Area & Perimeter

Exercise 11.4

Question 9: The adjoining figure represents a rectangular lawn with a circular flower bed in the middle.Find:
(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed
(iv) the circumference of the flower bed.

rectangle circle

Solution: Given, Length of rectangular lawn = 10m, Width of rectangular lawn = 5m, Radius of flower bed = 2m

(i) The area of whole land=Area of given rectangular lawn
= Length `xx` width
`= 10xx5 = 50` m2

(ii) Area of flower bed = Area of circle `= π r^2`
`= 3.14 xx 2 xx 2 = 12.56` sq m

(iii) The area of the lawn excluding the area of the flower bed = Area of lawn —area of flower bed
`= 50 —12.56 = 37.44` m2

(iv)Circumference of the flower bed = circumference of circle `= 2πr`
`= 2 xx 3.14 xx 2 = 12.56` m


Question 10: In the following figures, find the area of the shaded portions:

rectangle circle rectangle circle

Solution:Given; For rectangle ABCD: Length = 18cm, Width = 10cm,
For triangle AEF: Height = 10cm, Base = 6cm
For triangle EBC: Base = 8cm, Height = 10cm

Area of rectangle ABCD = Length `xx` Width
`= 18xx10= 180` cm2

Area of triangle AEF `= 1/2 xx \text(base) xx \text(height)`
`= 1/2 xx 6 xx 10 = 30` sq cm

Area of triangle EBC `= 1/2 xx \text(base) xx \text(height)`
`= 1/2xx8 xx 10 = 40` sq cm

Now, Area of shaded portion
= Area of Rectangle ABCD —(Area of ∆AEF + Area of ∆EBC)
= 180cm2 —(30cm2+40cm2) = 180cm2 — 70cm2 = 110cm2

Area of shaded portion = PQRS —(Area of ∆PQT+ Area of ∆SUT + Area of ∆RQU)

Given; SR = SU+UR = 10 cm + 10cm = 20cm = PQ = QR = SP

Therefore this figure is a square.


Area of square = side × side
Or, Area of PQRS = 20cm × 20cm = 400 cm2

For triangle PQT;

PQ = SR = 20cm = Height of triangle
PT = PS — TS
= 20cm — 10cm = 10cm = Base of triangle

Area of triangle PQT `= 1/2 xx \text(base) xx \text(height)`
`= 1/2 xx 20 xx 10 = 100` sq cm

For triangle SUT; Height = Base = 10cm

Area of triangle SUT `= 1/2 xx \text(base) xx \text(height)`
`= 1/2 xx 10 xx 10 = 50` sq cm

For triangle RQU; Height = QR = 20cm, Base = UR = 10cm

Are of triangle RQU `= 1/2 xx \text(base) xx \text(height)`
`= 1/2 xx 20 xx 10=100` sq cm

Area of shaded portion = PQRS —(Area of ∆PQT+ Area of ∆SUT + Area of ∆RQU)
= 400cm2 — (100cm2 + 50cm2 + 100cm2) = 400cm2 — 250cm2 = 150cm2


Question 11: Find the area of the quadrilateral ABCD.

rectangle circle

Solution: Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC

For ∆ABC, AC = Base = 22cm BM = Height = 3cm

Area of triangle ABC `= 1/2 xx \text(base) xx \text(height)`
`= 1/2 xx 22 xx 3 = 33` sq cm

For ∆ADC, AC = Base = 22cm, DN = Height = 3cm
Therefore, area of triangle ADC = area of triangle ABC = 33 sq cm

Now, Area of Quadrilateral ABCD = Area of ∆ABC + Area of ∆ADC
= 33cm2 + 33cm2 = 66cm2



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