find the area enclosed by the given area in the given figure , guys please please look at my question as soon as possible .....and I will mark you the brainest
Answers
Answer:
400cm^2
Step-by-step explanation:
area of the region = 2× area of triangle + area of rectangle
we know that
hypotenuse^2= side^2+side^2
FH=EC=12cm
length of rectangle =17cm
breadth of triangle = 8+12=20cm
area of region = 2×1/2×5×12+17×20
= 60+340
area of region = 400cm^2
Step-by-step explanation:
For the figure please refer to attachment
From the figure we can see that FEDG is a trapezium and CBAH is a rectangle
So to find the area of the whole figure we can add the area of the trapezium + area of the rectangle
We know that area of trapezium = (Sum of the two parallel sides)/2 x height between the two parallel sides.
We also know that area of rectangle = Length x Breadth.
From the figure we can also see that FEBA is also a rectangle,
Therefore,
AB = HC = FE = 17 cm
Area of the trapezium = (FE + GD)/2 x FH
We know that,
GD = GH + HC + CD = 5 + 17 + 5 = 27 cm
From the ∆FHG,
As ∆FHG is a right angled triangle,
Therefore,
According to pythagoras theorem,
FG² = GH² + FH²
13² = 5² + FH²
169 = 25 + FH²
FH² = 169 - 25
FH² = 144
FH = √144
FH = 12 cm
Therefore,
Area of trapezium = (17 + 27)/2 x 12 = 22 x 12 = 264 cm²
Area of the rectangle = HA x AB = 8 x 17 = 136 cm²
Total area of the figure = 264 + 136 = 400 cm²
From the figure we can see that,
Area of FEBA = FA x AB
FA = FH x HA
Above we have calucated the value of FH using pythagoras theorem, i.e 12 cm
FA = 12 + 8 = 20 cm
Area of FEBA = 20 x 17 = 340 cm²
Area of FHG = FH x GH/2 = 12 x 5/2 = 6 x 5 = 30 cm²
Area of EDC = EC x CD/2 = 12 x 5/2 = 6 x 5 = 30 cm²
Total area of the figure = 340 + 30 + 30 = 400 cm²