函数 二次方程1.The diagram shows the outdoor pool at Lakeview Park.A cement deck 2 m wide surrounds the outside edge of the pool.Calculate the area of the deck.2.Determine the maximum area of a triangle,in square centimetres,if the sum of its b

函数 二次方程1.The diagram shows the outdoor pool at Lakeview Park.A cement deck 2 m wide surrounds the outside edge of the pool.Calculate the area of the deck.2.Determine the maximum area of a triangle,in square centimetres,if the sum of its b
函数 二次方程
1.The diagram shows the outdoor pool at Lakeview Park.A cement deck 2 m wide surrounds the outside edge of the pool.Calculate the area of the deck.
2.Determine the maximum area of a triangle,in square centimetres,if the sum of its base and its height is 10 cm.
3.A photograph measuring 12cm by 8cm is to be surrounded by a mat before framing.The width of the mat is to be the same on all sides of the photograph.The area of the mta is to equal the area of the photograph.Find the width of the mat.

函数 二次方程1.The diagram shows the outdoor pool at Lakeview Park.A cement deck 2 m wide surrounds the outside edge of the pool.Calculate the area of the deck.2.Determine the maximum area of a triangle,in square centimetres,if the sum of its b
1.The diagram shows the outdoor pool at Lakeview Park.A cement deck 2 m wide surrounds the outside edge of the pool.Calculate the area of the deck.
Solution :
Pool area = 25×30 + 10×15 = 900 (m²)
Deck area = (25+4)×(30+10+4) - 5×10×2 - 900 = 276 (m²)
2.Determine the maximum area of a triangle,in square centimetres,if the sum of its base and its height is 10 cm.
Solution:
Let the base be x,so the height is (10 - x)
Area of triangle = ½x(10-x) = -½(x²-10x) = -½(x²-10x+25-25)
= 12.5 - ½(x-5)²
So,if x = 5,maximum area of triangle = 12.5 (cm²)
3.A photograph measuring 12cm by 8cm is to be surrounded by a mat before framing.The width of the mat is to be the same on all sides of the photograph.The area of the mat is to equal the area of the photograph.Find the width of the mat.
Solution:
Let x be the width of the mat.
Area of photograph = 12×8 = 96 (cm²)
Area of photograph with mat = (12+2x)×(8+2x) = 192
(12+2x)×(8+2x) = 192
(6+x)×(4+x) = 48
x²+10x-24=0
(x+12)(x-2)=0
x₁= 2 (cm); x₂= -12 (cm)(unresonable,rejected)
Ans:Width of the mat is 2 cm.

2. S=x(10-x)=<(10/2)^2=25
3. S=25*30+15*10=900
边长为根号900=30cm

比竞赛还难…1.一条边(棱)为2m，求面积应该是4m^2…2.三角形底边s与高h之和为10，求最大面积。由((s+h)/2)^2>=s*h知S=(1/2)*s*h=12.5(cm^2)…3.长12，宽8，最长的应该是4*((根号)13)cm