Mathematics Solutions Solutions for Class 8 Maths Chapter 4 Altitudes And Medians Of A Triangle are provided here with simple step-by-step explanations. These solutions for Altitudes And Medians Of A Triangle are extremely popular among Class 8 students for Maths Altitudes And Medians Of A Triangle Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

#### Question 1:

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

In ∆ LMN,  is an altitude and  is a median.

#### Question 2:

Draw an acute angled ∆ PQR. Draw all of its altitudes. Name the point of concurrence as 'O'.

Steps of construction :
(i) Draw any acute angled ∆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 intersect QR at L. So, PL is the altitude on side QR.
In the same manner, draw QM⊥PR and RN⊥PQ.
Hence, ∆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 ∆ STV. Draw its medians and show the centroid.

Steps of construction :
(i) Draw an obtuse angled ∆STV.
(ii) Draw the perpendicular bisector AB of side TV that intersect side TV at L. L is the mid point of TV.
(iii) Join SL, where SL is median to the side TV.
In the same manner, obtain the mid points M and N of sides SV and ST respectively.
(iv) Join TM and VN.
Hence, ∆STV is the required triangle in which the medians SL, TM and VN to the sides TV, SV and ST respectively intersect at point G.
The point G is the centroid of ∆STV.

#### Question 4:

Draw an obtuse angled ∆ LMN. Draw its altitudes and denote the orthocentre by 'O'.

Steps of construction :
(i) Draw an obtuse angled ∆LMN.
(ii) With L as centre and taking convenient radius, draw two arcs that intersect MN at P and Q.
(iii) With P as centre and taking radius more than half of PQ, draw an arc. With Q as centre and taking same radius, draw another arc that intersect the previous arc at R.
(iv) Join LR than intersect MN at I. LI is an altitude the side MN.
(v) Extend NL to a point V.
(vi) With M as centre and taking convenient radius, draw two arcs that intersect NL produced at D and E.
(vii) With D as centre and taking radius more than half of DE, draw an arc. With E as centre and taking same radius, draw another arc that intersect the previous arc at F.
(viii) Join MF that intersect NL produced at K. MK is an altitude to side NL produced.
(ix) Extend ML to point U.
(x) With N as centre and taking convenient radius, draw two arcs that intersect ML produced at A and B.
(xi) With A as centre and taking radius more than half of AB, draw an arc. With B as centre and taking same radius, draw another arc that intersect the previous arc at C.
(xii) Join NC that intersect ML produced at J. NJ is an altittude to the side ML produced.
Hence, ∆LMN is the required triangle in which the altitudes LI, MK and NJ to the sides MN, NL and ML respectively intersect at O.
The point O is the orthocentre of ∆LMN.

#### Question 5:

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

Steps of construction :
(i) Draw a right angled ∆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 YZ 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, ∆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 ∆XYZ.

#### Question 6:

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

Steps of contruction :
(i) Draw an isosceles ∆XYZ.
(ii) Draw the perpendicular bisector DE of the side YZ that intersect YZ at L.
(iii) Join XL. XL is the median to side YZ.
(iv) With X as centre and taking convenient radius, draw two arcs that intersect YZ at A and B.
(v) With A as centre and taking radius more than half of AB, draw an arc. With B as centre and taking same radius, draw another arc that cut the previous arc at C.
(vi) Join XC that intersect YZ at L. XL is the altitude to the side YZ.
(vii) Draw the perpendicular bisector IJ of the side ZX that intersect ZX at K.
(viii) Join YK. YK is the median to side ZX.
(ix) With Y as centre and taking convenient radius, draw two arcs that intersect ZX at Z and F.
(x) With Z as centre and taking radius more than half of ZF, draw an arc. With F as centre and taking same radius, draw another arc that cut the previous arc at H.
(xi) Join YH that intersect ZX at M. YM is the altitude to the side ZX.
(xii) Draw the perpendicular bisector ST of the side XY that intersect XY at U.
(xiii) Join ZU. ZU is the median to side XY.
(xiv) With Z as centre and taking convenient radius, draw two arcs that intersect XY at P and Q.
(xv) With P as centre and taking radius more than half of PQ, draw an arc. With Q as centre and taking same radius, draw another arc that cut the previous arc at R.
(xvi) Join ZR that intersect XY at N. ZN is the altitude to the side XY.
Hence, ∆XYZ is the required triangle in which the medians XL, YK and ZU to the sides YZ, ZX and XY respectively intersect at G and altitudes XL, YM and ZN to the sides YZ, ZX and XY respectively intersect at O. The point G is the centroid and point O is the orthocentre of ∆XYZ. We observe that, in an isosceles triangle, the point of concurrence of medians (centroid) and altitudes (orthocentre) lie on a same straight line.

#### Question 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) = ........

In ∆ABC, the medians AP, BQ and CR to the sides BC, CA and AB respectively intersect at G. Since, centroid of a triangle divides the medians in the ratio of 2 : 1, then AG : GP = BG : GQ = CG : GR = 2 : 1.
(1) We have, CG : GR = 2 : 1
$⇒\frac{\mathrm{GC}}{\mathrm{RG}}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{GC}}{2.5}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}⇒\mathrm{GC}=5$
(2) We have, BG : GQ = 2 : 1
$⇒\frac{\mathrm{BG}}{\mathrm{GQ}}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{BG}}{\mathrm{BQ}-\mathrm{BG}}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}⇒\frac{6}{\mathrm{BQ}-6}=2\phantom{\rule{0ex}{0ex}}⇒\mathrm{BQ}-6=3\phantom{\rule{0ex}{0ex}}⇒\mathrm{BQ}=9$
(3) We have, AG : GP = 2 : 1
$⇒\frac{\mathrm{AG}}{\mathrm{GP}}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{AG}}{\mathrm{AP}-\mathrm{AG}}=2\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{AG}}{6-\mathrm{AG}}=2\phantom{\rule{0ex}{0ex}}⇒2\left(6-\mathrm{AG}\right)=\mathrm{AG}\phantom{\rule{0ex}{0ex}}⇒12-2\mathrm{AG}=\mathrm{AG}\phantom{\rule{0ex}{0ex}}⇒3\mathrm{AG}=12\phantom{\rule{0ex}{0ex}}⇒\mathrm{AG}=\frac{12}{3}\phantom{\rule{0ex}{0ex}}⇒\mathrm{AG}=4$
Now, GP = AP − AG = 6 − 4 = 2.
Hence, we have,
(1) If l(RG) = 2.5 then l(GC) = 5
(2) If l(BG) = 6 then l(BQ) = 9
(3) If l(AP) =6 then l(AG) = 4 and l(GP) = 2

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