Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Additional Questions

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Additional Questions

Additional Questions and Answers

Exercise 4.1

Question 1.
“The sum of any two angles of a triangle is always greater than the third angle”. Is this statement true. Justify your answer.
Solution:
No, the sum of any two angles of a triangle is not always greater than the third angle. In an isosceles right angled triangles, the angle will be 90°, 45°, 45°.
Here sum of two angles 45° + 45° = 90°.

Question 2.
The three angles of a triangle are in the ratio 1:2:1. Find all the angles of the triangle. Classify the triangle in two different ways.
Solution:
Let the angles of the triangle be x, 2x, x.
Using the angle sum property, we have
x + 2x + x = 180°
4x = 180°
x = \(\frac{180^{\circ}}{4}\)
x = 45°
2x = 2 × 45° = 90°
Thus the three angles of the triangle are 45°, 90°, 45°.
Its two angles are equal. It is an isoscales triangle. Its one angle is 90°.
∴ It is a right angled triangle.

Question 3.
Find the values of the unknown x and y in the following figures
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry add 1
Solution:
(i) Since angles y and 120° form a linear pair.
y + 120° = 180°
y = 180° – 120°
y = 60°
Now using the angle sum property of a triangle, we have
x + y + 50° = 180°
x + 60° + 50° = 180°
x + 110° = 180°
x = 180° – 110°= 70°
x = 70°
y = 60

(ii) Using the angle sum property of triangle, we have
50° + 60° + y = 180°
110° + y = 180°
y = 180° – 110°
y = 70°
Again x and y form a linear pair
∴ x + y = 180°
x + 70° = 180°
x = 180° – 70°= 110°
∴ x = 110°; y = 70°

Question 4.
Two angles of a triangle are 30° and 80°. Find the third angle.
Solution:
Let the third angle be x.
Using the angle sum property of a triangle we have,
30° + 80° + x = 180°
x + 110° = 180°
x = 180° – 110° = 70°
Third angle = 70°.

Exercise 4.2

Question 1.
In an isoscleles ∆ABC, AB = AC. Show that angles opposite to the equal sides are equal.
Solution:
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry add 2
Given: ∆ABC in which \(\overline{A B}\) = \(\overline{A C}\).
To Prove: ∠B = ∠K.
Construction: Draw AD ⊥ BC.
Proof: In right ∆ADB and right ∆ADC.
we have side AD = side AD (common)
AB = AC (Hypoteneous) (given)
∆ADB = ADC (RHS criterion]
∴ Their corresponding parts are equal. ∠B = ∠C.

Question 2.
ABC is an isosceles triangle having side \(\overline{A B}\) = side \(\overline{A C}\). If AD is perpendicular to BC, prove that D is the mid-point of \(\overline{B C}\).
Solution:
In ∆ABD and ∆ACD, we have
∠ADB = ∠ADC [∵ AD ⊥ BC]
Side \(\overline{A D}\) = Side \(\overline{A D}\) [Common]
Side \(\overline{A B}\) = Side \(\overline{A C}\) [Common]
Using RHS congruency, we get
∆ABD ≅ ∆ACDc
Their corresponding parts are equal
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry add 3
∴ BD = CD
∴ O is the mid point of BC.

Question 3.
In the figure PL ⊥ OB and PM ⊥ OA such that PL = PM.
Prove that ∆PLO ≅ ∆PMO.
Solution:
Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry add 4
In ∆PLO and ∆PMO, we have
∠PLO = ∠PMO = 90° [Given]
\(\overline{O P}\) = \(\overline{O P}\) [Hypotenuse]
PL = PM
Using RHS congruency, we get
∆PLO ≅ ∆PMO