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 cm^{2} ........................................................................... (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 cm^{2} ............................................................. (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 cm^{2}.

ii. Area of Unshaded region = sum of Areas of 4 inner rectangles

∴ Area of Unshaded region = 200 cm^{2} .................................................................................... (From Equation x)

Final Answer:-

∴ In the given figure, Areas of the shaded region and unshaded region are 175 cm^{2} & 200 cm^{2}, respectively.

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