## Area Problems Solved Questions For IBPS, SBI and Other Bank Exams

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## Area Problems Solved Questions

- Area of a Rectangle =
**length * breadth**

Length = (Area/Breadth) and Breadth = (Area/Length)

- Perimeter of a Rectangle =
**2(Length + Breadth)** - Area of a Square =
**(side)**=^{2}**½ ( Diagonal)**^{2} - Area of four walls of a room =
**2(length + breadth) * height** - Area of a triangle =
**½ * base * height** - Area of a Triangle = (Heron’s formula)

where a,b,c are the sides of the triangle and s (semi-perimeter) = 1/2(a+b+c)

- Area of an equilateral triangle =
**(****/4)* (side)**^{2} - Area of a Parallelogram =
**(base * height)** - Area of Rhombus =
**1/2 (product of diagonals)** - Area of Trapezium =
**1/2 * (sum of parallel sides)* (distance between them)** - Area of a Circle where r is the radius
- Circumference of a Circle =
**2****r** - Length of an arc =
**θ/360*****2****r** - Area of a Sector =
**θ/360*** - Area of a Semicircle =
**/2**

Let us under the concept of Areas with the help of some examples:

**Eg: **Length of one side of a rectangular field is 15m and length of one of its diagonal is 17m. Find the area of the field.

**Sol: **Applying Pythagoras here, Length of another side = = = 8m

So, Area of the field = 15 * 8 =**120 sq. m**

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**Eg: **The length of a rectangle is twice its breadth & if its length is decreased by 5cm and breadth is increased by 5cm, the area of the rectangle is increased by 75 sq cm. Find the breadth of the rectangle.

**Sol:** let length = 2x and breadth = x then

(2x-5) (x+5) – (2x*x)=75

5x-25 = 75 => x=20

So, breadth of the rectangle = **20 cm**

** **

**Eg: **A rectangular plot 110m * 65cm has a gravel path 5cm wide all round it on the inside of the rectangular plot. Find the cost of graveling the path at 80 paise per sq.mt.

**Sol:** Area of the plot = 110 * 65 = 7150 sq m

Area of the plot excluding the path = (110-5)* (65-5) = 6300 sq m

Area of the path = 7150- 6300 =850 sq m

Cost of graveling the path = 850 * 80/100 = **680 Rs**

**Eg: **A room 5m 44cm long and 3m 74cm broad is to be paved with square tiles. Find the least number of square tiles that are required to cover the floor.

**Sol:** Area of the room = 544 * 374 sq cm

Size of largest square tile = H.C.F of 544cm and 374cm= 34cm

Area of 1 tile = 34*34 sq cm

No. of tiles required = (544*374) / (34 * 34) = **176**

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