CBSE Class 10 Mathematics chapter 6 - Triangle - Exercise 6.2

Q1.

In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Ans.

(i) Given, in △ABC, DE∥ BC We know that, as per Basic proportionality theorem - $$\frac{AD}{DB}=\frac{AE}{EC}$$ Putting value- $$\Rightarrow\frac{1.5}{3}=\frac{1}{EC}$$ $$\Rightarrow\frac{3}{1.5}$$ $$EC = 3×\frac{10}{15} = 2 cm$$ Hence, EC = 2 cm. (ii) Given, in △ ABC, DE∥BC We know that, as per Basic proportionality theorem - $$\frac{AD}{DB}=\frac{AE}{EC}$$ $$\Rightarrow\frac{AD}{7.2}=\frac{1.8}{5.4}$$ $$\Rightarrow\frac{1.8\times7.2}{5.4}$$ $$\Rightarrow\frac{18}{10}\times\frac{72}{10}\times\frac{10}{54}=\frac{24}{10}=2.4$$ Hence, AD = 2.4 cm.

Q2.

E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR : (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Ans.

Case (i) Given: PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm We know that Basic proportionality theorem - $$\frac{PE}{EQ}=\frac{PF}{FR}$$ By putting value- $$\Rightarrow\frac{PE}{EQ}=\frac{3.9}{3}=\frac{39}{30}=1.3$$ and $$\Rightarrow\frac{PF}{FR}=\frac{3.6}{2.4}=\frac{36}{24}=1.5$$ So, $$\frac{PE}{EQ}≠\frac{PF}{FR}$$ Hence, EF not parallel to QR. Case (ii) Given: PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm We know that Basic proportionality theorem - $$\frac{PE}{EQ}=\frac{PF}{FR}$$ By putting value- $$\Rightarrow\frac{PE}{EQ}=\frac{4}{4.5}=\frac{40}{45}=\frac{8}{9}$$ and $$\Rightarrow\frac{PF}{FR}=\frac{8}{9}$$ So, $$\frac{PE}{EQ}=\frac{PF}{FR}$$ Hence, EF is parallel to QR. Case (iii) Given: PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm As per figure EQ = PQ — PE = 1.28 — 0.18 = 1.10 cm And, FR = PR — PF = 2.56 — 0.36 = 2.20 cm We know that Basic proportionality theorem - $$\frac{PE}{EQ}=\frac{PF}{FR}$$ By putting value- $$\Rightarrow\frac{PE}{EQ}=\frac{0.18}{1.10}=\frac{18}{110}=\frac{9}{55}$$ and $$\Rightarrow\frac{PF}{FR}=\frac{0.36}{2.20}=\frac{36}{220}=\frac{9}{55}$$ So, $$\frac{PE}{EQ} =\frac{PF}{FR}$$ Hence, EF is parallel to QR.

Q3.	In Fig. 6.18, if LM || CB and LN || CD, prove that $$\frac{AM}{AB}=\frac{AN}{AD}$$
Ans.	Given: LM || CB and LN || CD As per given LM || CB and by using basic proportionality theorem, we get, $$\frac{AM}{AB}=\frac{AL}{AC}....................(i)$$ Similarly, as per given, LN || CD and by using basic proportionality theorem, $$\frac{AN}{AD}=\frac{AL}{AC}....................(ii)$$ From equation (i) and (ii), we get, $$\frac{AM}{AB}=\frac{AN}{AD}$$ Hence, proved.

Q4.	In Fig. 6.19, DE || AC and DF || AE. Prove that $$\frac{BF}{FE}=\frac{BE}{EC}$$
Ans.	In ΔABC, Given : DE || AC and DF || AE As per given DE || AC and by using basic proportionality theorem, we get, $$\frac{BD}{DA}=\frac{BE}{EC}....................(i)$$ Similarly, as per given, DF || AE and by using basic proportionality theorem, $$\frac{BD}{DA}=\frac{BF}{FE}....................(ii)$$ From equation (i) and (ii), we get, $$\frac{BE}{EC}=\frac{BF}{FE}$$ Hence, proved.

Q5.

In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

Ans.

In ΔPQO, Given : DE || OQ and DF || OR As per given DE || OQ and by using basic proportionality theorem, we get, $$\frac{PD}{DO}=\frac{PE}{EQ}....................(i)$$ Similarly, as per given, DF || OR and by using basic proportionality theorem, $$\frac{PD}{DO}=\frac{PF}{FR}....................(ii)$$ From equation (i) and (ii), we get, $$\frac{PE}{EQ}=\frac{PF}{FR}$$ Therefore, by converse of basic proportionality theorem, we get in ΔPQR EF || QR Hence, proved.

