8th Standard Syllabus & Materials
8th Standard
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NEW8th Standard
Tamilnadu 8th Standard роХрогро┐родроорпН роОрогрпНроХро│рпН Important Questions And Answers Study Material - QB365 Set B
NEW8th Standard
Tamilnadu 8th Standard роХрогро┐родроорпН роОрогрпНроХро│рпН Important Questions And Answers Study Material - QB365 Set A
NEW8th Standard
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NEW8th Standard
Tamilnadu 8th Standard Social Science роХрпБроЯро┐роорпИропро┐ропро▓рпН - роирпАродро┐родрпНродрпБро▒рпИ Important Questions And Answers Study Material - QB365
NEW8th Standard
Tamilnadu 8th Standard Social Science роХрпБроЯро┐роорпИропро┐ропро▓рпН - рокро╛родрпБроХро╛рокрпНрокрпБ рооро▒рпНро▒рпБроорпН ро╡рпЖро│ро┐ропрпБро▒ро╡рпБроХрпН роХрпКро│рпНроХрпИ Important Questions And Answers Study Material - QB365

Published on: 13/05/2022
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Questions + Answers key
Take MCQ Maths Test1.
In ΔPAT, the bisector of ∠P, meets AT at S. If ∠APT = 70o and ∠ASP = 100o, find ∠A and ∠T.
2.
(Illustrating RHS Congruence)
If TAP is an isosceles triangle with TA = TP and ∠TSA = 90°.
Is AS = PS? Why?
3.
(Illustrating SAS Congruence)
If CW and CT trisect OA and CO ≡ CA, prove that Δ COW ≡ Δ CAT

4.
(Illustrating SSS and SAS Congruence)
If ∠E =∠S and G is the midpoint of ES, prove that Δ GET ≡ Δ GST.
5.
(Illustrating AA Similarity)
In the given Fir, if ∠1 ≡ ∠3 and ∠2 ≡∠4 then, prove that ΔBIG я╜Ю ΔFAT . Also find FA.
1.
Now, PS is the bisector of ∠P
∴ ∠APS = ∠TPS = 35o
From ΔAPS,
∠A+∠APS+∠ASP=180o (angle sum property in Δ APS)
∠A + 35 + 100o = 180o
⇒∠A = 180o−135o = 45o
From ΔTPS,
∠T + ∠TPS + ∠TSP = 180o (angle sum property in Δ TPS)
⇒∠T + 35o + 80o = 180o
⇒∠T = 180o − 115 = 65o
2.
∠TSA = 90°
∠P = ∠A as TA = TP
∴ ∠ATS = ∠PTS
AS = PS (if sides then angles )
3.
Proof:
| Statements | Reasons | |
|---|---|---|
| 1 | CO ≡ CA | given |
| 2 | ∠A ≡ ∠O | if sides, then angles |
| 3 | CW and CT trisect OA | given |
| 4 | OW ≡ AT ≡ WT | trisection definition |
| 5 | Δ COW ≡ Δ CAT | by SAS (1,2,4) |
4.
Proof:
| Statements | Reasons | |
|---|---|---|
| 1 | ∠E ≡ ∠S | given |
| 2 | ET ≡ ST | if angles, then sides |
| 3 | G is the midpoint of ES | given |
| 4 | EG ≡ SG | follows from 3 |
| 5 | TG ≡ TG | reflexive property |
| 6 | Δ GET ≡ Δ GST | by SSS (2,4,5) & also by SAS (2,1,4) |
5.
Proof:
| Statements | Reasons | |
|---|---|---|
| 1 | ∠1 ≡ ∠3 | given |
| 2 | ∠IBG ≡ ∠AFT | supplements of congruent angles are congruent |
| 3 | ∠2 ≡ ∠4 | given |
| 4 | ∠IGB ≡ ∠ATF | supplement of congruent angles are congruent |
| 5 | ΔBIGя╜ЮΔFAT | by AA property (2, 4) |
Also, their corresponding sides are proportional
⇒ \(\frac { BI }{ FA } =\frac { BG }{ FT } \Rightarrow \frac { 10 }{ FA } =\frac { 5 }{ 8 } \)
⇒ FA = \(\frac { 10\times 8 }{ 5 } =\frac { 80 }{ 5 } \) = 16 cm
8th Standard Syllabus & Materials
8th Standard
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NEW8th Standard
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NEW8th Standard
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NEW8th Standard
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Tamilnadu Stateboard 8th Standard Subjects
Tamilnadu Stateboard Standards