Maths-
General
Easy
Question
In the figure given below, find the area of:
(i) the shaded portion and (ii) the unshaded portion.
The correct answer is: In the given figure, Areas of the shaded region and unshaded region are 175 cm2 & 200 cm2, respectively
Hint:-
i. Shaded portion = Outer rectangle - 4 Inner rectangles
ii. Unshaded portion = Sum of all 4 inner rectangles.
Step-by-step solution:-
In the adjacent figure, Let rectangle ABCD be the outer rectangle & rectangle AHQP, rectangle IBED, rectangle GFCJ & rectangle SRKD be the 4 inner rectangles.
For the outer rectangle-
length = AB = DC = 25 cm & breadth = AD = BC = 15 cm
For the inner rectangles-
We observe from the given diagram that the verticle and horizontal bars of the shaded region is in the exact middle of the outer rectangle rectangle ABCD.
∴ Distance between points A & H = Distance between points I & B
∴ AH = IB ......................................................................................... (Equation i)
Now, AB = AH + HI + IB
∴ AB = AH + HI + AH ..................................................................... (From Equation i)
∴ 25 = 2 AH + 5 ........................................................................... (From given information)
∴ 25 - 5 = 2 AH
∴ 20 = 2 AH
∴ AH = 10 ................................................................................... (Dividing both sides by 2)
∴ AH = IB = 10 cm .............................................................................. (Equation ii)
Also, Distance between points D & K = Distance between points J & C
∴ DK = JC ......................................................................................... (Equation iii)
Now, DC = DK + KJ + JC
∴ DC = DK + KJ + DK ..................................................................... (From Equation iii)
∴ 25 = 2 DK + 5 ........................................................................... (From given information)
∴ 25 - 5 = 2 DK
∴ 20 = 2 DK
∴ AH = 10 ................................................................................... (Dividing both sides by 2)
∴ DK = JC = 10 cm .............................................................................. (Equation iv)
Also, Distance between points A & P = Distance between points S & D
∴ AP = SD ......................................................................................... (Equation v)
Now, AD = AP + PS + SD
∴ AD = AP + PS + AP ..................................................................... (From Equation v)
∴ 15 = 2 AP + 5 ........................................................................... (From given information)
∴ 15 - 5 = 2 AP
∴ 10 = 2 AP
∴ AP = 5 ................................................................................... (Dividing both sides by 2)
∴ AP = SD = 5 cm .............................................................................. (Equation vi)
Also, Distance between points B & E = Distance between points F & C
∴ BE = FC ......................................................................................... (Equation vii)
Now, BC = BE + EF + FC
∴ BC = BE + EF + BE ..................................................................... (From Equation vii)
∴ 15 = 2 BE + 5 ........................................................................... (From given information)
∴ 15 - 5 = 2 BE
∴ 10 = 2 BE
∴ BE = 5 ................................................................................... (Dividing both sides by 2)
∴ BE = FC = 5 cm .............................................................................. (Equation viii)
Now,
i. Area of outer rectangle = Area of rectangle ABCD
∴ Area of outer rectangle = (AB × BC) …......... (Area of a rectangle = length × breadth)×
∴ Area of outer rectangle = 25 × 15
∴ Area of outer rectangle = 375 cm2 ........................................................................... (Equation ix)
Sum of areas of 4 inner rectangles = Area of rectangle AHQP + Area of rectangle IBED + Area of rectangle GFCJ + Area of rectangle SRKD
∴ Sum of areas of 4 inner rectangles = (AH × AP) + (IB × BE) + (JC × FC) + (DK × SD) …......... (Area of a rectangle = length breadth)
∴ Sum of areas of 4 inner rectangles = (10 × 5) + (10 5) + (10 * 5) + (10 * 5) ..................... (From given information & Equations ii, iv, vi & viii)
∴ Sum of areas of 4 inner rectangles = 50 + 50 + 50 + 50
∴ Sum of areas of 4 inner rectangles = 200 cm2 ............................................................. (Equation x)
Area of Shaded region = Area of outer rectangle - sum of areas of 4 inner rectangles
∴ Area of Shaded region = 375 - 200 .................................................... (From Equations ix & x)
∴ Area of Shaded region = 175 cm2.
ii. Area of Unshaded region = sum of Areas of 4 inner rectangles
∴ Area of Unshaded region = 200 cm2 .................................................................................... (From Equation x)
Final Answer:-
∴ In the given figure, Areas of the shaded region and unshaded region are 175 cm2 & 200 cm2, respectively.
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