Uncategorized

# triangles 6.2 – class 10

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

(i) (ii) Solution:

(i) Let EC = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

1.5/3 = 1/x

x = (3 x 1)/1.5

x = 2

EC = 2cm

(ii) It is given that DE || BC.

By using basic proportionality theorem, we obtain

x/7.2 = 1.8/5.4

x = (1.8×7.2)/(5.4)

x = 2.4 cm

2. E and F are points on the sides PQ and PR respectively of a ∆PQR. 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 = 0.63 cm

Solution: Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

PE/EQ  = 3.9/3 = 1.3

PF/FR = 3.6/2.4 = 1.5

Hence, PE/EQ ≠  PF/FR

Therefore, EF is not parallel to QR. PE = 4cm, QE = 4.5 cm, PF = 8cm , RF = 9cm

PE/EQ = 4/4.5 = 8/9

PF/FR = 8/9

Hence PE/EQ = PF/FR

Therefore, EF is  parallel to QR

(ii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

PE/PQ = 0.18/1.28 = 18/128 = 9/64

PF/PR = 0.36/2.56  =  9/54

Hence, PE/PQ = PF/PR

Therefore, EF is parallel to QR

3. In the following figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD Solution:

In the given figure, LM || CB

By using basic proportionality theorem, we obtain

AM/AB = AL/AC    —————–(1)

Similarly, LN||CD

From (1) and (2), we obtain

4. In the following figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC Solution: In ∆ABC, DE || AC

BD/DA = BE/EC (Basic proportionality theorem) In ∆BAE, DF || AE

BD/DA = BF/EF (Basic proportionality theorem)

5. In the following figure, DE || OQ and DF || OR, show that EF || QR Solution: In ∆ POQ, DE || OQ

Therefore, PE/EQ = PD/DO —————–(i) n ∆ POQ, DF || OR

PF/FR = PD/DO (Basic proportionality theorem) ———–(ii)

From (i) and (ii), we obtain

PE/EQ = PF/FR

Therefore, EF||QR (converse of basic proportionality theorem) 6. In the following figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR Solution: In ∆ POQ, AB || PQ

OA/AP = OB/BQ (Basic proportionality theorem) ———–(i) In ∆ POR, AC || PR

OA/AP = OC/CR (Basic proportionality theorem) ———–(ii)

From (i) and (ii)

OB/BQ = OC/CR

Therefore, BC||QR 7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Solution: Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ||BC

By using proportionality theorem, we obtain,

AQ/QC = AP/PB

AQ/QC = 1/1 (P is the midpoint of AB, AP = PB)

Therefore, AQ = QC

Or, Q is the mid-point of AC.

8. Using Converse of basic proportionality theorem, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Solution: Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively. i.e., AP = PB and AQ = QC It can be observed that

AP/PB = 1/1

and AQ/Qc = 1/1

AP/PB = AQ/QC

Hence, by using basic proportionality theorem, we obtain

PQ||BC

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO

Solution: Draw a line EF through point O, such that EF||CD

By using basic proportionality theorem, we obtain

AE/ED = AO/OC ——–(1)

In ΔABD, OE||AB

So, by using basic proportionality theorem, we obtain

ED/AE = OD/BO

Therefore, AE/ED = BO/OD ——(2_

From equations (1) & (2(, we obtain,

AO/OC = BO/OD

Therefore, AO/BO = OC/OD

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Solution:

Let us consider the following figure for the given question. Draw a line OE || AB In ∆ABD, OE || AB

By using basic proportionality theorem, we obtain

AE/ED = BO/OD ———–(1)

However, it is given that

AO/OC = OB/OD —————(2)

From equations (1) and (2), we obtain,

AE/ED = AO/OC

⇒ EO || DC [By the converse of basic proportionality theorem]

⇒ AB || OE || DC

⇒ AB || CD

∴ ABCD is a trapezium.