find area pls solve itttttt
Answers
Step-by-step explanation:
On the bottom you have a triange whose base is (34+6+10)cm = 50cm and height is 16cm. So its area will be 16*50/2= 400cm².
On the top there is a trapezium with parallel side 12 and 14 cm and perpendicular distance between them as 34cm. So its area will be (12+14)(34)/2 = 442 cm²
And you also have two triangles, one on left and other on the right of the trapezium.Their bases and heights are given so calculate the area.
Question:
Find the area of the given figure.
Answer:
948 cm^2
Note:
• Area of parallelogram = Base×Height
• Area of rectangle = Length×Breadth
• Area of square = Side×Side = (1/2)×(diagonal)^2
• Area of triangle = (1/2)×Base×Height
• Area of circle = π×(radius)^2
• Area of rhombus = (1/2)×D1×D2
where D1 and D2 are its two diagonals.
• Area of trapezium = (1/2)×(a+b)×h
where a & b are its parallel sides and h is the perpendicular length between a & b.
Solution:
• Area of ∆ABF = (1/2)×(AF)×(BF)
= (1/2)×6×12
= 3×12
= 36 cm^2
• Area of ∆CDH = (1/2)×(HD)×(CH)
= (1/2)×10×14
= 10×7
= 70 cm^2
• Area of ∆ADE = (1/2)×(AD)×(GE)
= (1/2)×(AF+FH+HD)×(GE)
= (1/2)×(6+34+10)×16
= (1/2)×50×16
= 50×8
= 400 cm^2
• Area of trap.BCHF = (1/2)×(BF+CH)×FH
= (1/2)×(12+14)×34
= (1/2)×26×34
= 13×34
= 442 cm^2
Now ,
• Area of the given figure
= Area of ∆ABF + Area of ∆CDH
+ Area of ∆ADE + Area of trap.BCHF
= 36 + 70 + 400 + 442
= 948 cm^2
