The Class 8 is an important year in a student’s life and Maharashtra State Board Maths is one of the subjects that require dedication, hard work, and practice. It’s a subject where you can score well if you are well-versed with the concepts, remember the important formulas and solving methods, and have done an ample amount of practice. Worry not! Home Revise is here to make your Class 8 journey even easier. It’s essential for students to have the right study material and notes to prepare for their board examinations, and through Home Revise, you can cover all the fundamental topics in the subject and the complete Maharashtra State Board Class 8 Maths Book syllabus.
Practice set 4.1 PAGE NO: 22
1. In ∆LMN, ……is an altitude and ……is a median. (Write the names of appropriate segments.)
Solution:
In ∆LMN, LX is the altitude (since it makes a 90° angle) and LY is a median (since it divides the base into two equal halves i.e., MY = NY).
2. Draw an acute-angled ∆PQR. Draw all of its altitudes. Name the point of concurrence as ‘O’.
Solution:
Here, is the acute-angled ∆PQR
Seg PA, seg BQ, seg CR are the altitudes of ∆PQR. The point of concurrence is denoted by the point O.
3. Draw an obtuse-angled ∆STV. Draw its medians and show the centroid.
Solution:
Here, is the obtuse-angled ∆STV.
Seg SP, seg UT and seg RV are medians of ∆STV.
Their point of concurrence is denoted by G.
4. Draw an obtuse-angled ∆LMN. Draw its altitudes and denote the orthocenter by ‘O’.
Solution:
Here, is the obtuse-angled ∆ LMN.
The orthocenter of the obtuse triangle lies outside the triangle.
The point O denotes the orthocenter of the obtuse-angled ∆LMN.
5. Draw a right angled ∆XYZ. Draw its medians and show their point of concurrence by G.
Solution:
Here, is the right angled ∆XYZ.
Their point of concurrence is denoted by G.
6. Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.
Solution:
Here, is the isosceles triangle.
About the points of concurrence:
The medians i.e. G and altitudes i.e. O lie on the same line on PS which is the perpendicular bisector of seg QR.
7. Fill in the blanks.
Point G is the centroid of ∆ABC.
(1) If l(RG) = 2.5 then l(GC) = ……
(2) If l(BG) = 6 then l(BQ) = ……
(3) If l(AP) = 6 then l(AG) = ….. and l(GP) = …..
Solution:
(1) If l(RG) = 2.5 then l(GC) = 5.
We know that the centroid divides each median in the ratio 2:1.
So, CG/RG = 2/1
CG/2.5 = 2/1
CG = 2 × 2.5
= 5
(2) If l(BG) = 6 then l(BQ) = 9.
We know that the centroid divides each median in the ratio 2:1.
So, BG/QG = 2/1
6/QG = 2/1
6 × 1 = 2 × QG
6 = 2 × QG
6/2 = QG
QG = 3.
Since we have to find I(BQ), and from the figure it can be seen that,
(BQ) = I(BG) + I(QG)
So, I(BQ) = 6 + 3
I(BQ) = 9.
(3) If l(AP) = 6 then l(AG) = 4 and l(GP) = 2.
We know that the centroid divides each median in the ratio 2:1.
Here both I(AG) and I(GP) are unknown so,
Let I(AG), I(GP) be 2x and x respectively, from equation (i)
I(AP) = I(AG) + I(GP)
6 = 2x + x
6 = 3x
6/3 = x
x = 2.
I(AG) = 2x = 2×2 = 4.
I(GP) = x = 2.