AREA OF SHADED REGION
Example 1 :
Find the area of the shaded portion
Solution :
To find the area of shaded portion of given composite figure, first let
us draw a line
Area of given figure = Area of ABGE + Area of GCFD
Area of ABGE :
Area of rectangle = Length ⋅ Width
length BE = 6 m, width GE = 2 m
Area of ABGE = 6(2) = 12 m2
Area of GCFD :
Area of rectangle = Length ⋅ Width
length CD = 6 m, width FD = 2 m
Area of GCFD = 6(2) = 12 m2
Area of shaded portion = 12 + 12 = 24 m2
Example 2 :
Find the area of the shaded portion
Solution :
To find the area of shaded portion, we have to subtract area of GEHF
from area of rectangle ABCD.
Area of shaded portion
= Area of rectangle ABCD - Area of square GEHF
Area of ABCD :
Area of rectangle = Length ⋅ Width
Length AB = 20 cm
Width AC = 16 cm
Area of ABCD = 20 (16) = 320 cm2
Area of GEHF :
Area of square (GEHF) = side ⋅ side
Length GE = 6 cm
Area of square GEHF = 6 ⋅ 6
Area of square (GEHF) = 36 cm2
Area of shaded region = 320 - 36 = 284 cm2
Example 3 :
Find the area of the shaded portion
Solution :
To find the area of shaded region, we have to subtract area of
semicircle with diameter CB from area of semicircle with diameter
AB and add the area of semicircle of diameter AC.
Area of shaded portion
= Area of AEB - Area of semicircle with diameter BC + area of
semicircle with diameter AC
Area of semicircle AEB = (1/2) Πr2
= (1/2) ⋅ (22/7) ⋅ (14)2
= (1/2) ⋅ (22/7) ⋅ 14 ⋅ 14
= 22 ⋅ 14
= 308 cm2
Example 4 :
Find the area of the shaded region
Solution :
To find the area of shaded portion, we have to subtract area of
semicircles of diameter AB and CD from the area of square ABCD.
Area of shaded portion
= Area of square ABCD - (Area of semicircle AEB + Area of
semicircle DEC)
= a2 - [ (1/2) Πr2) + ((1/2) Πr2) ]
= 72 - Πr2
= 49 - (22/7) ⋅ (7/2)2
= 49 - (22/7) ⋅ (7/2) ⋅ (7/2)
= 49 - 38.5
Cổng đăng ký trực tuyến Kỳ thi
= 10.5 cm2
Olyfgeefgrfgfenjhrtrmpic Toán học quốc tế
TIMO 2024-2025
🎯 THÔNG BÁO ĐÓNG ĐĂNG KÝ THAM GIA TIMO 2024-2025 🎯
💥 Kỳ thi Olympic Toán học quốc tế TIMO 2024-2025 đã
đónggbbkjvshbnjnbvkjnbkjbkjbkjbhjbbjkbjhbfvdsvfdsfvfsfdvfvfvsfdssfd đăng ký.
gfrnbgc bhgnhbnhbfgcfgnfgcnhbfgnhbftgnhbftgnhb
🔖 Quý phụ huynh và thí sinh vui lòng tham khảo các kỳ thi Olympic uy tín khác dành cho học sinh
tại: https://bit.ly/TonghopOlympic
💎 Để thêm thông tin tư vấn, hỗ trợ, vui lòng liên hệ qua email: timo@fermat.edu.vn hoặc
fanpage: https://bit.ly/FacebookTimoVietnam hoặc qua các số hỗ trợ: 0917830455 / 0961 603003
trong khung giờ từ 08h30 đến 17h30 hàng ngày.
--
Ban Tổ chức Olympic Việt Nam
1. Địa chỉ: Số 6A1, Tiểu khu Ngọc Khánh, Ba Đình, Hà Nội
2. Tel: 0961 603 003 - 0917 830 455
3. Email: timo@fermat.edu.vn
4. Website: https://olympic.fermat.edu.vn
Find the area of the shaded region to the nearest 10th
Solution :
Area of shaded region = Area of circle - Area of triangle
Area of circle = πr2
Let R be the radius of larger circle and r be the radius of smaller
circle.
Factoring π, we get
= 3.14[122 - 62]
= 3.14(144 - 36)
= 339.12 square units
Find the area of the shaded region to the nearest whole number :
Solution :
Area of shaded region = Area of square - area of L shape
= side2 - area of L shape ----(1)
Side length of square = 10 m
By drawing the horizontal line, we get the shapes square and
rectangle.
= 2 x 2 + (8 x 3)
= 4 + 24
= 28 m2
Applying the value in (1), we get
= 102 - 28
= 100 - 28
= 72 square units.
Find the area of the shaded region.
Solution :
Area of shaded region = Area of trapezium - area of rectangle
Area of trapezium = (1/2) x h (a + b)
= (1/2) x 12 x (26 + 20)
= 6(46)
= 276 square units.
Area of rectangle = length x width
=8x5
= 40 square units
Area of shaded region = 276 - 40
= 236 square units.
Determine the area of the shaded region.
Solution :
Area of shaded region = Area of semicircle
Angle in a semicircle is right angle, diameter of the circle is
hypotenuse.
Let x be the hypotenuse.
x2 = 8 2 + 6 2
= 100
x = 10
diameter of the circle = 10 cm
radius = 5 cm
Area of semi circle = (1/2) πr2
= (1/2) x 3.14(5)2
= 1.57(25)
Find the area of the shaded region to the nearest 10th of a
centimeter:
Solution :
Area of shaded region = Area of semicircle - area of triangle.
Diameter of the circle = base = 25 cm
height of the triangle = radius = 12.5 cm
= (1/2) πr2 - (1/2) x base x height
= (1/2) x 3.14 x (12.5)2 - (1/2) x 12.5 x 25
= 1.57 x 156.25 - 156.25
= 245.31 - 156.25
Example 10 :
Find the area of the shaded region to the nearest 100th of a
centimeter:
Solution :
Area of shaded region = area of circle - area of right triangle
= πr2 - (1/2) x base x height
Diameter of the circle = 25 cm, radius = 12.5 cm
= 3.14(12.5)2 - (1/2) x 9 x 16
= 490.625 -72
= 418.625 cm2
