Solutions
Practice Set 4.1
Question 1. In Δ LMN, ...... is an altitude and ...... is a median. (write the names of appropriate segments.)
In Δ LMN, seg LX is an altitude and seg LY is a median.
Question 2. Draw an acute angled Δ PQR. Draw all of its altitudes. Name the point of concurrence as ‘O’.
Steps of construction:
(i) Draw any acute angled Δ PQR.
(ii) With P as centre, draw an arc that cut the side QR at X and Y.
(iii) With X as centre and radius more than half of XY, draw an arc below QR. With Y as centre and same radius draw another arc that cut the previous arc at A.
(iv) Join PA that intersects QR at L. So, PL is the altitude on side QR.
In the same manner, draw QM ⊥ PR and RN ⊥ PQ.
Hence, Δ PQR is the required triangle with altitudes PL, QM and RN on sides QR, RP and PQ respectively, with O as the point of concurrence of all the three altitudes.
Question 3. Draw an obtuse angled Δ STV. Draw its medians and show the centroid.
G is the centroid of Δ STV.
Question 4. Draw an obtuse angled Δ LMN. Draw its altitudes and denote the orthocenter by ‘O’.
Question 5. Draw a right angled Δ XYZ. Draw its medians and show their point of concurrence by G.
Steps of construction:
(i) Draw a right angled Δ XYZ.
(ii) Draw the perpendicular bisector PQ of side YZ that intersect YZ at L.
(iii) Join XL. XL is the median to the side YZ.
(iv) Draw the perpendicular bisector TU of side ZX that intersect XZ at M.
(v) Join YM. YM is the median to side ZX.
(vi) Draw the perpendicular bisector RS of side XY that intersect XY at N.
(vii) Join ZN. ZN is the median to the side XY.
Hence, Δ XYZ is the required triangle in which medinas XL, YM and ZN to the sides YZ, ZX and XY respectively, intersect at G.
The point G is the centroid of Δ XYZ.
Question 6. Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.
The medians AM, BM and CS intersect at point G and the altitudes CR, BT and AM intersect at point O.
Observation : The centroid G and the orthocentre O lie on the line which is the perpendicular bisector of base BC.
Question 7. Fill in the blanks.
Point G is the centroid of Δ 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) = .....
(1) If l(RG) = 2.5 then l(GC) = ?
The centroid divides each median in the ratio 2 : 1
∴ \(\frac{l(GC)}{l(RG)}=\frac{2}{1}\)
∴ \(\frac{l(GC)}{2.5}=\frac{2}{1}\)
l(GC) = 2.5 × 2 = 5
If l(RG) = 2.5 then l(GC) = 5
(2) If l(BG) = 6 then l(BQ) = ?
\(\frac{l(BG)}{l(BQ)}=\frac{2}{1}\)
∴ \(\frac{6}{l(BQ)}=\frac{2}{1}\)
l(BQ) = 6/2 = 3
If l(BG) = 6 then l(BQ) = 3
(3) If l(AP) = 6 then l(AG) = ? and l(GP) = ?
\(\frac{l(AG)}{l(GP)}=\frac{2}{1}\)
∴ l(AG) = 2l(GP)
Let l(GP) = x
∴ l(AG) = 2x
l(AG) + l(GP) = l(AP)
∴ x + 2x = 6
∴ 3x = 6
∴ x = 2
∴ l(GP) = x = 2
∴ l(AG) = 2x = 2 x 2 = 4