>> 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

11). The ratio of length of each equal side and the third side of an isosceles triangle is 3:4. If the area of the triangle is \( \Large \Large 18\sqrt{5} \) sq units, the third side is
Let sides of isosceles triangle are 3x,3x and 4x. | ||||

12). Three sides of a triangular field are of length 15 m, 20 m and 25 m long, respectively. Find the cost of sowing seeds in the field at the rate of RS. 5 per Sq m.
Since,\( \Large AC^{2}=AB^{2}+BC^{2} \) => \( \Large (25)^{2}=(15)^{2}+(20)^{2} \) | ||||

13). A \( \Large \Large \triangle DEF \) is formed by joining the mid points of the sides of \( \Large \Large \triangle ABC\). Similarly, a \( \Large \Large \triangle PQR \) is formed by joining the mid-points of the sides of the \( \Large \Large \triangle DEF \). If the sides of the PQR are of lengths 1, 2 and 3 units, what is the perimeter of the \( \Large \Large \triangle ABC \)?
Perimeter of \( \Large \triangle PQR \) = 1 + 2 + 3 = 6 units
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14). Find the area of a rectangle having 15 m length and 8 m breadth.
Requnred area = \( \Large Length\times Breadth \) = \( \Large 15\times 8 \)=120 sq m | ||||

15). The length of a rectangular field is 100 m and its breadth is 40 m. What will be the area of the field?
Requnred area = \( \Large Length\times Breadth \) = \( \Large 100\times 40 \)=4000 sq m = \( \Large (4\times 10^{3}) \) sq m | ||||

16). The area of a rectangular playground is 300 sq m. If the breadth of the field is 15 m, find the length of the field.
According to the question. \( \Large L\times B \)=Area => \( \Large L\times 15 \)=300 [L = length and B = breadth] L= \( \Large \frac{300}{15} \)=20m | ||||

17). The ratio of length and breadth of a rectangle is 5 : 3. If length is 8 m more than breadth, find the area of the rectangle.
Let the length of rectangle = 5x and breadth of triangle = 3x According to the question, | ||||

18). The perimeter of a rectangle having area equal to \( \Large \Large 144 cm^{2} \) and sides in the ratio 4 : 9 is
Let l=4x and b = 9x Area of rectangle = \( \Large l\times b \) 144=\( \Large 4x\times 9x \)=> \( \Large x^{2}=\frac{144}{36} \) => \( \Large x^{2} \)=4 x=2 l=8 cm and b=18 cm Perimeter of rectangle = 2 (l + b) = 2 (8 + 18) = \( \Large 2\times 26 \) = 52 cm | ||||

19). The area of a rectangle lies between \( \Large \Large 40 cm^{2} \) and \( \Large \Large 45 cm^{2} \). If one of the sides is 5 cm, then its diagonal lies between
Area of rectangle lies between \( \Large 40 cm^{2}\) and \( \Large 45 cm^{2}\). Now, one side = 5 cm Since, area cannot be less than \( \Large 40 cm^{2}\) Hence, other side cannot be less than = \( \Large \frac{40}{5} \) = 8 cm Since, area cannot be greater than \( \Large 45 cm^{2}\) Hence, other side cannot be greater than = \( \Large \frac{45}{5} \) = 9 cm Minimum value of diagonal = \( \Large \sqrt{8^{2}+5^{2}}=\sqrt{89} \) =9.43cm Maximum value of diagonal = \( \Large \sqrt{9^{2}+5^{2}}=\sqrt{106} \) = 10.3cm | ||||

20). The area of a rectangular field is 15 times the sum of its length and breadth. If the length of that field is 40 m, what is the breadth of that field?
Length of rectangle = 40 m Let breadth of = x Then, according of the question, (40 + x)15=\( \Large 40\times x \) => 600+ 15x=40x => 25x=600 x=24 m |