# Area problems with solutions for ssc and bank exams

Are you puzzled while solving area problems in your preparation? Learn here how to solve area problems with solutions for ssc and bank exams because most of the students have doubts regarding how to solve area problems.

With the help of this blog, you can easily resolve your doubts which you face while solving area problems. So, improve your performance by practicing area problems with solutions and learn the correct way to solve area equations.

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## Area Problems with Solutions for Competitive Exams

**Q.1. A lawn is in the form of a rectangle having its sides in the ratio 2 : 3. The are of the lawn is 1/6 hectares. Find the length and breadth of the lawn.**

**Solution **

Let length = 2x metres and breadth = 3x metres.

$$Now, area = \left( {1\over6}×1000\right)m^2=\left( {5000\over3}\right)m^2$$

$$so, 2x×3x={5000\over3}⟺x^2={2500\over9}⟺x=\left({50\over3}\right)$$

$$∴ Length = 2x={100\over3}m=33{100\over3}m \ and \ Breadth =3x=\left(3×{50\over3} \right)m=50 \ m $$

**Q.2. If the diagonal of a rectangle is 17 cm long its perimeter is 46 cm, find the area of the rectangle.**

**Solution **

Let length = x and breadth = y. Then,

2(x+y)= 46 or x+y = 23 and x^{2}+y^{2}=(17)^{2}=289.

Now, (x+y)^{2} = (23)^{2}⟺(x^{2}+y^{2})+2xy = 529

⟺ 289+2xy=529⟺ xy = 120.

∴ Area = xy = 120 cm^{2}

**Q.3. In measuring the sides of a rectangle, one side is taken 5% in excess, and the other 4% in deficit. Find the length of the rectangle.**

Let x and y be the sides of the rectangle. Then, Correct area o xy.

$$Calculated \ area = \left({105\over100}x \right)×\left({96\over100}y \right)={504\over500}xy $$

$$ Error \ in \ measurement = \left({504\over500}xy \right)-xy={4\over500}xy$$

$$ ∴ Error \% =\left({4\over500}xy ×{1\over xy} ×100\right)\% = {4\over 5}\%=0.8 \% $$

**Q.4.The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares.**

**Solution.**

Side of firs square = (40/4) cm = 10 cm

Side of second square = (32/4) cm = 8 cm.

Area of third square = [(10)^{2} – (8)^{2}] cm^{2} = (100 – 64) cm^{2} = 36 cm^{2}

Side of third square = √36 cm = 6 cm.

∴ Required perimeter = ( 6 × 4) cm = 24 cm.

**Q.5. Find the area of a square, one of whose diagonals is 3.8 m long.**

**Solution **

$$Area \ of \ the \ square = {1\over2}×(diagonal)^2=\left({1\over2}×3.8×3.8 \right)m^2=7.22 \ m^2$$

**Q.6.One side of a rectangular field is 15 m and one of its diagonals is 17 m. Find the are of the field. **

**Solution**

$$Other\ side = \sqrt{(17)^2-(15)^2} \ =\sqrt{289-225} \ = \sqrt {64} \ = 8 \ m. $$

$$∴ Area=(15×8)m^2=120 \ m^2$$

**Q.7. Find the cost of carpeting a room 13 m long and 9 m broad with a carpet 75 cm wide at the rate of Rs. 12.40 per square metre.**

**Solution **

Area of the carpet = Area of the room = (13×9) m^{2} = 117 m^{2}.

$$Length \ of \ the \ carpet = \left({Area\over Width} \right)=\left(117×{4\over3} \right)m=156.$$

∴ Cost of carpeting = Rs.(156 × 12.40) = Rs. 1934.40.)

**Q.8.The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. Find the length of the rectangle.**

**Solution **

Let breadth = x. Then, length = 2x. Then,

(2x-5)(x+5) – 2x × x =75⟺ 5x – 25 = 75 ⟺ x = 20.

∴ Length of the rectangle = 20 cm.

**Q.9. A rectangular grassy plot 110 m by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. metre.**

**Solution **

Area of the plot = (110×65)m^{2}=7150m^{2}

Area of the plot excluding the path = [(110-5)× (65-5)]m^{2} = 6300 m^{2}

∴ Area of the path = (7150 – 6300) m^{2 }= 850 m^{2}

Cost of graveling the path $$ = Rs.\left(850× {80\over100} \right)=Rs.680. $$

**Q.10. A room 5m 55cm long 3m 74 cm broad is to be paved with square tiles. Find the least number of square this tiles required to cover the floor.**

**Solution **

Area of the room = (544 × 374) cm^{2}

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

Area of 1 tile = (34×34) cm^{2}

∴ Number of tiles required $$ = \left( {544×374\over34×34} \right)=176. $$

I hope these area problems with solutions are helpful for you. If you have any doubt or want to ask anything related area problems, ask me in the comment section.

All the best for your exam.