>> Elementary Mathematics >> Area and perimeter

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Contents:

- Elementary Mathematics
- Area and perimeter
- Circles
- Clocks
- Factorisation
- Geometry
- Height and Distance
- Indices and Surd
- LCM and HCF
- Loci and concurrency
- Logarithms
- Polynomials
- Quadratic Equations
- Quadrilateral and parallelogram
- Rational expression
- Real Analysis
- Rectangular and Cartesian products
- Set theory
- Simple and Decimal fraction
- Simplification
- Statistics
- Straight lines
- Triangle
- Trigonometric ratio
- Trigonometry
- Volume and surface area

21). The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is
Let the length of rectangle = L m Breadth of rectangle = B m using conditions from the question, L - B = 23 ....(i) 2(L + B) = 206 L + B = 103 ....(ii) On adding Eqs. (i) and (ii), we get 2L = 126 => L = 63 m => B = 103-63=40m Then, area of rectangle = \( \Large L\times B \) = \( \Large 63\times 40 \) = \( \Large 2520 m^{2} \) | ||||

22). If the length of a rectangle decreases by 5 m and breadth increases by 3 m, then its area reduces 9 sq m. If length and breadth of this rectangle increased by 3 m and 2 m respectively, then its area increased by 67 sq m. What is the length of rectangle?
Let length and breadth of a rectangle is x and y. Then, as per first condition, | ||||

23). The area of a rectangle whose length is 5 more than twice its width is 75 sq units. What is the perimeter of the rectangle?
Let the width of the rectangle = x units | ||||

24). Area of a rectangular field is \( \Large \Large 3584 m^{2} \) and the length and the breadth are in the ratio 7 : 2, respectively. What is the perimeter of the rectangle?
Let length of the rectangular field = 7x m and breadth of the rectangular field = 2x m According to the question, Area of rectangular field = \( \Large Length\times Breadth \) => 3584 = \( \Large 7x\times 2x \) => \( \Large 14x^{2} \) = 3584 => \( \Large x^{2} \) = \( \Large \frac{3584}{14} \) = 256 => \( \Large x^{2} \)=256 => x=16m Length of rectangular field = \( \Large 7\times 16 \) = 112 m and breadth of rectangular field = \( \Large 2\times 16 \)= 32 m Perimeter of rectangle = 2 (Length + Breadth) = 2(112+ 32)=\( \Large 2\times 144 \)=288m . | ||||

25). The length and perimeter of a rectangle are in the ratio of 5:18. What will be the ratio of its length and breadth?
According to the question, \( \Large \frac{l}{2(l+b)}=\frac{5}{18} \) => 10 l+ 10 b=18 l => 8 l=10 b => \( \Large \frac{l}{b}=\frac{10}{8}=\frac{5}{4} \) l : b = 5 : 4 Hence, ratio of length and breadth of a rectangle is 5 : 4. | ||||

26). A rectangle has 20 cm as its length and 200 sq cm as its area. If the area is increased by \( \Large \Large 1\frac{1}{5} \) times the original area by increasing its length only then the perimeter of the rectangle so formed (in cm) is
\( \Large l_{1} \) = 20 cm, \( \Large A_{1} \) = 200 sq cm | ||||

27). A ground \( \Large \Large 100\times 80 m^{2} \) has, two cross roads in its middle. The road parallel to the length is 5 m wide and the other road is 4 m wide, both roads are perpendicular to each other. The cost of laying the bricks at the rate of RS. 10 per \( \Large \Large m^{2} \) , on the roads, will be
Area to be paved with bricks | ||||

28). The breadth of a rectangle is 25 m. The total cost of putting a grass bed on this field was RS. 12375, at the rate of RS. 15 per sq m. What is the length of the rectangular field?
Area of the rectangular field | ||||

29). The length and the breadth of a rectangular plot are in the ratio of 5:3. The owner Spends RS. 3000 for surrounding it from all the sides at the rate of RS. 7.5 per metre. What is the difference between the length and breadth of the plot?
Perimeter of the field = \( \Large \frac{3000}{7.5} \) = 400 m | ||||

30). If side of a square is 20 cm, find its area.
Required area =\( \Large a^{2} \) = \( \Large (20)^{2} \) [a = side] = 400 sq cm |