find the perimeter and area of the adjoining figure in being given that AB is parallel to CD ab = 12 CM CD = 15 cm and BD = 4 cm and one end of the figure is a semicircle with BD as diameter .(take Pi =3.14)
Answers
NICE QUESTION!
GIVEN:
- AB = 12 cm.
- CD = 15 cm.
- BD = 4 cm.
- BD is the diameter of the semicircle ◠BD. Thus, radius = ½Diametre, i.e., radius = ½×4 = 2 cm.
- AB║CE.
CONSTRUCTION:
- Draw BE║AC [It is already drawn in the figure]. Then, we can see that if BE║AC and AB║CE, ABEC is a ║gm. [In case you didn't know, ║gm is a symbol for a parallelogram].
SOLUTION:
PERIMETER:
Now, CD is 15 cm. And CE is 12 cm because AB=CE [ABCD is a ║gm and opposite sides of a parallelogram are equal] and AE=12 cm.
We know that CE+ED=CD. [CD=15 cm, CE=12 cm.]
12+ED=15
ED=15-12
ED=3cm.
Since BD is a diameter, It is perpendicular to the base. In other words, BD⊥CD. Thus, BD can be considered the height of ΔBDE, ΔBDE is a right-angled triangle.
We can use Pythagoras theorem:
BE²=BD²+ED²
BE²=4²+3²
BE²=16+9
BE²=25
BE=√25
BE=5cm.
Here, BE=AC because ABEC is a ║gm and opposite sides of a ║gm are equal.
∴ AC=5cm.
Then,
We know that the circumference of a semicircle is (πr)
Radius=2 cm,
π=3.14.
Circumference (BD) = 2×3.14 = 6.28 cm.
Now, we know BD=6.28 cm, CD=15cm, AC=5cm, AB=12cm.
Add them to get the perimeter.
Perimeter = 6.28+15+5+12 = 38.28 cm.
AREA:
Now, consider trapezium ABDC.
Since BD is a diameter, It is perpendicular to the base. In other words, BD⊥CD. Thus, BD can be considered the height of ABDC.
Area of trapezium = ½(sum of ║ sides)×(height)
Ar(ABDC)=½(12+15)(4)
Ar(ABDC)=½(108)
Ar(ABDC)=54 cm².
Then, Area of semicircle=½×(πr²)
Radius=2, π=3.14
Ar(BD)=½×3.14×2²
Ar(BD)=6.28cm².
Total area = Area of the semicircle + area of the trapezium
A = (6.28+54) cm²
A = 60.28 cm²
CONCLUSION:
- Area = 60.28 cm²
- Perimeter = 38.28 cm.
THANK YOU (＾▽＾)
HOPE THIS HELPS :D
we have to add all the areas according to their formulas
Then the perimeter of the semicircle and add all the boundaries of the figure
