Insecticides and Pesticides

Pesticides are chemical compounds which are used to control or kill the pets. This note contains an information about uses of insecticides and their usefulness and negative impact on nature.

Summary

Pesticides are chemical compounds which are used to control or kill the pets. This note contains an information about uses of insecticides and their usefulness and negative impact on nature.

Things to Remember

  • Pesticides are chemical compounds which are used to control or kill the pets. 
  • Insecticides are used for controlling insects.
  • Herbicides are used for destroying weeds.
  • Fungicides are used for killing fungi.
  • Miticides are used to destroy mites.

MCQs

No MCQs found.

Subjective Questions

Q1:

In the given diagram, PQRS is a square. If PO \(\parallel\) ST and SQ = \(3 \sqrt{2}\) cm then find the area of \( \Delta PQT. \) 

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Type: Short Difficulty: Easy

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Answer: <p>Solution:</p>
<p>SQ = \(3 \sqrt{2}\) cm<br />Let, PO = SP = x cm<br />In right angle \( \Delta PQS, \)</p>
<p>\begin{align*} SQ^2 \ &nbsp;&amp;= PQ^2 + SP^2\\ or,(3 \sqrt{2})^2 &amp;= x^2+ x^2\\ or, x^2 &amp;= \frac{18}{2}\\ or,x^2&amp;=9\\ \therefore x &amp;=3cm \end{align*}</p>
<p>\begin{align*}Area \: of \: \Delta PQS &amp;= \frac{1}{2}\times b \times h \\ &amp;= \frac{1}{2}\times 3 \times 3\ cm^2 \\&amp;=4.5\: cm^2 \end{align*}</p>
<p>\begin{align*}Area \: of \: \Delta PQT &amp;= Area \: of \: \Delta PQT = 4.5 \: cm^2 \end{align*}</p>
<p>&nbsp;</p>

Q2:

Find the area of the given quadrilateral PQRS where PA and RB are perpendicular to QS and 3RB = 2PA = QS = 6 cm

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Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p>
<p>QS = 6 cm, 2PA = 6cm \( \therefore PA = \frac{6}{2} = 3cm\)</p>
<p>3RB = 6cm \(\therefore RB = \frac{6}{3} = 2cm \)</p>
<p>Area of quad. PQRS = ?</p>
<p>\begin{align*} Area \: of \: PQRS &amp;= \frac{1}{2} QS (PA + RB)\\ &amp;=\frac{1}{2}\times 6 (3+2)\\ &amp;= 3 \times 5\\ &amp;= 15 \: cm^2 \: _{ans} \end{align*}</p>

Q3:

In the given figure, ABCD in a square. If a diagonal BD = \(3 \sqrt{2}cm\) find the area of ABCD.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>BD = \(3 \sqrt{2}cm\)<br>BC = CD = AD = AB [\(\therefore\) Sides of square]</p> <p>In right angled \(\Delta\)BCD</p> <p>\begin{align*} (\sqrt{BD} \ )^2 &amp;= (\sqrt{BC} \ )^2 +(\sqrt{CD} \ )^2\\ 3\sqrt{2} &amp;= (\sqrt{BC} \ )^2 +(\sqrt{CD} \ )^2\\ 18 &amp;= 2BC^2\\ BC^2 &amp;= \frac{18}{2} \\ &amp;= 9 \\ &amp;= 3^2\: cm \end{align*}</p> <p>Area of ABCD = 9 cm<sup>2</sup></p>

Q4:

PQRS is a quadrilateral in which PR = 10cm, perpendicular from S and Q on PR are 3.4 cm and 4.6cm respectively. Calculate the area of the quadrilateral.

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Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>PR = 10 cm<br>SM = 3.4 cm<br>QN = 4.6 cm</p> <p>We know,</p> <p>\begin{align*} Area \: of \: quad. PQRS &amp;= \frac{1}{2}PR (SM + QN)\\&amp;=\frac{1}{2}\times 10 (3.4+4.6)\\&amp;=5\times8\\&amp;=40 \:cm^2\:\:\: _{ans} \end{align*}</p>

Q5:

Find the area of the given quadrilateral where AM and CN are perpendicular to BD and 4AM = 2CN = BD = 12cm.

