Basic Proportionality Theorem Triangles NCERT Exercise 6.2 part two Class Ten Mathematics

Triangle

Exercise 6.2 Part 2

Question 5: In the given figure, DE || OQ and DF || OR. Show that EF || QR.

similar triangles exercise solution

Solution: In Δ PQO and Δ PED;

`(PE)/(EQ)=(ED)/(QO)=(PD)/(DO)`

(Because these are similar triangles, as per Basic Proportionality theorem.)

Similarly, in Δ PRO and Δ PFD;

`(PF)/(FR)=(FD)/(RO)=(PD)/(DO)`

From above two equations, it is clear;

`(PE)/(EQ)=(PF)/(FR)`

Hence;

`(PE)/(EQ)=(EF)/(QR)`

Hence, EF || QR proved.


Question 6: In the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

similar triangles exercise solution

Solution: In Δ OPQ and Δ OAB;

`(OA)/(AP)=(OB)/(BQ)`

(Because these are similar triangles as per BPT.)

Similarly, in Δ OPR and Δ OAC;

`(OA)/(OP)=(OC)/(CR)`

From above two equations, it is clear;

`(OB)/(BQ)=(OC)/(CR)`

Hence; BC || QR proved

Question 7: 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.

similar triangles exercise solution

Solution: PQR is a triangle in which DE || QR. Line DE intersects PQ at D so that PD = DQ

To Prove: PE = ER

In Δ PQR and Δ PDE;

`(PD)/(DQ)=(PE)/(ER)`

(Because as per BPT, these are similar triangles.)

`(PD)/(DQ)=1` given

Hence, `(PE)/(ER)=1`

Hence; E is the midpoint of PR proved.


Question 8: 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.

Solution: The figure from the previous question can be used to solve this question.

ABC is a triangle in which D and E are mid points of PQ and PR respectively.

To Prove: DE || QR

In Δ PQR and Δ PDE;

`(PD)/(DQ)=(PE)/(ER)=1` given similar triangles exercise solution

Hence, as per BPT these triangles are similar.

Hence; DE || QR proved.

Question 9: ABCD is a trapezium in which AB || CD and its diagonals intersect each other at the point O. show that; `(AO)/(BO)=(CO)/(DO)`

similar triangles exercise solution

Solution: Draw a line EF || CD which is passing through O.

In Δ ABC and Δ EOC;

These are similar triangles as per BPT.

`(AE)/(EC)=(BO)/(OC)`

Similarly, in Δ BOD and Δ FOD;

`(BF)/(FD)=(AO)/(OD)`

In Δ ABC and Δ BAD;

`(BO)/(OC)=(AO)/(OD)`

(Because diagonals of a trapezium divide each other in same ratio)

From above three equations, it is clear;

`(AE)/(EC)=(BF)/(FD)`

Hence, Δ ABC ~ Δ BAD

Using the third equation;

`(BO)/(OC)=(AO)/(OD)`

Or, `(AO)/(BO)=(CO)/(DO)`

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

Solution: This question can be proved by using the figure in previous question. Since diagonals are dividing each other in the same ratio hence, ABCD is a trapezium; proved.



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