12. Chapter 12. Concurrent Lines NIOS Class 10 Mathematics Textbook Solutions
Chapter 12. Concurrent Lines NIOS Class 10 Mathematics Textbook Solutions for Exam Preparation
Chapter 12. CONCURRENT LINES
CHECK YOUR PROGRESS 12.1
1. In the given figure BF = FC , ∠BAE = ∠CAE and ∠ADE = ∠GFC = 90° , then name a median, an angle bisector, an altitude and a perpendicular bisector of the triangle.
Fig. 12.25
Solution: Since, AF is a median
So, BF = FC
AE is an angle bisector of ∠BAC .
So, ∠BAE = ∠CAE 
Again, AD⊥BC 
and GF⊥BC 
∠ADE = ∠GFC = 90° 
So, AD and GF are an altitude of triangle ABC .
F is a mid-point of BC and also ∠GFC = 90° 
So, ∠GFB =∠GFC = 90° 
Therefore,GF is a perpendicular bisector .
2. In an equilateral triangle show that the incentre, the circumcentre, the orthocentre and the centroid are the same point.
Solution: Let's consider an equilateral triangle ABC with AB=BC=CA .
The Median is also the Angle Bisector and Altitude:
Draw a median from vertex A to side BC. Let D be the midpoint of BC.
In △ABD and △ACD we have,
AB=AC ( given)
BD=CD (D is the midpoint of BC)
AD=AD (Common side)
Therefore, △ABD≅△ACD [ SSS ]
∠BAD=∠CAD [ CPCT]
So, AD is the angle bisector of ∠A.
∠ADB=∠ADC = 90°
Centroid (G): The point where all three medians meet. Since AD is a median, the centroid lies on AD.
Incentre (I): The point where all three angle bisectors meet. Since AD is an angle bisector, the incentre lies on AD.
Orthocentre (H): The point where all three altitudes meet. Since AD is an altitude, the orthocentre lies on AD.
Circumcentre (O): The point where all three perpendicular bisectors meet. Since AD is a perpendicular bisector, the circumcentre lies on AD.
3. In an equilateral ΔABC (Fig. 12.26). G is the centroid of the triangle. If AG is 4.8 cm, find AD and BE.
Solution: Here, AG = 4.8 cm
We know that, medians of a triangle pass through the same point, which divides each of the medians in the ratio 2 : 1.
∴ AGGD=21 
⇒AG=2GD 
⇒GD=AG2=4.82=2.4 cm 
⇒GD=2.4 cm 
AD=AG+GD=4.8+2.4=7.2 cm 
AD=BE=CF=7.2 cm 
Therefore, AD=7.2 cm and BE=7.2 cm 
4. If H is the orthocentre of ΔABC, then show that A is the orthocentre of the ΔHBC.
Solution: Since H is the orthocentre of ΔABC .
BH is an altitude, so BH ⊥ AC.
CH is an altitude, so CH ⊥ AB.
In ΔHBC, we have
AB is perpendicular to CH.
So, AB is an altitude from vertex B.
AC is perpendicular to BH.
So, AC is an altitude from vertex C.
These two altitudes (AB and AC) meet at A.
Therefore, A is the point of intersection of altitudes of ΔHBC.
Hence, A is the orthocentre of ΔHBC.
5. Choose the correct answers out of the given alternatives in the following questions:
(i) In a plane, the point equidistant from vertices of a triangle is called its
(a) centroid (b) incentre (c) circumcentre (d) orthocenter
Answer: (c) circumcentre .
[ The point equidistant from the vertices is the circumcentre. ]
(ii) In the plane of a triangle, the point equidistant from the sides of the triangle is called its
(a) centroid (b) incentre (c) circumcentre (d) orthocenter
Answer: (b) incentre .
[ The point equidistant from the sides of a triangle is always the incentre.]
TERMINAL EXERCISE
1. In the given Fig. 12.27, D, E and F are the mid points of the sides of ΔABC. Show that BE+CF>32BC 
.
Solution: Given, D, E and F are the mid points of the sides of ΔABC .
We show that BE+CF>32BC 
Proof: We know that, the sum of two sides must exceed the third side .
∴ BG+CG>BC………i
Since, medians of a triangle pass through the same point, which divides each of the medians in the ratio 2 : 1.
So, BG=23BE……… (ii)
and CG=23 CF………(iii)
From (i) , (ii) and (iii) , we have
23BE+23CF<BC 
⇒BE+CF<32 BC 
Hence proved .
2. ABC is an isoceles triangle such that AB = AC and D is the midpoint of BC. Show that the centroid, the incentre, the circumcentre and the orthocentre, all lie on AD.
Photo Fig. 12.28
Solution: Given, ABC is an isoceles triangle such that AB = AC and D is the midpoint of BC.
In ∆ABD and ∆ACD 
we have,
AB=AC (given)
AD=AD (common side)
BD=DC (D is a midpoint of BC)
∆ABD≅∆ACD 
[SSS]
∠ADB=∠ADC=90° 
[CPCT]
So, AD⊥BC and AD bisects BC.
Hence, AD is the perpendicular bisector as well as the median and altitude.
AD is a median, then the centroid lies on AD.
In ∆
ABC , we have
AB = AC , the angles at B and C are equal.
The angle bisector of ∠A and also bisects BC perpendicularly.
So, it coincides with AD.
Thus, the incentre lies on AD.
AD is the perpendicular bisector of BC, the circumcentre lies on AD.
AD is the altitude from A to BC, the orthocentre lies on AD.
Therefore, all four centres lie on AD
Hence proved.
3. ABC is an isoceles triangle such that AB = AC = 17 cm and base BC = 16 cm. If G is the centroid of ΔABC, find AG.
Solution: Given, AB = AC = 17 cm and BC = 16 cm
Let, M be the midpoint of BC.
BM=MC=BC2=162=8 cm 
In ∆ABM 
, we have,
AB2=AM2+BM2
⇒AM2=AB2-BM2
⇒AM=172-82=289-64=225
⇒AM=15 cm 
We have, 
⇒GMAG=12 
⇒GMAG+1=12+1 
⇒GM+AGAG=1+22 
⇒AMAG=32 
⇒AGAM=23 
⇒AG=23×AM=23×15=2×5 
⇒AG=10 cm 
4. ABC is an equilateral triangle of side 12 cm. If G be its centroid, find AG.
Solution: Given, ABC is an equilateral triangle .
So, AB=BC=AC=12 cm 
and G is the centroid .
Let, AD , BE and CE be the altitude .
i.e., AD=BE=CE 
The altitude AD=32×AB=32×12=63 cm 
Since, the centroid divides each median in the ratio 2 : 1, with the longer part towards the vertex.
AGGD=21 
⇒GDAG=12 
⇒GDAG+1=12+1=1+22=32 
⇒GD+AGAG=32 
⇒ADAG=32 
⇒AGAD=23 
⇒AG=23×AD 
⇒AG=23×63=2×23=43 cm 
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