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Answer:
1. Fungi cause decay by releasing enzymes onto the dead animal or plant. These break down complex compounds into simple soluble ones that can be absorbed by decomposers. Organisms that feed on dead material in this way are called saprophytes.
Explanation:
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18 4,224 (a) 1113 (b) 12 13 C (d) 112 1 1. What is 384 x 11? ( (a) 3,224 (d) 3,234 (c) 4,234 2. If a dozen eggs cost * 48, then the cost of 20 eggs is (a) 80 (b) 375 (c) ₹ 82 (d) ₹ 70 3. 614 = 15 gives: (a) Q = 4, R = 4 de = 4, R = 14 (c) Q = 40, R = 4 Q = R (d) Q = 40, R = 14 4. The value of 90 - 18 X 16+ 20 - 75 is: (a) 80 (b) 50 (c) 25 (d) 15 5. Which of the following is not a multiple of 7? (a) 35 (b) 43 (c) 77 (d) 84 6. Which of the following is not a prime number? (a) 17 (b) 19 (c) 39 (d) 41 7. The improper fraction 85 can be written in mixed fraction as: 7 7 (c) 12. ) 8. The value of 3 12 / is: 8 17 (b) (c) 19 40 5 (d) 5 40 40 40 II. Solve mentally and answer. 1. 46 X 132 = B2 x 46 354 2. 13,085 X0 = 3. 7,98,000 - 100 = _798000 4. 999 + 0 is 0. (Irde/False) 5. The smallest multiple of a number is 6. A number can have limited 6number of factors. (True/False) 7. The prime factorization of 24 is 8. x 72 = 9 9. Anupam had 12 pieces of chocolate. He ate 7 of them. What fraction of pieces of chocolate did he not eat? 10. Put the correct sign >, < or =. 3 11 (b) 4.5 33 001w (a) 37 으 b) O 7 118True or false: a substance that brings about reduction or reduces another substance is called a reducing agent.Question :-
The area of a square is equal to the area of the rectangle. If the side of the square is 40 cm and the breadth of the rectangle is 25 cm, find the length of the rectangle and also find the perimeter of the rectangle .
\large\underline{\sf{Solution-}}
Solution−
Given that,
Side of the square, a = 40 cm
Breadth of rectangle, b = 25 cm
Let assume that Length of rectangle = L cm
Further given that,
\begin{gathered} \red{\rm \: Area_{(Square)} \: = \: Area_{(Rectangle)} \: } \\ \end{gathered}
Area
(Square)
=Area
(Rectangle)
\begin{gathered}\rm \: {a}^{2} = L\times b \\ \end{gathered}
a
2
=L×b
On substituting the values, we get
\begin{gathered}\rm \: {(40)}^{2} = 25 \times L \\ \end{gathered}
(40)
2
=25×L
\rm \: L \: = \: \dfrac{40 \times 40}{25}L=
25
40×40
\begin{gathered}\bf\implies \: L \: = \: 64 \: cm \\ \\ \end{gathered}
⟹L=64cm
Now, we have to find the perimeter of rectangle.
So,
\begin{gathered} \red{ \bf \: Perimeter_{(Rectangle)}} \\ \end{gathered}
Perimeter
(Rectangle)
\begin{gathered}\rm \: = \: 2 \times (L \: + \: b) \\ \end{gathered}
=2×(L+b)
\begin{gathered}\rm \: = \: 2 \times (64 \: + \: 25) \\ \end{gathered}
=2×(64+25)
\begin{gathered}\rm \: = \: 2 \times 89 \\ \end{gathered}
=2×89
\begin{gathered}\rm \: = \: 178 \: cm \\ \end{gathered}
=178cm
Hence,
\begin{gathered} \red{ \rm\implies \:\bf \: Perimeter_{(Rectangle)} \: = \: 178 \: cm} \\ \\ \end{gathered}
⟹Perimeter
(Rectangle)
=178cm
\begin{gathered}\rule{190pt}{2pt} \\ \end{gathered}
{ \red{ \mathfrak{Additional\:Information}}}AdditionalInformation
\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}
†
FormulasofAreas:−
⋆Square=(side)
2
⋆Rectangle=Length×Breadth
⋆Triangle=
2
1
×Base×Height
⋆Scalene△=
s(s−a)(s−b)(s−c)
⋆Rhombus=
2
1
×d
1
×d
2
⋆Rhombus=
2
1
d
4a
2
−d
2
⋆Parallelogram=Base×Height
⋆Trapezium=
2
1
(a+b)×Height
⋆EquilateralTriangle=
4
3
(side)
2