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
NEW8th Standard
TN 8th Tamil இயல் 2 - ஈடில்லா இயற்கை -பட்டமரம் Important Questions And Answers Study Material - QB365 Set A
NEW8th Standard
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.
In ΔABC, S is the circumcentre, BC = 72 cm and DS = 15 cm. Find the radius of its circumcircle.
2.
XYZ is a triangle and G is its centroid. If GP = 2.5 cm, GY = 6 cm and ZR = 12 cm, find XP, QY and GR.
3.
In the figure G is the centroid of the triangle XYZ.
(i) If GL = 2.5 cm, find the length XL.
(ii) If YM = 9.3 cm, find the length GM.
4.
In the figure, ABC is a triangle and CD is one of its medians.
If AD = 9x − 13 and BD = 4x + 2, find the length of the side AB.
5.
In the figure, ABC is a triangle and AM is one of its medians. If BM = 3.5 cm, find the length of the side BC.
1.
As S is the circumcentre of ΔABC, it is equidistant from A,B and C. So AS = BS = CS = radius of its circumcircle. As AD is the perpendicular bisector of BC \(BD=\cfrac { 1 }{ 2 } \times BC=\cfrac { 1 }{ 2 } \times 72=36cm\)
In right-angled triangle BDS, by Pythagoras theorem
BS2 = BD2 + SD2 = 362 + 152 = 1521 = 392 ⇒ BS = 39cm.
ஃThe radius of the circumcircle of ΔABC is 39 cm.
2.
Centroid G divides each median in the ratio 2:1
ஃ XG = 2GP ⇒ XG = 2(2.5) = 5 cm and XP = XG + GP = 5 + 2.5 = 7.5 cm.
Similarly, GY = 2GQ ⇒ 6 = 2GQ ⇒ GQ = 3 cm and QY = 3 + 6 = 9 cm.
Now, ZR = 12 cm ⇒ GR + GZ = 12 cm ⇒ GR + 2GR = 12 cm ⇒ GR = 4 cm
3.
(i) Since G is the centroid, XG : GL = 2 : 1 which gives XG : 2.5 = 2 : 1.
Therefore, we get 1 x (XG) = 2 x (2.5) ⇒ XG = 5 cm.
Hence, length XL = XG + GL = 5 + 2.5 = 7.5 cm.
(ii) If YG is of 2 parts then GM will be 1 part. (Why?)
This means YM has 3 parts.
3 parts is 9.3 cm long. So GM (made of 1 part) must be 9.3 ÷ 3 = 3.1 cm.
4.
AD = DB, since D is the midpoint of AB. (Why?)
Therefore, 9x − 13 = 4x + 2. Solving this (try!) simple equation, we get x = 3.
Hence, AB = 2(9x − 13) = 2(9(3) − 13)=28 units.
5.
AM is median ⇒ M is the midpoint of BC.
Given that, BM = 3.5 cm, hence BC = twice the length BM = 2 x 3.5 cm = 7 cm.
8th Standard Syllabus & Materials
8th Standard
TN 8th Tamil இயல் 3 - கல்வி கரையில - பாடறிந்து ஒழுகுதல் Important Questions And Answers Study Material - QB365 Set A
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Tamilnadu Stateboard 8th Standard Subjects
Tamilnadu Stateboard Standards