# Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.1

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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.1

Question 1.
Fill in the blanks:

1. The altitudes of a triangle intersect at………..
2. The medians of a triangle cross each other at………..
3. The meeting point of the angle bisectors of a triangle is………..
4. The perpendicular bisectors of the sides a triangle meet at………..
5. The centroid of a triangle divides each medians in the ratio………..

Solution:

1. Orthocentre
2. Centroid
3. Incentre
4. Circumcentre
5. 2 : 1 Question 2.
Say True or False:
(i) In any triangle the Centroid and the Incentre are located inside the triangle.
(ii) The centroid, orthocentre, and incentre of a triangle are collinear.
(iii) The incentre is equidistant from all the vertices of a triangle.
Solution:
(i) True
(ii) True
(iii) False

Question 3.
(a) Where does the circumcentre lie in the case of
(i) An acute angled triangle.
Solution:
Inside the triangle.

(ii) An obtuse-angled triangle.
Solution:
Exterior of the triangle.

(iii) A right angled triangle.
Solution:
On the hypotenuse.

(b) Where does the orthocentre lie in the case of
(i) An acute-angled triangle.
Solution:
Interior of the triangle.

(ii) An obtuse-angled triangle.
Solution:
Exterior of the triangle.

(iii) A right angled triangle.
Solution:
On the vertex containing 90°.

Question 4.
Fill in the blanks:
In the triangle ABC, (i) The angle bisector is……….
(ii) The altitude is………..
(iii) The median is…………
Solution:
(i) BE
(iii) CF

Question 5.
In right triangle ABC, what is the length of altitude drawn from the vertex A to BC? Solution:
In this right angled triangle ΔABC, length of the altitude drawn from vertex A is the leg AB itself. By Pythagoras theorem.
AC² = AB² + BC²
13² = AB² + 12²
169 = AB² + 144
AB² = 169 – 144 = 25
AB² = 52
AB = 5cm Question 6.
In triangle XYZ, YM is the angle bisector of ∠Y and ∠Y is 100°. Find ∠XYM and ∠ZYM. Solution:
Given YM is the angle bisector of ∠Y.
also ∠Y = 100°
Angle -bisector divides the angle into two congruent angles.
∠XYM = ∠ZYM
∠Y = ∠XYM + ∠ZYM
100° = ∠XYM + ∠ZYM [∴ ∠XYM = ∠ZYM]
2 ∠XYM = 100°
∠XYM = $$\frac{1}{2}$$ (100°)
∠XYM = 50°
∴ ∠ZYM = 50°

Question 7.
In triangle PQR, PS is a median and QS = 3.5 cm, then find QR? Solution:
Given PS is the median and QS = 3.5 cm
Median is the line drawn from a vertex to the midpoint of the opposite side.
∴ QS = RS
so QS = RS = 3.5 cm
∴ QR = QS +SR = 3.5 + 3.5 = 7 cm
QR = 7 cm

Question 8.
In triangle ABC, line is a perpendicular bisector of BC. If BC = 12 cm, SM = 8 cm, find CS. Solution:
Given l1 is the perpendicular bisector of BC.
∴ ∠SMC = 90° and BM = MC
BC = 12 cm
⇒ BM + MC = 12 cm
MC + MC = 12 cm [∴ BM = MC]
2MC = 12
MC = $$\frac{12}{2}$$
MC = 6 cm
Given SM = 8 cm
By Pythagoras theorem SC² = SM² + MC²
SC² = 8² + 6²
SC² = 64 + 36
CS² = 100
CS² = 10²
CS = 10 cm 