 

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Type: Short Difficulty: Easy

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Answer: <p>Solution:<br>BD = 12 cm<br>\(2CN = 12 cm\\ CN = \frac{12}{2} = 6\:cm \)<br>\(4AM = 12 cm\\ AM = \frac{12}{4} = 3\:cm \)<br>Area of quad. ABCD = ? <br>We know,<br>\begin{align*}Area\:of\:quad.ABCD &amp;= \frac{1}{2}BD(AM+CN)\\ &amp;=\frac{1}{2}\times12(3+6)\\&amp;= 6 \times 9 \\ 6 \times 9\\ &amp;= 54\:cm^2 \end{align*}</p>

Q6:

Calculate the area of the give quadrilateral.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:<br>BD = 8 cm<br>AN = 3 cm<br>CM = 6 cm<br>Area of quad. ABCD = ? <br>We know,<br>\begin{align*}Area\:of\:quad.ABCD &amp;= \frac{1}{2}BD(AN+CM)\\ &amp;=\frac{1}{2}\times8(3+6)\\&amp;= 4 \times 9 \\ &amp;= 36\:cm^2 \end{align*}</p>

Q7:

In the given figure, WXYZ is a square. If diagonal WZ = \(2 \sqrt{2}\:cm\), find the area of WXYZ.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Soution:</strong></p> <p>WZ =\(2 \sqrt{2}\:cm\)<br>WX = XY = YZ = WZ [\(\therefore\) Sides of squares ]</p> <p>In right angled \(\Delta \)WXY.</p> <p>\begin{align*} (WX)^2 + (XY)^2 &amp;=(WY)^2\\ (WX)^2 + (WX)^2 &amp;=( 2 \sqrt{2} )^2\\ 2(WX)^2&amp;=8\\(WY)^2&amp;=\frac{8}{2}\\WY &amp;=\sqrt{4}\\WY&amp;=2\end{align*}</p> <p>Now,</p> <p>\begin{align*}\text{Area of square WXYZ} &amp;= (WX)^2\\&amp;=(2)^2 \\&amp;=4\:cm^2 \:\:\:_{ans} \end{align*}</p> <p></p>

Q8:

Find the area of the given parallelogram.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>QR = 10cm<br>RT = 6cm<br>Area of parallelogram, PQRS = ?</p> <p>\begin{align*} PQRS &amp;= base \times height \\&amp;=QR \times RT \\&amp;=10\times 6 \\&amp;=60\:\:cm^2\:\:\:\:_{ans} \end{align*}</p>

Q9:

Calculate the area of the given quadrilateral PQRS.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p><strong>PM= 10cm<br>PS= 5.7<br>QR= 6.3cm<br>Area of quad. PQRS = ?</strong></p> <p><strong>\begin{align*} \text{Area of quad. PQRS} &amp;= \frac{1}{2}h(a+b)\\ &amp;= \frac{1}{2}\times 10(5.7+6.3)\\ &amp;= 5 \times 12\\ &amp;=60cm^2 \:\:\:\:_{}ans \end{align*}</strong></p>

Q10:

Find the base of a parallelogram whose area is \(56 cm^2\) and height to its base is 7 cm.


Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>Area of parallelogram = \(56\:cm^2\)<br>Height of parallelogram = 7 cm<br>Base of parallelogram = ?</p> <p>\begin{align*} \text{Area of parallelogram} &amp;= base \times height\\ 56&amp;=base \times 7\\base &amp;= \frac{56}{7}\\&amp;= 8 \: cm \end{align*}</p>

Q11:

In the given rhombus, PQ = 6 cm and PT = 4 cm, find the area of the rhombus PQRS.


Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>PQ = QR = RS = PS = 6cm \([\because sides\:of\: rhombus ]\)<br>PT = 4 cm<br>Area of rhombus PQRS = ?</p> <p>We know,</p> <p>\begin{align*} Area \: of \: Rhombus\: PQRS &amp;= base \times height\\ &amp;= QR \times PT\\ &amp;= 6 \times 4\\ &amp;= 24\:cm^2 \:\: _{Ans} \end{align*}</p>

Q12:

Adjoining the figure MNRP, is a quadrilateral in which MN\(\parallel \) PR. \(\angle MNR=\angle PRN = 90 º\), NR = 8 cm, and PM = 10 cm, If the area of the quadrilateral MNRP is 72 sq. cm. Calculate the length of MN.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>NR = 8 cm<br>PM = 10 cm<br>MN = ?<br>Area of the MNRP = 72 \(cm^2\)<br>MA= NR = 8 cm<br>In the right angled triangle PMA,</p> <p>\begin{align*} PA &amp;= \sqrt{PM^2 -MA^2}\\&amp;= \sqrt{10^2 - 8^2}\\ &amp;=\sqrt{100-64}\\ &amp;= \sqrt{36}\\ &amp;= 6 cm \end{align*}</p> <p>\begin{align*} \text{Area of quadrilateral MNRP} &amp;= \frac{1}{2}h(a+b)\\ 72 &amp;= \frac{1}{2}\times MA (MN + PR)\\ or, 72 &amp;= \frac{1}{2}\times 8 (MN + PA+AR)\\ or, 72 &amp;= 4(MN + 6 + MN)\\ or, \frac{72}{4}&amp;= 2MN +6 \\ or, 18-6&amp;=2MN\\ MN &amp;= \frac{12}{2}\\ \therefore MN &amp;= 6 cm\:\; _{ans} \end{align*}</p> <p></p> <p></p> <p></p>

Q13:

In the adjoining figure, PQRS is a rhombus. If PR = 4 cm and QS = 6 cm, find the area of PQRS.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>PR = 4cm<br>QS = 6 cm<br>Area of rhombus PQRS = ?</p> <p>\begin{align*} \text{Area of rhombus PQRS} &amp;= \frac{1}{2}d_1 \times d_2 \\ &amp;= \frac{1}{2}\times PR \times QS\\ &amp;= \frac{1}{2} \times 4 \times 6\\ &amp;= 12 \: cm^2 \end{align*}</p>

Q14:

Area of a rhombus and its diagonal are 24 sq. cm and 6 cm respectively. Find the length of its another diagonal.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>Area of rhombus = 24 \(cm^2\)<br>Length of the diagonal \( d_1= 6 \: m\)<br>Length of other diagnal \(d_2 = ? \)<br>We know, <br>\begin{align*} Area \: of\: rhombus &amp;= \frac{1}{2} \times d_1 \times d_2\\ or, 24 &amp;= \frac{1}{2} \times 6 \times d_2 \\ or, 24 &amp;= 3 \times d_2\\ or, d_2 &amp;= \frac{24}{3}\\ \therefore d_2 &amp;= 8\: cm \end{align*}</p>

Q15:

M is the Mid point of the LN of \(\Delta KLN\) . If the area of \(\Delta KLN \: is 30 \: cm^2 ,\) what will be the area of \(\Delta KML\)

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>Area of the \(\Delta KMN = 30\: cm^2\)<br>\(Area \: of\: the\: \Delta KMV = \frac {1}{2} \Delta KLN \\ [\because Median \: KM\: bisect\: the \: \Delta KLM ]\\ Area \: of \: \Delta KMN = \frac{1}{2}\times 30 = 15 \: cm^2 \:\:_{Ans} \)</p>

Q16:

In the given figure, ABCD is parallelogram in which E is the mid point of DC. If the area of \(\Delta ADE \) is 5 sq. cm. Find the area of the quadrilateral ABCE.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>\( Area \: of \: \Delta ADE = 5 \: cm^2 \\ Area\: of\: \Delta ACE = Area \: of \: \Delta ADE = 5 \: cm^2 [\because The \: median \: AE \: bisect\: the \: \Delta ACD]\\ Area \: of \: \Delta ABC = Area \: of\: \Delta ACD \: [\because \text{The diagonal AC bisect the parallelogram ABCD}] \\ Area \: of \: \Delta ABC = \Delta ADE + \Delta ACE = 5+5 = 10\: cm^2 \\ The \: quad. ABCE = \Delta ABC + \Delta ACE = 10+ 5 = 15 \: cm^2 \:\: _{Ans.} \)</p>

