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# (Online Course) CSAT Paper - II : Basic Numeracy: Area of Plane Figures

## Basic Numeracy

## Area of Plane Figures

## Area

The area of a plane figure is the measure of the surface enclosed by its boundary. The area of a triangle or a polygon is the measure of the surface enclosed by its sides.## 1. Triangle

## 2. Right-Angles Triangle

Hypotenuse,

Perimeter = b + d + h

## 3. Isosceles Right-Angled Triangle

## 4. Equilateral Triangle

Perimeter = 3a

Radius of incircle of an equilateral triangle of side **a = a/2
Ö3**Radius of circumcircle of an equilateral triangle of side a =

**a/Ö3**

## 5. Square

## 6. Rectangle

7. Parallelogram

Opposite sides are parallel and equal and opposite angles are equal.

Area = Base × Height = b × h

Perimeter = 2 (a + b)

## 8. Rhombus

All sides are equal and opposite angles are equal. Diagonals bisect each other at right angles.

## 9. Trapezium

## 10. Qudrilateral

## 11. Circle

## 12. Semi-Circle

## 13. Sector

**r = Radius, l = Length of the arc
q = Central angle**Area of the sector =1/2 (Arc length* r )=

**Πr**

^{2}q/360where are length=

**2Πrq/360**

**
**

**Example 1:** One side of a rectangular field is 16 m and one of its
diagonal is 20 m. Find the area of the field. Solution. BC^{2} = AC^{2
}– AB^{2} = (20)^{2} – (16)^{2 }= 400 – 256 = **
144**

**Example 2: **The length and breadth of a room are in the
ratio of 3 : 2. Its area is 864 m2. Find its perimeter.**Solution. **Let the length be 3x m and breadth be 2x m.

**Area = (3x × 2x) = 6x ^{2} = 864= x^{2} = 144
= x = 12**

Length = 3x = (3 × 12) m = 36 m

Breadth = 2x = (2 × 12) m = 24 m

Perimeter = 2 (l + b) = 2 (36 + 24) = 120 m

**Example 3:**Find the circumference of a circle whose diameter is 14 m.

**Solution.**r =diameter/2=14/2 = 7m

Area = Πr2=22 /7 *7 *7 =154m2

Circumference = 2Πr == 44m

**Example 4:
**The areas of two concentric circles are 154m2 and 308m2 respecively. Find
the width of the ring made by them.**Solution. **The radius of inner circle r = Ö154/Π
= 7m

Another circle is outside this circle.

If the width of the ring be w then area of the ring

Πw (2r + w) = 308 – 154

22/7*w (14 +w)= 154

w = 289 m

**Example 5:**Find the area of a regular hexagon inscribed in a circle whose radius is 15 cm.

**Solution:**OC = r = 15 cm

Since, regular hexagon is inscribed in the circle. D BOC is an equilateral triangle

So, BC = a = 15 cm

**Example 6:** A rectangular grassy plot 120m × 80m has a gravel path 3m
wide all round it on the inside.

Find the cost of gravelling the path at Rs. 2 per m^{2}.**Solution.** Area of the plot= (120 × 80)m^{2} = 9600 m^{2}Area of the plot (excluding path)= [(120 – 6) × (80 – 6)] m^{2}
=(114 × 74) m^{2} = 8436 m^{2}

Area of the path= (9600 – 8436) m^{2} = 1164 m^{2}

Cost of gravelling the path= Rs. (1164 × 2) = Rs.** 2328**

**Example 7:**Find the length of the diagonal of a square whose area is 441 m

^{2}

**Solution.**Solution. Area of the square (side)2= 441m2

Side = 441 = 21 m

Diagonal of a square= Ö2*side =

**21Ö 2m**

**Example 8: **Find the area of a square that can be
inscribed in a circle of radius r.**Solution. **Let ABCD be a square inscribed in a circle of radius r.

Diagonal AC = 2(OC) = 2r

Let AB = a, then AC =Ö2a

Ö2a = 2r , a = 2/Ö2r = Ö2r

Hence, Area = a^{2}= ( 2r)^{2}= 2r^{2}

**Example 9:**Find the area of a triangle whose sides are 6m, 11m and 15m in length.

**Solution:**

**Example 10: **Find the area of a right angled triangle which is
inscribed in a circle of radius 4 cmand altitude drawn to the hypotenuse is 3
cm.**Solution. **

**Example 11: **Find the percentage increase in the area of a
triangle, if its each side is doubled.**Solution:** Let a, b, c be the sides of the given triangle

**Example 12: **The perimeter of a rhombus is 48 m and the
sum of the lengths of its diagonals is 26 m. Find the area of rhombus.**Solution. **The perimeter of rhombus = 4a = 48

Þ a = 12m

If the diagonals of rhombus are d_{1} and d_{2}, then d_{1}
+ d_{2} = 26m (given)

Now, by the relation between diagonals and sides of rhombus