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In DABC and DCDA, we have A 2 B
1
∠2 = ∠4 (Alternate angles as AB||CD)
∠3 = ∠1 (Alternate angles as BC ||AD) 4 3
D C
AC = AC (Common)
So, DABC ≅ ∠CDA (By ASA Criterion)
\ AB = DC, BC = AD

and ∠B = ∠D (By CPCT)
If we have drawn the diagonal BD, we can also prove the following:

AB = DC, BC = AD and ∠A = ∠C
Thus, by combining the above two results, we get that in a parallelogram, opposite sides and
opposite angles are equal.
Also, each diagonal of a parallelogram divides it into two congruent triangles.

Example 15: Find the perimeter of the parallelogram PQRS.
Solution: In a parallelogram, the opposite sides have same length. S R

Therefore, PQ = SR = 12 cm and QR = PS = 7 cm 7 cm
So, Perimeter of parallelogram PQRS = PQ + QR + RS + SP P 12 cm Q

= 12 cm + 7 cm + 12 cm + 7 cm = 38 cm
Example 16: The ratio of two adjacent sides of a parallelogram is 4 : 3 and its perimeter is 56 cm,
find the lengths of its adjacent sides.
Solution: Let the lengths of two adjacent sides of the parallelogram be 4x cm and 3x cm respectively.

Therefore, the perimeter of the parallelogram = 2(4x + 3x)cm = 8x + 6x = 14x cm
Since, the perimeter of the parallelogram is 56 cm.

Therefore, 14x = 56
56
x = = 4
14
Thus, the length of adjacent sides of the parallelogram are: (4 × 4) cm = 16 cm and (3 × 4) cm = 12 cm.
Example 17: Find the angles a, b and c in the parallelogram TUVW T U
as shown in the adjoining figure. a
Solution: In a parallelogram, the opposite angles are equal, that is,
∠W = ∠U.
95°
So, 95° = ∠a W b c
V
Now ∠a = ∠c (Alternate angles as TU || WV and UV is a transversal)

So, ∠c = 95°
Now, ∠b + ∠c = 180° (Linear pair of angles)
So, ∠b + 95° = 180°

⇒ ∠b = 180° – 95° = 85°
Thus, ∠a = 95°, ∠b = 85° and ∠c = 95°.

Mathematics-8 68