Q17:

Calculate the area of the given quadrilateral.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>AB = 12 cm<br>CD = 21 cm<br>MC = 15 cm<br>AD = 10 cm<br>\begin{align*} Here , DM &amp;= CD - MC \\&amp;= 21cm - 15cm \\ &amp;= 6 cm \end{align*}</p> <p>\(In \: Right \: Angled \: \Delta ADM, \)</p> <p>\begin{align*} AM &amp;= \sqrt{(AD)^2 - (DM^2)} \\ &amp;= \sqrt{10^2 - 6^2 }\\ &amp;= \sqrt{100 - 36}\\ &amp;= \sqrt{64} \\ &amp;= 8 cm \end{align*}</p> <p>\begin{align*} Area \: of \: quadrilateral\:\: ABCD &amp;= \frac{1}{2} \times h (a+b ) \\ &amp;= \frac{1}{2} AM\:(AB +CD)\\ &amp;= \frac{1}{2} \times 8(12+21)\\ &amp;= 4 \times 33 \\ &amp;= 132\:cm^2 \:\: _{Ans} \end{align*}</p>

Q18:

Find the area of Trapezium ABCE

 

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>DE = 12 cm<br>CE = 13 cm<br>AB = BC = CD = AD [\(\because\) Sides of square]</p> <p>In right angled \(\Delta CDE, \)</p> <p>\begin{align*} CD &amp;= \sqrt {(CE)^2 - (DE)^2} \\ &amp;= \sqrt{13^2 -12^2 }\\ &amp;= \sqrt{169-144}\\ &amp;= \sqrt{25} \\&amp;=5 \:cm \\ \\ AE &amp;= AD+DE\\ &amp;= 5 +12\\ &amp;= 17 \: cm \\ BC &amp;= 5 cm \end{align*}</p> <p>\begin{align*} Area \: of\: Trap. ABCE &amp;= \frac{1}{2} h(a+b)\\&amp;= \frac{1}{2} \times CD \:(AE + BC ) \\ &amp;= \frac{1}{2} \times 5 (17 + 5)\\ &amp;= 2.5 \times 22 \\ &amp;= 55 \: cm^2 \: _{Ans} \end{align*}</p>

Q19:

In the given figure, ABCD is a square and EBCF is a parallelogram. If AB = 4 cm, Calculate the area of the parallelogram EBCF.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>AB = 4 cm<br>Area of EBCF = ?</p> <p>\begin{align*} Area \: of \: square ABCD &amp;= (AB)^2 \\ &amp;= 4^2 \\ &amp;= 16 \: cm^2 \\ \\ Area \: of \: parallelogram \: EBCF &amp;= Area \: of \: square \: ABCD \\ &amp;= 16 \: cm^2 \:\: [\because \text{Standing n same base and between same parallel lines}] \end{align*}</p>

Q20:

In the given figure, PQRS is a parallelogram in which T is the middle of PS. If the area of the \(\Delta PTQ \: is \: 6 \:cm^2,\) what will be the area of the quadrilateral QTSR.

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Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>Area of \(\Delta PTQ = 6 \: cm^2\)<br>Area of quadrilateral QTSR = ?</p> <p>Area of \(\Delta QTS \) =Area of \(\Delta PTQ = 6 \: cm^2[\because \text{ Median line QT bisect the} \: \Delta PQS]\)<br>\begin{align*}Area \: of\: \Delta QRS &amp;= Area \:of \: \Delta PQT +Area \:of\: \Delta QTS \\&amp;= 6 + 6 \\ &amp;= 12 \: cm ^2 \end{align*}</p> <p>\begin{align*}Area \: of \: quad \: QTSR &amp;= Area \: of \: \Delta QTS + Area\: of\: \Delta QSR\\ &amp;= 6 + 12 \\&amp;= 18 \: cm^2 \end{align*}</p>

