Respuesta :
The line segments AC and CE break this quadrilateral into three triangles. The area of a triangle is given by A=1/2bh. The base of ΔADC is 6 and the height is 4.5; A=1/2(6)(4.5)=13.5m².
To find the area of ΔACE, we must first find the length of segment AE. To do this, we will need to know both EC, which we are given, and AC, which we are not. We will use the Pythagorean theorem for ΔADC to find the length of AC.
6²+(4.5)²=(AC)²
36+20.25=(AC)²
56.25=(AC)²
√56.25=√(AC)²
7.5=AC
Now that we know the length of AC, we will use the Pythagorean theorem on ΔACE to find the length of AE:
7²+(AE)²=(7.5)²
49+(AE)²=56.25
49+(AE)²-49=56.25-49
(AE)²=7.25
√(AE)²=√7.25
AE≈2.69
Now we can find the area of ΔACE: A=1/2(7)(2.69)=9.415m²
To find the area of ΔBCE, we need to find the length of EB. We know that AB=9 and AE=2.69. Subtracting those two, we find the length of EB to be 6.31m. Therefore the area of ΔBCE=1/2(6.31)(7)=22.085m²
The total area is given by adding these three together:
13.5+9.415+22.985=45m².
It would take 15 rows of 20 tiles each to cover the rectangular portion of this patio (4.5m*100 =450cm; 450cm/30-cm tiles = 15 rows of tiles; 6m*100=600cm; 600cm/30-cm tiles = 20 tiles per row) which comes out to 300 tiles. However, adding an extra 10% to that:
300*0.1+300=30+300=330 tiles.
To find the area of ΔACE, we must first find the length of segment AE. To do this, we will need to know both EC, which we are given, and AC, which we are not. We will use the Pythagorean theorem for ΔADC to find the length of AC.
6²+(4.5)²=(AC)²
36+20.25=(AC)²
56.25=(AC)²
√56.25=√(AC)²
7.5=AC
Now that we know the length of AC, we will use the Pythagorean theorem on ΔACE to find the length of AE:
7²+(AE)²=(7.5)²
49+(AE)²=56.25
49+(AE)²-49=56.25-49
(AE)²=7.25
√(AE)²=√7.25
AE≈2.69
Now we can find the area of ΔACE: A=1/2(7)(2.69)=9.415m²
To find the area of ΔBCE, we need to find the length of EB. We know that AB=9 and AE=2.69. Subtracting those two, we find the length of EB to be 6.31m. Therefore the area of ΔBCE=1/2(6.31)(7)=22.085m²
The total area is given by adding these three together:
13.5+9.415+22.985=45m².
It would take 15 rows of 20 tiles each to cover the rectangular portion of this patio (4.5m*100 =450cm; 450cm/30-cm tiles = 15 rows of tiles; 6m*100=600cm; 600cm/30-cm tiles = 20 tiles per row) which comes out to 300 tiles. However, adding an extra 10% to that:
300*0.1+300=30+300=330 tiles.
The line segments AC and CE break this quadrilateral into three triangles. The area of a triangle is given by A=1/2bh. The base of ΔADC is 6 and the height is 4.5; A=1/2(6)(4.5)=13.5m².
To find the area of ΔACE, we must first find the length of segment AE. To do this, we will need to know both EC, which we are given, and AC, which we are not. We will use the Pythagorean theorem for ΔADC to find the length of AC.
6²+(4.5)²=(AC)²
36+20.25=(AC)²
56.25=(AC)²
√56.25=√(AC)²
7.5=AC
Now that we know the length of AC, we will use the Pythagorean theorem on ΔACE to find the length of AE:
7²+(AE)²=(7.5)²
49+(AE)²=56.25
49+(AE)²-49=56.25-49
(AE)²=7.25
√(AE)²=√7.25
AE≈2.69
Now we can find the area of ΔACE: A=1/2(7)(2.69)=9.415m²
To find the area of ΔBCE, we need to find the length of EB. We know that AB=9 and AE=2.69. Subtracting those two, we find the length of EB to be 6.31m. Therefore the area of ΔBCE=1/2(6.31)(7)=22.085m²
The total area is given by adding these three together:
13.5+9.415+22.985=45m².
It would take 15 rows of 20 tiles each to cover the rectangular portion of this patio (4.5m*100 =450cm; 450cm/30-cm tiles = 15 rows of tiles; 6m*100=600cm; 600cm/30-cm tiles = 20 tiles per row) which comes out to 300 tiles. However, adding an extra 10% to that:
300*0.1+300=30+300=330 tiles.
