# Area & Perimeter NCERT Exercise 11.4

## Part 3

**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.

**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` m^{2}

(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` m^{2}

(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:

**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` cm^{2}

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)

= 180 cm^{2} —(30 cm^{2} + 40 cm^{2}) = 180cm^{2} — 70cm^{2} = 110cm^{2}

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 cm^{2}

**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)

= 400cm^{2} — (100cm^{2} + 50cm^{2} + 100cm^{2}) = 400cm^{2} — 250cm^{2} = 150cm^{2}

**Question 11:** Find the area of the quadrilateral ABCD.

**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

= 33cm^{2} + 33cm^{2} = 66cm^{2}