm, Length of the diagonal = 17 cm

In a rectangle,

(Diagonal)² = (Length)² + (Breadth)²

⇒ (Breadth)² = (Diagonal)² - (Length)² = (17)² - (8)² = 289 - 64 = 225

⇒ Breadth = √225 = 15

Formula for area of the rectangle = Length × Breadth = 8 × 15 = 120 cm²

AREA of tent = Curved SURFACE area of CYLINDER + Area of cone

Curved surface area of cylinder = 2πrh

Area of cone = πrl

Area of tent = 2πrh + πrl

Total area = 2(22/7) × (7) × (8) + (22/7) × (7) × (12) = 616 m²

Let breadth of rectangle be B cm.

⇒ 2(40 + B) = 104

⇒ B = 52 – 40 = 12

If SHEET is folded ALONG its length to form a RIGHT circular CYLINDER, the length will be equal to circumference of base and breadth will be equal to height of cylinder.

Let R be radius of base of cylinder.

⇒ 40 = 2π R

⇒ R = 40/6.28 = 6.37 cm

Height of cylinder = B = 12 cm

A room of DIMENSIONS 4 m × 9 m is to be CARPETED leaving a margin of 5 cm form each wall.

4 m = 400 cm and 9 m = 900 cm.

So, new length of the carpet leaving margin = 900 – 5 – 5 = 890 cm = 8.9 m

New WIDTH of the carpet leaving margin = 400 – 5 – 5 = 390 cm = 3.9 m

The area of the room to be carpeted = 8.9 × 3.9 = 34.71 m²

The cost of the carpet is Rs. 15 per m².

As per the given DATA,

Let ABCD be the parallelogram

Area of parallelogram ABCD = 2 × (area of triangle ABC)

Now let us consider a = 20 m, b = 7m and c = 21 m

Area of triangle ABC = √s(s - a)(s - b)(s - c), where s = (a + b + c)/2 = 20 + 7 + 21/2 = 24 m

= √24(24 - 20)(24 - 7)(24 - 21)

= 12√34 sq.cm

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

Area of equilateral triangle = (√3/4) × a²

Cost to PAVE = 14 × area = 14 × (√3/4) × a² = 6.06 a²

Perimeter of triangle = 3a

Cost to fence = 30 × 3a = 90a

? Cost to pave = cost to fence

⇒ 6.06 a² = 90a

LET the lengths of the LINE segments be x and x + 3 cm then,

⇒ (x + 3)² – x² = 36

⇒ x² + 6X + 9 – x² = 36

⇒ 6x = 27

⇒ x = 27/6 = 4.5 cm

Length of LONGER segment = 4.5 + 3 = 7.5 cm

Let the inner radius of the path be r₁ and OUTER be r₂.

⇒ Width = r₂ – r₁

⇒ 2πr₂ – 2πr₁ = 132(? given in question)

⇒ r₂ – r₁ = 132/2π = 132/(2 × 22/7) = 132 × 7/44 = 21 m

∴ Width = 21m

As,

⇒ Perimeter = 2(length + breadth)

⇒ 46 = 2(length + 8)

⇒ Length = 23 – 8 = 15 cm

Also, EVERY angle of RECTANGLE is 90°

AREA of rectangle = l × b

357 = 17 × b

Other side = 21 cm

Diameter of LARGEST circle POSSIBLE = 17 cm

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

GIVEN, ratio of radius and height is 4:7

Let them be 4A and 7a respectively. Where a is any constant

Given, volume = 1008π cc

$(\therefore \pi\TIMES {(4a)^2} \times \frac{{7a}}{3} = 1008\pi)$

⇒ a³ = 27

⇒ a = 3 cm

SLANT height = √( radius² + height² )

⇒ Slant height = √( (4a)² + (7a)²)

Let the initial radius and length of the wire be R and l respectively.

Now, radius changes to r/3, let the new length be $(\;{l_1})$.

? Volume does not change,

$(\PI {r^2}l = \pi {\left( {\frac{r}{3}} \RIGHT)^2}{l_1})$

⇒ $({r^2}l = \frac{{{r^2}}}{9} \times {l_1})$

⇒ l₁ = 9l

Area of a square = side × side

Perimeter of a square = 4 × side

Given, wire when bent in the form of a square encloses an area of 484 sq.CM

Side = √484 = 22 cm

Perimeter = 88 cm

Same wire is USED to make a circle.

Perimeter of a circle of RADIUS r = 2πr

⇒ 2πr = 88

⇒ 44r/7 = 88

⇒ r = 14 cm

Area of a circle of radius r = πr2

Area enclosed by the circle $(= \frac{{22}}{7} \times {14^2})$

⇒ Radius = diameter/2 = 28/2 = 14 cm

Now,

Total surface area of HEMISPHERE (TSA) = 3πr²

Curved surface area (CSA) = 2πr²

CSA = 2/3 × TSA

= 2/3 × 462

= 308 cm²