8th Standard Syllabus & Materials
8th Standard
TN 8th Tamil இயல் 3 - கல்வி கரையில - வினைமுற்று Important Questions And Answers Study Material - QB365 Set A
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TN 8th Tamil இயல் 2 - ஈடில்லா இயற்கை - மயங்கொலிகள் Important Questions And Answers Study Material - QB365 Set A
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TN 8th Tamil இயல் 2 - ஈடில்லா இயற்கை - இயற்கையை போற்றுவோம் Important Questions And Answers Study Material - QB365 Set A
NEW8th Standard
TN 8th Tamil இயல் 1 - தமிழ் இன்பம் - ஆழிக்கு இணை Important Questions And Answers Study Material - QB365 Set A
NEW8th Standard
TN 8th Tamil இயல் 2-ஈடில்லா இயற்கை - திருக்குறள் Important Questions And Answers Study Material - QB365 Set A
Published on: 15/06/2021
QB365 provides detailed and simple solution for every book back questions in class 8 Maths subject.It will helps to get more idea about question pattern in every book back questions with solution.
Download Tamil Nadu 8th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Maths Test1.
(Illustrating RHS similarity)
The height of a man and his shadow form a triangle similar to that formed by a nearby tree and its shadow. What is the height of the tree?
2.
In Fig, if ΔPEN ～ ΔPAD, then find x and y.
3.
In the Fig if ΔPQR ～ ΔXYZ, find a and b.
4.
Find the unknowns in the following figures
.JPG)
5.
If PQ || RS and ∠ONR = 30o find ∠MON and hence ∠MOX .
1.
Here, ΔABC ～ ΔADE (given)
∴ Their corresponding sides are proportional (by RHS similarity).
∴ \(\frac { AC }{ AE } =\frac { BC }{ DE } \)
⇒ \(\frac { 12 }{ 96 } =\frac { 54 }{ h } \)
⇒ h = \(\frac { 5\times 96 }{ 12 } \) = 40 feet
∴ The height of the tree is 40 feet.
2.
Given that ΔPEN ～ ΔPAD,
∴ \(\frac { PE }{ PA } =\frac { EN }{ AD } \Rightarrow \frac { 4 }{ 7 } =\frac { 6 }{ x } \Rightarrow x=\frac { 42 }{ 4 } \)
Also, \(\frac { PE }{ PA } =\frac { PN }{ PD } =\frac { y }{ y+5 } \)
i.e. \(\frac { 4 }{ 7 } =\frac { y }{ y+5 } \) ⇒ 4y + 20 =7y ⇒ 7y - 4y = 20
3y = 20 ⇒ \(y=\frac { 20 }{ 3 } \)cm
3.
Given that ΔPQR ～ ΔXYZ
∴ Their corresponding sides are proportional
⇒ \(\frac { PQ }{ XY } =\frac { QR }{ YZ } =\frac { PR }{ XZ } \)
⇒ \(\frac { 8 }{ a } =\frac { 14 }{ b } =\frac { 10 }{ 16 } \)
⇒ \(\frac { 8 }{ a } =\frac { 10 }{ 16 } \)
⇒ \(a=\frac { 8\times 16 }{ 10 } =\frac { 128 }{ 10 } \)
a = 12.8 cm
Also, \(\frac { 14 }{ b } =\frac { 10 }{ 16 } \)
⇒ \(\frac { 14\times 16 }{ 10 } =\frac { 224 }{ 10 } \)
∴ b = 22.4 cm
4.
Now, from Fig \(\angle\)140o+ \(\angle\)z = \(\angle\)180o (linear pair)
⇒ \(\angle\)z = 180o −140o = 40o
Also \(\angle\)x+ \(\angle\)z= \(\angle\)70 + \(\angle\)z (exterior angle property)
⇒\(\angle\)x=70o
Also \(\angle\)z+ \(\angle\)y + 70o = 180o (angle sum property in ΔABC)
⇒ 40o + \(\angle\)y + 70o = 180o
⇒ \(\angle\)y = 180o − 110o = 70o
5.
Extend MO and let it meet RS at Y. Extend NO and let it meet PQ at X.
As PQ || RS,
∠OYN = ∠OMX = 55o (alternate angles are equal)
∴ ∠MON = ∠OYN +∠ONY (exterior angle of ΔOYN = sum of interior opposite angles)
= 55o + 30o = 85o
⇒∠MOX = 180o− 85o = 95o (∠MON, ∠MOX are linear pair)
8th Standard Syllabus & Materials
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Tamilnadu Stateboard 8th Standard Subjects
Tamilnadu Stateboard Standards