Q6.

In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Ans.

Given : In ΔOPQ, AB || PQ and In ΔOPR, AC || PR In ΔOPQ, As per given AB || PQ and by using basic proportionality theorem, we get, $$\frac{OA}{AP}=\frac{OB}{BQ}....................(i)$$ Similarly in ΔOPR, as per given, AC || PR and by using basic proportionality theorem, $$\therefore\frac{OA}{AP}=\frac{OC}{CR}....................(ii)$$ From equation (i) and (ii), we get, $$\frac{OB}{BQ}=\frac{OC}{CR}$$ Therefore, by converse of basic proportionality theorem, we get in ΔOQR BC || QR Hence, proved.

Q7.

Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Ans.

Given : In ΔABC, D is midpoint of AB, so AD = DB. As per figure shown above, line BC intersect AC at E such that DE || BC. And We have to prove that E is the midpoint for AC. Since, D is the mid-point of AB. $$\therefore AD = DB$$ $$\Rightarrow\frac{AD}{DB}=1..................... (i)$$ In ΔABC, DE || BC, By using Basic Proportionality Theorem, $$\Rightarrow\frac{AD}{DB}=\frac{AE}{EC}$$ From equation (i), we can write, $$\Rightarrow1=\frac{AE}{EC}$$ $$\therefore AE = EC$$ Hence, proved that E is the midpoint of AC.

Q8.

Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Ans.

Given : In ΔABC, D and E are midpoint of AB and AC, so AD = DB and AE = EC. We have to prove that DE || BC As per given, D is the mid-point of AB. $$\therefore AD = DB$$ $$\Rightarrow\frac{AD}{BD}=1..................... (i)$$ As per given, E is the mid-point of AC. $$\therefore AE = EC$$ $$\Rightarrow\frac{AE}{EC}=1.................... (ii)$$ From equation (i) and (ii), we can say, $$\frac{AD}{BD}=\frac{AE}{EC}$$ Therefore, by converse of basic proportionality theorem, DE || BC Hence, proved that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Q9.

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $$\frac{AO}{BO}=\frac{CO}{DO}$$

Ans.

Given : In trapezium ABCD, AB || DC and Diagonal AC and BD intersect each other at point O We have to prove that $$\frac{AO}{BO}=\frac{CO}{DO}$$ For this we darw a line EO which touch AD at point E, in such a way that EO || DC || AB So, In ΔADC, We have EO || DC so, by using basic proportionality theorem, we get, $$\frac{AE}{ED}=\frac{AO}{CO}....................(i)$$ Similarly, In ΔABD, We have OE || AB so, by using basic proportionality theorem, $$\frac{DE}{EA}=\frac{DO}{BO}....................(ii)$$ From equation (i) and (ii), we get, $$\frac{AO}{CO}=\frac{DO}{BO}$$ $$\Rightarrow\frac{AO}{BO}=\frac{CO}{DO}$$ Hence, proved.

Q10.

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $$\frac{AO}{BO}=\frac{CO}{DO}$$. Show that ABCD is a trapezium.

Ans.

Given : In Quadrilateral ABCD, AC and BD intersect each other at point O, such that $$\frac{AO}{BO}=\frac{CO}{DO}$$ We have to prove that ABCD is a trapezium. For this we darw a line EO which touch AD at point E, in such a way that EO || DC || AB So, In ΔDAB, We have EO || AB so, by using basic proportionality theorem, we get, $$\frac{DE}{EA}=\frac{DO}{OB}....................(i)$$ As per given $$\frac{AO}{BO}=\frac{CO}{DO}$$ $$\Rightarrow\frac{AO}{CO}=\frac{BO}{DO}$$ $$\Rightarrow\frac{CO}{AO}=\frac{DO}{BO}$$ $$\Rightarrow\frac{DO}{OB}=\frac{CO}{AO}........................(ii)$$ from equation (i) and (ii) we get $$\frac{DE}{EA}=\frac{CO}{AO}$$ Therefore, by converse of basic proportionality theorem, EO || DC and also EO || AB ⇒ AB || DC Hence, proved that quadrilateral ABCD is a trapezium with AB || DC.