Can anyone help me with this qn. Thnks

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# Question

# Answer

Alternate way.

**Using Chief’s diagram,**

*Area Z,*

Using base as AB, the height is only 2/3 of H. So area is 2/3 of triangle with full height. The area of the full height triangle is half of the rectangle,i.e. 36 cm^{2}

**So area Z is 2/3 X 36 = 24 cm ^{2}**

*Area X,*

Going by the same logic as above. Using AD as base, height is only 1/3

**Area X = 1/3 X 36 = 12 cm ^{2}**

*Area Y,*

**Has, 1/3 height and 2/3 base. So area is 1/3 X2/3 X 36 = 8 cm**^{2 }

**Shade area = 72 – 24 – 12 – 8 = 28 cm ^{2 }**

Let b = breadth of rectangle

Let h = height of rectangle

Area of X = 1/2 (1/3b)h = 1/6 bh

Area of Y = 1/2 (2/3b)(1/3h) = 1/2 (2/9)bh = 1/9 bh

Area of Z = 1/2 b (2/3h) = 1/3 bh

Total unshaded area = Area of X + Y + Z

= (1/6 + 1/9 + 1/3) bh

= (3/18 + 2/18 + 6/18) bh

= 11/18 bh

Area of shaded area = 7/18 bh = 7/18 x 72 cm^{2} = 28 cm^{2}

Let b = breadth of rectangle

Let h = height of rectangle

Area of X = 1/2 (1/3b)h = 1/6 bh

Area of Y = 1/2 (2/3b)(1/3h) = 1/2 (2/9)bh = 1/9 bh

Area of Z = 1/2 b (2/3h) = 1/3 bh

Total unshaded area = Area of X + Y + Z

= (1/6 + 1/9 + 1/3) bh

= (3/18 + 2/18 + 6/18) bh

= 11/18 bh

Area of shaded area = 7/18 bh = 7/18 x 72 cm^{2} = 28 cm^{2}