>> 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

31). What is the area of a square having perimeter 68 cm?
According to the question, 4a = 68 [a = side] a =\( \Large \frac{68}{4} \) = 17 cm Required area =\( \Large a^{2} \) =\( \Large (17)^{2} \)= 289 sq cm | ||||

32). The diagonal of a square is \( \Large \Large 4\sqrt{2} \) cm. The diagonal of another square whose area is double that of the first square is
Diagonal of square = \( \Large \sqrt{2} a \) [a = side] . \( \Large 4\sqrt{2} \)=\( \Large \sqrt{2} a \) a = 4 cm Now, area of square = \( \Large a^{2} \)=\( \Large (4)^{2} \)=16 Side of a square whose area is \( \Large 2\times 16 \) \( \Large a^{2}_{1} \)=32 => \( \Large a_{1}=\sqrt{32} \) => \( \Large a_{1}= 4\sqrt{2} \) Now, diagonal of new square = \( \Large \sqrt{2} a \)=\( \Large \sqrt{2}\times 4\sqrt{2} \)=8 cm | ||||

33). If the sides of a square is increased by 25%, then the area of the square will be increased by
Required increment | ||||

34). The perimeter of two squares is 12 cm and 24 cm. The area of the bigger square is how many times that of the Smaller?
We know that, Perimeter of square = \( \Large 4\times Side \) = \( \Large 4\times a \) = 12 => a = 3 Area of smaller square = \( \Large 3\times 3=9 cm^{2} \) ...(i) Now, \( \Large 4\times b \) = 24 => b = 6 Area of bigger square = \( \Large 6\times 6=36 cm^{2}=4\times 9 cm^{2} \) = \( \Large 4\times Area \) of smaller square [from Eq. (ii] Heance area of bigger square is 4 times that of smaller square | ||||

35). The perimeters of two squares are 68 cm and 60 cm. Find the perimeter of the third Square whose area is equal to the difference of the areas of these two squares.
\( \Large a_{1} \)=\( \Large \frac{68}{4} \)=17cm and \( \Large a_{2} \)=\( \Large \frac{60}{4} \)=15cm According to the question, Area of the third square = [\( \Large (17)^{2}-(15)^{2} \)] = (17 + 15) (17 - 15) = \( \Large 32\times 2 \) = 64 sq cm Let \( \Large a_{3} \) = Side of the third square According to the question. \( \Large (a_{3})^{2} \) = 64 sq cm \( \Large a_{3} \) = \( \Large \sqrt{64} \) = 8 cm Perimeter of the third square = \( \Large 4\times a_{3}=4\times 8 \)=32cm | ||||

36). In the figure, side of each square is 1 cm. The area, in sq cm of the shaded part is
Shaded part ABCD can be visualised as \( \Large \triangle ABD \) + \( \Large \triangle BCD \) | ||||

37). Diagonals of a rhombus are 1 m and 1.5 m in lengths. The area of the rhombus is
Area of rhombus = \( \Large \frac{1}{2}\times d_{1}\times d_{2} \) = \( \Large \frac{1}{2}\times 1\times 1.5=0.75 m^{2} \) | ||||

38). If the diagonals of a rhombus are 4.8 cm and 1.4 cm, then what is the perimeter of the rhombus?
Perimeter of rhombus | ||||

39). Angles of a quadrilateral are in the ratio 3: 4: 5: 8. The smallest angle is
Let First angle = 3x | ||||

40). Find the distance between the two parallel sides of a trapezium if the area of the trapezium is 500 sq m and the two parallel sides are equal to 30 m and 20 m, respectively.
According to the question, |