Tuesday, April 26, 2011

6. The Triangle and its Properties

Exercise 6.1

Question 1:
In ΔPQR, D is the mid-point of .
is __________.
PD is __________.
Is QM = MR?

(i) Altitude
(ii) Median
(iii) No

Question 2:
Draw rough sketches for the following:
(a) InΔABC, BE is a median.
(b) InΔPQR, PQ and PR are altitudes of the triangle.
(c) InΔXYZ, YL is an altitude in the exterior of the triangle.

(a)

(b)

(c)

Here, it can be observed that for ΔXYZ, YL is an altitude drawn exterior to side XZ which is extended up to point L.

Question 3:
Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.

Draw a line segment AD perpendicular to BC. It is an altitude for this triangle. It can be observed that the length of BD and DC is also same. Therefore, AD is also a median of this triangle.

Exercise 6.2

Question 1:
Find the value of the unknown exterior angle x in the following diagrams:

(i) x= 50° + 70° (Exterior angle theorem)
x = 120°
(ii) x= 65° + 45° (Exterior angle theorem)
= 110°
(iii) x= 40° + 30° (Exterior angle theorem)
= 70°
(iv) x= 60° + 60° (Exterior angle theorem)
= 120°
(v) x= 50° + 50° (Exterior angle theorem)
= 100°
(vi) x= 30° + 60° (Exterior angle theorem)
= 90°

Question 2:
Find the value of the unknown interior angle x in the following figures:

(i) x+ 50° = 115° (Exterior angle theorem)
x = 115° − 50° = 65°
(ii) 70° + x = 100° (Exterior angle theorem)
x = 100° − 70° = 30°
(iii) x+ 90° = 125° (Exterior angle theorem)
x = 125° − 90° = 35°
(iv) x+ 60° = 120° (Exterior angle theorem)
x = 120° − 60° = 60°
(v) x+ 30° = 80° (Exterior angle theorem)
x = 80° − 30° = 50°
(vi) x+ 35° = 75° (Exterior angle theorem)
x = 75º − 35º = 40°

Exercise 6.3

Question 1:
Find the value of the unknown x in the following diagrams:

The sum of all interior angles of a triangle is 180°. By using this property, these problems can be solved as follows.
(i) x+ 50° + 60° = 180°
x + 110° = 180°
x = 180° − 110° = 70°
(ii) x+ 90° + 30° = 180°
x + 120° = 180°
x = 180° − 120° = 60°
(iii) x+ 30° + 110° = 180°
x + 140° = 180°
x = 180° − 140° = 40°
(iv) 50° + x + x = 180°
50° + 2x = 180°
2x = 180° − 50° = 130°

(v) x+ x + x = 180°
3x = 180°

(vi) x+ 2x + 90° = 180°
3x = 180° − 90° = 90º

Question 2:
Find the value of the unknowns x and y in the following diagrams:

(i) y+ 120° = 180° (Linear pair)
y = 180° − 120º = 60º
x + y + 50° = 180° (Angle sum property)
x + 60° + 50° = 180°
x + 110° = 180°
x = 180° − 110° = 70°
(ii) y= 80° (Vertically opposite angles)
y + x + 50° = 180° (Angle sum property)
80° + x + 50° = 180°
x + 130º = 180°
x = 180° − 130º = 50°
(iii) y+ 50° + 60° = 180° (Angle sum property)
y = 180° − 60° − 50° = 70°
x + y = 180° (Linear pair)
x = 180° − y = 180° − 70° = 110°
(iv) x= 60º (Vertically opposite angles)
30° + x + y = 180°
30° + 60° + y = 180°
y = 180° − 30° − 60° = 90°
(v) y= 90° (Vertically opposite angles)
x + x + y = 180° (Angle sum property)
2x + y = 180°
2x + 90° = 180°
2x = 180° − 90° = 90°

(vi)

y = x (Vertically opposite angles)
a = x (Vertically opposite angles)
b = x (Vertically opposite angles)
a + b + y = 180° (Angle sum property)
x + x + x = 180°
3x = 180°

y = x = 60°

1. Your blog is very informative but a little bit complicated and I am here to give the definition and properties of quadrilaterals as they are figures with four sides and four angles. There are many different types of quadrilaterals, which are classified by their sides and angles here are some examples-rectangle, parallelogram, trapezoid, rhombus, square and kite.
What is the Midpoint Formula

2. your blog is very useful .... thanq...

3. Its very good.
thanks man for doing this for us.

4. Little bit difficult to understand some questions. Overall its GOOD

5. Now i will get good marks thanks my friend

6. man tommmorow is review u helped me very much thank u i hope you have a better future thank u for your kindness...

7. man tommmorow is review u helped me very much thank u i hope you have a better future thank u for your kindness...

8. This comment has been removed by the author.

9. Thank you so much. It's so useful