10th CBSE math solution for exercise 6.2 part 2

This page 10th CBSE math solution for exercise 6.2 part 2 is going to provide you solution for every problems that you find in the exercise no 6.2

10th CBSE math solution for exercise 6.2

(5) In fig 6.20 DE ∥ OQ and DF ∥ OR .show that EF ∥ QR

In triangle PQO

DE ∥ OQ

(PE/EQ) = (PD/DO)----(1)

In triangle POR

 DE ∥ OQ

(PF/FR) = (PD/DO)  ------(2)

(PE/EQ) = (PF/FR) by using inverse of BPT theorem

EF is parallel to QR

(6) 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

Solution:

In triangle OPQ

AB is parallel to PQ

(OA/AP) = (OB/BQ) ---(1)

 In triangle OPR

 AC is parallel to PR

 (OA/AP) = (OC/CR) ---(2)

(1) = (2)

(OB/BQ) =(OC/CR)

by using converse of BPT theorem

BC is parallel to QR


(7) Using theorem 6.1,prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.

Solution:

Draw OM parallel to AB meeting BC at M.

In triangle ACB

OM is parallel to AB

(OC/OA) = (CM/MB) ----(1)

 similarly in triangle BDC

OM is parallel to CD

(BM/MC) = (OB/OD)

taking reciprocal on both sides 

 (CM/MB) = (OD/OB) ----(2)

 (OC/OA) =(OD/OB)

 (OC/OD) = (OA/OB)

 hence proved.


(8) Using theorem 6.2 prove that the line joining the midpoint of any two sides of a triangle is parallel to the third side

∴ AD=DB

⇒ AD/BD = 1 ---- (i)

Also, E is the mid-point of AC (Given)
∴ AE=EC

⇒AE/EC = 1 [From equation (i)]

From equation (i) and (ii), we get

AD/BD = AE/EC

Hence, DE || BC [By converse of Basic Proportionality Theorem]


(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)


In ΔADC, we have
OE || DC (By Construction)

∴ AE/ED = AO/CO  ...(i) [By using Basic Proportionality Theorem]

In ΔABD, we have
OE || AB (By Construction)

∴ DE/EA = DO/BO ...(ii) [By using Basic Proportionality Theorem]

From equation (i) and (ii), we get
AO/CO = BO/DO

⇒  AO/BO = CO/DO





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