class 8 :maths : Altitudes and Medians of a triangle : Practice Set :4.1

class 8 :maths : Altitudes and Medians of a triangle : Practice Set :4.1

Practice Set :4.1 

1 .In  ∆LMN, ...... is an altitude and ...... is a median. (write the names of appropriate segments.)

ANSWER :

In ∆LMN, seg LX is an altitude and seg LY is a median.

Question 2.
Draw an acute angled ∆PQR. Draw all of its altitudes. Name the point of concurrence as ‘O’.

ANSWER :

Question 3.
Draw an obtuse angled ∆STV. Draw its medians and show the centroid.

ANSWER :

Question 4.
Draw an obtuse angled ∆LMN. Draw its altitudes and denote the ortho centre by ‘O’.

ANSWER :

Question 5.
Draw a right angled ∆XYZ. Draw its medians and show their point of concurrence by G.

ANSWER :

Question 6.
Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.

ANSWER :

The point of concurrence of medians i.e. G and that of altitudes i.e. O lie on the same line PS which is the perpendicular bisector of seg QR.

Question 7.
Fill in the blanks.
Point G is the centroid of ∆ABC.
i. If l(RG) = 2.5, then l(GC) = ___
ii. If l(BG) = 6, then l(BQ) = ____
iii. If l(AP) = 6, then l(AG) = ___ and l(GP) = ___.

Solution:
The centroid of a triangle divides each median in the ratio 2:1.
i. Point G is the centroid and seg CR is the median.
∴ l(GC)l(RG)=21
∴ l(GC)2.5=21 ……[∵ l(RG) = 2.5]
∴ l(GC) × 1 = 2 × 2.5
∴ l(GC) = 5

ii. Point G is the centroid and seg BQ is the median.
∴ l(BG)l(GQ)=21
∴ 6l(GQ)=21 …..[∵ l(BG) = 6]
∴ 6 × 1 = 2 × l(GQ)
∴ 62 = l(GQ)
∴ 3 = l(GQ)
i.e. l(GQ) = 3
Now, l (BQ) = l(BG) + l(GQ)
∴ l(BQ) = 6 + 3
∴ l(BQ) = 9

iii. Point G is the centroid and seg AP is the median.
∴ l(AG)l(GP)=21
∴ l(AG) = 2 l(GP) …..(i)
Now, l(AP) = l(AG) + l(GP) … (ii)
∴ l(AP) = 2l(GP) + l(GP) … [From (i)]
∴ l(AP) = 3l(GP)
∴ 6 = 3l(GP) ..[∵ l(AP) = 6]
∴ 63 = l(GP)
∴ 2 = l(GP)
i.e. l(GP) = 2
l(AP) = l(AG) + l(GP) …[from (ii)]
∴ 6 = l(AG) + 2
∴ l(AG) = 6 – 2
∴ l(AG) = 4

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