Q21:

In the given figure, QR = RS and PM = RM. If area of \(\Delta MQR = 6.5 \: cm^2 .\) Find the area of \(\Delta PQS.\) 

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Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>Area of \(\Delta MQR = 6.5 \: cm^2\)<br>Area of \(\Delta PQM = Area\: of\: \Delta MQR = 6.5 \: cm^2 \:\:[\because \text{Median line QM bisect the } \Delta PQR]\)<br>\begin{align*} Area \: of \: \Delta PRS &amp;= Area \: of \: \Delta PRS\\ &amp;= 6.5+6.5 \\ &amp;= 13 \:cm^2\\ Area \: of \: \Delta PQS &amp;= Area \: of \: \Delta PQR + area \: of\: \Delta PRS \\&amp;= 13 + 13 \\ &amp;= 26 \:\: cm^2 \:\:\: _{Ans.} \end{align*}</p>

Q22:

Find the area of the quadrilateral PQRS given in the adjoining figure in which 3RB = 2PA = QS = 6 cm.

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Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>QS = 6 cm<br>\(2PA = 6 cm\\ PA = \frac{6}{2}\\ \therefore PA = 3 cm\)<br>\(3RB = 6 cm\\ PA = \frac{6}{3}\\ \therefore PA = 2 cm\)</p> <p>\begin{align*} Area \: of\: quadrilateral \: PQRS &amp;= \frac{1}{2} QS (PA + RB) \\ &amp;= \frac{1}{2} \times 6 (3 + 2 )\\ &amp;= 3 \times 5 \\ &amp;= 15 \: cm^2 \:\:\:\: _{Ans} \end{align*}</p>

Q23:

In the adjoining figure, find the area of quadrilateral BMNC.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p>MN = 10 cm<br>NT = 8 cm<br>MN \(\parallel\) BC and BC = 2MN \([\because \text{ M and N are midpoints of sides AB and AC}]\)<br>\(BC = 2 \times 10 = 20\:cm \)</p> <p>\begin{align*} Area \: of \: quad. \: BMNC &amp;= \frac{1}{2} h (a+b)\\ &amp;= \frac{1}{2} \times NT (MN + BC)\\&amp;= \frac{1}{2} \times 8 (10 + 20)\\ &amp;= 4 \times 30 \\ &amp;= 120 \: cm^2 \:\:\: _{Ans} \end{align*}</p>

Q24:

Find the area of quadrilateral ABCD in which BD = 8 cm, AN = 6 cm and CM = 3 cm.

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Type: Short Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p>BD = 8 cm<br>AN = 6 cm<br>CM = 3 cm<br>Area of ABCD = ?</p> <p>\begin{align*} Area \: of\: ABCD &amp;=\frac{1}{2} BD (AN + CM)\\ &amp;= \frac{1}{2} \times 8 (6 + 3) \\ &amp;= 4 \times 9 \\ &amp;= 36 \: cm^2 \end{align*}</p>

Q25:

Prove that area of a triangle is equal to one half of the area of a parallelogram on the same base and between the same parallels.

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Type: Long Difficulty: Easy

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Answer: <p><strong>Solution:</strong></p> <p><strong>Given:</strong> \(\Delta ABC \) and parm. EBCF are on the same base and between the same parallels.</p> <p><strong>To prove:</strong>\(\Delta ABC = \frac{1}{2} \: parm. \:EBCF \)</p> <p><strong>Construction:</strong> Draw CD parallel to BA and to meet AF at D</p> <table width="454"><tbody><tr><td><strong>S.N</strong></td> <td><strong>Statements</strong></td> <td><strong>Reasons</strong></td> </tr><tr><td>1</td> <td>ABCD is a parallelogram.</td> <td>By construction and given.</td> </tr><tr><td>2</td> <td>\(\Delta ABC = \frac{1}{2}\)parm ABCD</td> <td>Diagonal AC bisects parm ABCD.</td> </tr><tr><td>3</td> <td>Parm ABCD = Parm EBCF</td> <td>They stand on the same base BC and between the same parallels AF and BC.</td> </tr><tr><td>4</td> <td>\(\Delta ABC = \frac{1}{2} parm. EBCF \)</td> <td>From statement (2) and (3)</td> </tr></tbody></table><p>\( \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: _{Proved}\)</p>

Q26:

Prove that the parallelograms on the same base and between the same parallels are equal in area.

