KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.2

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.2

Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangle Exercise 2.2

Question 1.
In the following fig. (i) and (ii). DE || BC.
Find EC in (i) and AD in (ii).

Question 8.
Using Theorem 2.2, 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)

Solution:
Solution: In ∆ABC, D and E are mid-points of AB and AC.
∴ AD = DB
AE = EC
AB = 2AD
AC = 2AE
AD/AB=AE/AC=1/2
As per S.S.S. Postulate.
∆ADE ~ ∆ABC
∴ They are equiangular triangles.
∴ ∠A is common.
∠ADE = ∠ABC
∠ADE = ∠ACE
These are pair of corresponding angles
∴ DE || BC

Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/CO=CO/DO
Solution:
Data : ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.
To Prove: AO/CO=CO/DO
Proof: In Trapezium ABCD. AB || DC.
∴ In ∆AOB and ∆DOC,
∠OCO = ∠OAB (alternate angles)

∠ODC = ∠OBA (alternate angles)
∠DOC = ∠AOB (vertically opposite angles)
∴ ∆AOB and ∆DOC are equiangular triangles.
∴ ∆AOB ||| ∆DOC
Similar triangles divides sides in ratio.
AO/CO=CO/DO

Question 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:
Data: In the quadrilateral ABCD. the diagonals intersect at 0’ such that AO/BO=CO/DO
To Prove: ABCD is a trapezium.
Solution: In the qudrilateral ABCD, AO/BO=CO/DO
AO/CO=OB/OD
It means sides of ∆AOB and ∆DOC divides proportionately.
∴ ∆AOB ||| ∆DOC.
Similarly. ∆AOD ||| ∆BOC.
Now, ∆AOB + ∆AOD = ∆BOC + ∆DOC
∆ABD = ∆ABC.
Both triangles are on the same base AB and between two pair of lines and equal in area.
∴ AB || DC.

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