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Published on: 16/06/2021
QB365 provides detailed and simple solution for every book back questions in class 7 Maths subject.It will helps to get more idea about question pattern in every book back questions with solution.
Download Tamil Nadu 7th 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.
From the given figure, find the missing angle
2.
If the three angles at a point are in the ratio 1 : 4 : 7, find the value of each angle?
3.
Find the angle \(\angle\)JIL from the given figure.
4.
Name the pairs of adjacent angles.
5.
Given that AB is a straight line. Calculate the value of x° in the following cases.
1.
From \(\angle \mathrm{PQR}=\angle \mathrm{SOP}\) (vertically opposite angels)
105 = x
(ie) x = 105o
2.
Let the three angles be x, 4x and 7x .
The sum of three angles at a points is supplementary
∴ x + 4x + 7x = 360°
12x = 360°
xo = \(\frac{{360}^o}{12}\)
x = 30°
If x = 30o then
4x = 4 x 30° = 120°
7x = 7 x 30° = 210°.
∴ The three angles are 30°, 120° and 210°.
3.
From the figure
∴ ∠JlL = ∠JIK + ∠KIL
= 38° + 27°
= 65°
∴ ∠JlL = 65°
4.
From the figure the pair ofadjacent angles are
i) \(\angle\)ABG and, \(\angle\)CBG
ii) \(\angle\) DCE and \(\angle\)ECF
iii) \(\angle\)ECE and \(\angle\)ACF
iv) \(\angle\)ACF and \(\angle\)ECD.
v) \(\angle\)BCF and \(\angle\)FCE
5.
(i) From the figure
∠AOC + ∠BOC = 180°
72° + xo = 180°
72° + xo - 72° = 180° - 72°
xo = 108°
(ii) From the figure
∠AOC + ∠BOC = 180°
3x + 42° - 180°
3xo + 42° - 42°= 180o - 42°
3xo = 138°
xo = \(\frac{{138}^o}{3}\) = 46o
xo = 46°
(iii) From the figure
∠AOC + ∠BOC = 180°
4xo + 2xo = 180°
6xo = 180°
xo = \(\frac{{138}^o}{6}\)
xo = 30°
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Tamilnadu Stateboard 7th Standard Subjects
Tamilnadu Stateboard Standards
