Length = 12m
Breadth = Xm
Area = 360m = L×B = 12×X
= 360m = 12x
X = 360/12
X = 30m

so, perimeter = 2(L+B)
= 2(12+30)
= 2(42)
= 84m.

Length of the rectangular ground = 30 m

Breadth of the rectangular ground = 20 m

Area of the rectangular ground = L*B

⇒ 30*20

= 600 sq m

Given that the lane of uniform width 2 m bordering the ground.

Then,

Length of the inner rectangular portion = 30 – (2 + 2)

= 30 – 4

= 26 m

Breadth of the inner rectangular portion = 20 – (2 + 2)

= 20 – 4

= 16 m

Area of the inner rectangular portion = 26*16

= 416 sq m

Area of the lane of uniform width of 2 m = Area of the rectangular ground – Area of the inner rectangular portion

⇒ 600 – 416

= 184 sq m

Let the altitude of the triangle be y
Then the base will be 3y

Area of triangle is 1/2 x base x altitude
=1/2 x 3y x y
= 1/2 x
= /2

Cost of sowing the field at rs 58 per hectare = 58 × 3×2/2 = 87×2
Therefore, 87×2 = 783
= 783/87 = 9
⇒ y = 3
Base = 3(3) = 9
Hence base is 9 and altitude is 3.

Distance=2002m
Radius=49cm
Circumference=2r
=2*22/7*49
=308 cm
Distance covered by wheel in one rotation=Circumference of wheel
total rotations=total distance/total dist. covered in one rotation
=200200/308
=650 revolutions

LET THE ORIGINAL LENGTH BE L

AND

ORIGINAL BREATH BE B.

ORIGINAL AREA = LB

LENGTH INCREASED BY 50% = 150 L / 100

BREATH INCREASED BY 20% = 120 B / 100

INCREASED AREA = 150 L X 120 B / 100 X 100

= 18000 LB / 10000

= 1.8 LB

NOW,

1.8 LB / LB = 1.8

CHANGING 1.8 INTO FRACTION.

1.8 = 18/10 = 9/5

SO,

THE INCREASED AREA WILL BE 9/5 TIMES OF THE ORIGINAL AREA

Perimeter of semicircle= πr+2r
perimeter is 36cm
36=22/7r+2r
36=22r+14r/7
36=36r/7
r=36×7/36
r=7
Diameter=2×7
=14

Let a circle with center O And radius R.
let
another circle inside the first circle  with center o’ and radius r .
A/Q,
Area of 1st circle + area of 2nd circle = 116π cm²
⇒ πR² + πr² = 116π
⇒ π(R² + r²)  = 116π
⇒ R² + r² =116 ——————–(i)
Now,
Distance between the centers of circles = 6 cm
i.e, R – r = 6
⇒  R = r + 6  ——————-(ii)
From Eqn (i) & (ii),
(r + 6)² + r² = 116
⇒ r² + 12r +36 + r² =116
⇒ 2r² +12r +36 -116 = 0
⇒ 2r² +12r – 80 = 0
⇒ r² +6r – 40 = 0
⇒ r² +10r – 4r – 40 = 0
⇒ r(r + 10) – 4(r + 10) = 0
⇒ (r + 10)(r – 4) = 0
hence  r = 4 cm
r ≠ -10 cm        {∵ length can’t be -ve}
Therefore radii of the circles are
r = 4 cm ,
R = 4 + 6 = 10 cm.

Let the outer rectangle be ABCD and the inner rectangle be abcd.
Length and Breadth of the outer rectangle = 120 meter and 85 meter.
The uniform width of the inside path is 3.5 meter,
So, the length of the inner rectangle = 120 – 7 = 113 m
and the breadth of the inner rectangle = 85 – 7 = 78 m
The area of the inner rectangle = 113 × 78 = 8814 m²
Cost of covering the garden by grass = Rs 0.5 per square meter.
The total cost of covering the garden by grass = 0.5 × 8814
= Rs 4407

An isosceles right triangle has area 32 CM
Let the length of each sides of isosceles triangle be x (cm)
Area of triangle = 1/2 base x altitude
Area = 32cm² (given)
∴ , 32 = 1/2 x²
x² = 64
x = 8
length of base and height = 8
now, length of hypotenuse = √ (base)² + (altitude)²
length of hypotenuse = √ (8)²  + (8)²
length of hypotenuse = √ 64 + 64
= √128 = 6√2 cm

Let original length = x and original breadth = y.

Original area = xy.

New breadth = 3y.

New area =		x	x 3y		=	3	xy.
		2				2

Increase % =		1	xy x	1	x 100	%	= 50%.
		2		xy

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = (41 x 41) cm².

Required number of tiles =		1517 x 902		= 814.
		41 x 41

Area of the park = (60 x 40) m² = 2400 m².

Area of the lawn = 2109 m².

 Area of the crossroads = (2400 – 2109) m² = 291 m².

Let the width of the road be x metres. Then,

60x + 40x – x² = 291

 x² – 100x + 291 = 0

 (x – 97)(x – 3) = 0

 x = 3.

Here is your answer.
Length of the rectangular plot is 90 m
Breadth of the rectangular plot is 50 m

Perimeter is 2( L + B)

If the poles of the fence are 5 m apart,
Then Number of poles = 280/5=56

Lenght of the room=15m 17cm=
15×100+17=1517
breadth=9m2cm=902 cm
the HCF of the 1517 and 902 will be size of saure tiles
HCF 1517 and 902 = 41cm
area of room= leanth×breadth=1517×902cm²
area of tiles=41×41cm²1
so num. of tiles required(1517×902)(41×41)
=814 tiles

Area $=5.5\times 3.75$ sq. metre.

Cost for $1$ sq. metre. = Rs.$800$

Hence, total cost
$=5.5\times 3.75\times 800=5.5\times 3000=\text{Rs}.16500$

Consider a square plot as shown above.

Please refer the diagram given above.

The field is a rectangle whose length is 25 meters and width is 12 meters. The area is L × W = 25 × 12 = 25 × 4 × 3 = 100 × 3 = 300 m²
The field covered by the house is a square with the side of 9 meters. The area is 9 × 9 = 81 m²
The garden’s area will be 300 – 81 = 219 m²

The area of the first rectangle is L × W = 6 × 4 = 24 cm²
The new length is 6 + 2 = 8 cm
8 × W = 24 then W = 24 ÷ 8 = 3 cm

The sides of the box are equal in pairs. Two of the sides have edges of 10 cm and 6 cm.
Their area is (10 × 6) × 2 = 60 × 2 = 120 cm²
Two of the sides have edges of 10 cm and 8 cm.
Their area is (10 × 8) × 2 = 80 × 2 = 160 cm²
Two of the sides have edges of 8 cm and 6 cm.
Their area is (8 × 6) × 2 = 48 × 2 = 96 cm²
The total area is 120 + 160 + 96 = 376 cm²

The area of the first rectangle is L × W = 6 × 4 = 24 cm²
The new length is 6 + 2 = 8 cm
8 × W = 24 then W = 24 ÷ 8 = 3 cm