8th Standard Syllabus & Materials
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NEW8th Standard
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NEW8th Standard
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NEW8th Standard
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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Maths Test1.
In the given fi gure, if SW ≡ SE and ∠NWO≡∠NEO . then, prove that NS bisects WE and ∠NOW = 90°
2.
In the given figure, UB || AT and CU ≡ CB Prove that ΔCUBя╜ЮΔCAT and hence ΔCAT is isosceles
3.
In the given figure, if ΔEATя╜ЮΔBUN, find the measure of all angles.
4.
In the given fi gure prove that ΔGUM я╜Ю ΔBOX
5.
From the given figure, prove that ΔABCя╜ЮΔDEF
1.
| S.No. | Statements | Reasons |
| 1. | SW \(\equiv \\ \) SE ....(1) | Given |
| 2. | NW = NE ....(2) | тИ╡ \(\angle \)NWO \(\equiv \\ \) \(\angle \)NEO given opposite sides of equal angles. |
| 3. | PQ = PQ | Common side of \(\triangle \)NWS and \(\triangle \)NES |
| 4. | \(\triangle \)NWS \(\equiv \\ \) \(\triangle \)NES | SSS criteria |
| 5. | \(\angle \)ONW \(\equiv \\ \) \(\angle \)LONE | By (4) |
| 6. | PO \(\equiv \\ \) PO \(\triangle \)NWO \(\equiv \\ \) \(\triangle \)NEO |
Common side 4, 5, 6 using SAS criteria |
| 7. | WO \(\equiv \\ \) EO | By CPCTC |
| 8. | \(\angle \)NOW \(\equiv \\ \) \(\angle \)NOB | By CPCTC |
| 9. | \(\angle \)WON +\(\angle \)EON = 180o | тИ╡ AOC is line |
| 10. | \(\angle \)WON = \(\angle \)EON = 90o | By 8 |
| 11. | Also WO \(\equiv \\ \) EO | By 7 |
| ∴ | NS bisects WE and \(\angle \)NOW = 90o | By 10 and 11 |
2.
\(\text { In } \triangle \mathrm{CUB} \text { and } \Delta \mathrm{CAT}\)
CU = CB (given)
\(\angle \mathrm{CUB}=\angle \mathrm{CBU}\)
\(
\mathrm{UB} \| \mathrm{AT}
\)
\(\angle \mathrm{CUB} =\angle \mathrm{CAT}
\)
\(\angle \mathrm{CBU} =\angle \mathrm{CTA}
\)
\(\angle \mathrm{C} =\angle \mathrm{C}
\)
\(\therefore \Delta \mathrm{CUB} \sim \Delta \mathrm{CAT}
\)
Also \(\triangle\) CAT is isosceles.
Hence proved.
3.
\(Given \triangle \mathrm{EAT} \sim \triangle \mathrm{BUN}
\)
\(\therefore \angle \mathrm{E}=\angle \mathrm{B}, \angle \mathrm{A}=\angle \mathrm{U}, \angle \mathrm{T}=\angle \mathrm{N}\)
Sum of the angles of a triangle is 180o
\(\angle \mathrm{E}+\angle \mathrm{A}+\angle \mathrm{T}=180^{\circ}\)
x + 2x + x + 40 = 180o
4x = 180o - 40o = 140o
\(
x =\frac{140^{\circ}}{4}=35^{\circ}
\)
\(x\therefore \angle \mathrm{E} =\angle \mathrm{B}=x=35^{\circ}
\)
\(\angle \mathrm{A} =\angle \mathrm{U}=2 x
\)
\(=2 \times 35^{\circ}=70^{\circ}
\)
\(\angle \mathrm{T} =\angle \mathrm{N}=x+40^{\circ}
\)
\(=35^{\circ}+40^{\circ}=75^{\circ}
\)
4.
\(\frac { GU }{ BO} \)= \(\frac { 32 }{ 8 } \) = \(\frac { 4 }{ 1 } \)
\(\frac { UM }{ OX} \)= \(\frac { 48 }{ 12 } \)= \(\frac { 4 }{ 1 } \)
\(\frac { GM }{ BX} \)= \(\frac {52 }{ 13 } \)=\(\frac { 4 }{ 1 } \)
We find that \(\frac { GU }{ BO} \)= \(\frac { UM }{ OX} \)= \(\frac { GM }{ BX} \)= \(\frac { 4 }{ 1 } \)
That is their corresponding sides are proportional.
:. By SSS similarity \(\triangle \)GUM ~ \(\triangle \)BOX.
5.
From the \(\triangle \)ABC, AB = AC
It is an isosceles triangle
Angles opposite to equal sides are equal
∴ \(\angle \)B =\(\angle \)C = 65o
∴ \(\angle \)B +\(\angle \)C = 65o + 65o
= 130o
We know that sum of three angles is a triangle = 180o
\(\angle \)A+\(\angle \)B+\(\angle \)C = 180o
\(\angle \)A + 130o = 180o
\(\angle \)A = 180o -130o
\(\angle \)A = 50o
From \(\triangle \)DEF, \(\angle \)D = 50o
∴ Sum of Remaining angles 180o - 50o = 130o
DE = FD
∴ \(\angle \)D = \(\angle \)F
From \(\triangle \)ABC and \(\triangle \)DEF∴ \(\angle \)E = \(\frac { 130 }{ 2 } \)= 65o
\(\angle \)A = \(\angle \)D = 50o
\(\angle \)B =\(\angle \)E = 65o
\(\angle \)C =\(\angle \)F = 65o
∴ By AAA criteria \(\triangle \)DEF ~ \(\triangle \)ABC
8th Standard Syllabus & Materials
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Tamilnadu Stateboard 8th Standard Subjects
Tamilnadu Stateboard Standards