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Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p><strong>Solution:</strong></p> <p><strong>Given:</strong> Parallelograms ABCD and EBCF on the same base BC and between the same parallels AF and BC.<br><strong>To prove:</strong> Parm. ABCD = Parm. EBCF<br><strong>Proof</strong></p> <table width="571"><tbody><tr><td><strong>S.N</strong></td> <td><strong>Statements</strong></td> <td><strong>Reasons</strong></td> </tr><tr><td>1</td> <td>In \(\Delta ABE \:and\: \Delta CDF \\ (i) \angle BAE = \angle CDF\\(ii) \angle BEA = \angle CFD\\ (iii) AB = DC \)</td> <td>\(\\ (i) Corresponding \:angles\\Corresponding\:angles\\ (iii) Opposite \: sides \: of \: para \: ABCD \)</td> </tr><tr><td>2</td> <td>\(\Delta ABE&cong; \Delta CDF\)</td> <td>by A.A.S axiom</td> </tr><tr><td>3</td> <td>\(\Delta ABE = \Delta CDF\)</td> <td>Area of congruent triangles</td> </tr><tr><td>4</td> <td>\(Trapezium\: ABCF - \Delta CDF = Trapezium\: ABCF - \Delta ABE \)</td> <td>Subtracting equal triangles from same figure.</td> </tr><tr><td>5</td> <td>Parm. ABCD = Parm. EBCF</td> <td>From Statement (4)</td> </tr></tbody></table>

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Insecticides and Pesticides

Insecticides and Pesticides

Pesticides

Pesticides are chemical compounds which are used to control or kill the pets. Depending upon the types of organism, pesticides are classified as follows:

  1. Insecticides: These pesticides are used for controlling insects.

  2. Herbicides:These pesticides are used for destroying weeds.
  3. Fungicides: These pesticides are used for killing fungi.
  4. Miticides: They destroy mites.

Insecticides:

The poisonous chemicals which are made by humans and used to control or kill the insects are called insecticides.

  1. Organic insecticides: Those insecticides which are obtained from the living source are called organic insecticides. Mainly, there are three types of organic insecticides. They are:
    1. Organochloride

      Example: DDT (Dichloro Diphenyl Trichloroethane)

      BHC ( Benzene Hexachloride)

    2. Organophosphorous

      Example: Malathion, Parathion

    3. Carbonate insecticides
      • Baygon
      • Temik
      • Dinetan
  2. Inorganic insecticides

    Those insecticides derived from minerals are called inorganic insecticides.

    Example: calcium arsenate, lead arsenate

Advantages of using insecticides

  1. They kill or destroy harmful insects.
  2. They can destroy all the stages of the lifecycle of harmful insects. As a result, there is increase in crop yield.
  3. They help to control several diseases by killing germs.

Disadvantages of using insecticides

  1. Most of the insecticides are non biodegradable. Hence, they are harmful to insects and plants.
  2. Insecticides leave harmful deposits on food crops. The use of such food crops has adverse effects on the health of human beings and all other animals.
  3. They can pollute lakes, ponds, streams, etc.
  4. They destroy useful insects also.
  5. The spray of insecticides can affect plants and animals too.

Precautions in using insecticides:

  1. They should be kept away from the reach of children and ignorant persons
  2. The name of the insecticides should be labelled clearly.
  3. While applying it, the mouth, nose and eyes should be well covered.
  4. It should be used in little amount.
  5. We should be careful while using it.
  6. After spraying or dusting it, hands and other body parts must be properly cleaned.

Lesson

Materials Used In Daily Life

Subject

Science

Grade

Grade 10

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