A square flower bed should be built in the park, and a 2m wide lawn should be laid beside the flower bed. The lawn area is 96 square meters. What is the total area of the flower bed and lawn? In the simplest way. No equations

A square flower bed should be built in the park, and a 2m wide lawn should be laid beside the flower bed. The lawn area is 96 square meters. What is the total area of the flower bed and lawn? In the simplest way. No equations

The width of the lawn is 2M and the area is 96m2, which indicates that the length of the lawn is 96 △ 2 = 48m
This 48 is the side length of the square flower bed, so the area of the flower bed is 48 × 48 = 2304 square meters
So the total area of flower bed and lawn is 2304 + 96 = 2400 square meters

The length of the rectangular flower bed is 2m more than the width, and there is a half meter wide lawn around it, with a total area of 24m ^ 2

If the width of the flowerbed is x meters, the length is (x + 2) meters
(X+1)(X+2+1)=24
The solution is X1 = 3, X2 = - 7
3+2=5
3*5=15
The area of flower bed is 15 square meters

It is planned to build a square flower bed and lay a 3 meter wide lawn around the flower bed. The lawn area is 300 square meters
How many square meters will it take to build this flower bed

The area of lawn is a big square minus a small square, which can be regarded as four identical rectangles
The rectangle is 3 meters wide and 3 meters long
300 △ 4 = 75 square meters
75 △ 3 = 25m
25-3 = 22m
22 × 22 = 484 square meters

Lay a 2-meter-wide lawn around the square, with a total area of 40 square meters. How many square meters do the lawn and flower beds occupy

2 × 2 × 4 = 16 (square meters). Four squares (4 corners) with side length of 2 are 40-16 = 24 (square meters). First subtract four squares (4 corners) with side length of 2 from lawn, and the remaining four rectangular lawn areas are 24 △ 4 = 6 (square meters). Find a rectangular lawn area of 6 △ 2 = 3 (meters). The length of rectangle is 3 + 2 + 2 = 7 (meters)

In a rectangular lawn, there are three small square flower beds with side length of 2 meters, and the total area of the lawn is calculated

The total area of flower bed is 3x2x2 = 12
Lawn 12

There is a square pool. Around it, a flower bed with a width of 8 meters is built. The area of the flower bed is 480 square meters. How many meters is the side length of the pool

480÷4÷8-8
=15-8
=7 meters

There is a rectangular flower bed. If the side length is increased by 8 decimeters and the area is increased by 144 square decimeters, how many square decimeters is the original area of the flower bed?

As shown in the figure: let the side length of the original square be x decimeter, & nbsp; 8x + 8x + 8 × 8 = 144, & nbsp; 16x + 64-64 = 144-64, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 16x = 80, & nbsp; & nbsp; & nbsp; 16x △ 16 = 80 △ 16, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 5,5 × 5 = 25 (square decimeter), a: the original area of this flowerbed is 25 square decimeter

Build a 1-meter-wide stone road around a pool with a diameter of 8 meters. How many square meters is the area of this stone road?

The radius of inner circle is: 8 △ 2 = 4 (m); 3.14 × [(4 + 1) 2-42] = 3.14 × [25-16] = 3.14 × 9 = 28.26 (M2). Answer: the area of stone road is 28.26 m2

The side length of a square flower bed is 6 meters, and there is a 1 meter wide path around the periphery. Please find the area of the path

Arithmetic solution:
6＋1＝7（m）
7×7＝49（㎡）
49-6×6=13（㎡）
A: the area of the path is 13 square meters
Equation solution:
Set a path of X m2
x＝（6＋1）² -6×6
x＝49－36
x＝13
A: the area of the path is 13 square meters
It's not guaranteed to be all right

If the image of the function y = sin2x + acos2x is symmetric with respect to the line x = - π 8, then a is equal to ()
A. 2B. 1C. −2D. -1

From the meaning of the question, we know that y = sin2x + acos2x = A2 + 1sin (2x + φ) when x = - π 8, the function y = sin2x + acos2x takes the maximum value ± A2 + 1, and substituting x = - π 8, we can get: sin [2 × (− π 8)] + ACOS [2 × (− π 8)] = 22 (a − 1) = ± A2 + 1, the solution is a = - 1, so we choose D