draw a circle of radius 3.5 centimetre using pthread find its circumference verify the relation circumference =3.14 X diametres (
approximately)
Answers
Answer:Given :- ABCD is a trapezium
AB || CD
AD = BC
To proof :-
(i)∠A = ∠B
(ii)∠C = ∠D
(iii)∆ ABC ≅ ∆ BAD
(iv)Diagonal AC = Diagonal = BD
Construction :- Draw DA || CE
Solution :-
(i) Since it's given ABCD is a trapezium
AB || CD
DA || CE ( By construction)
Therefore, ADCE is a parallelogram
So, DA = CE &
DC = AE ( Opposite side of parallelogram are equal )
But, AD = BC
Therefore, BC = CE ( Given )
∠CEB = ∠CBE ( In ∆ CBE angles opposite to equal sides are equal )
180° - ∠DAB = 180° - ∠ABC
[ ADCE is a parallelogram and ∠A + ∠E = 180° ∠B & ∠CBE form a linear pair ]
∠A = ∠B ( Cancelling 180° from both sides)
(ii) Co interior angles on the same side of a transversal are supplementary
∠A + ∠D = 180° & ∠B + ∠C = 180°
∠A + ∠D = ∠B + ∠C
∠B + ∠D = ∠B + ∠C ( ∠A = ∠B proved above)
∠D = ∠C
(iii) In ∆ ABC & ∆ BAD
AB = BA
∠B = ∠A ( proved above )
BC = BD ( Given )
∆ ABC ≅ ∆ BAD ( By SAS criteria)
(iv) AC = BD ( CPCT )
Step-by-step explanation:Given :- ABCD is a trapezium
AB || CD
AD = BC
To proof :-
(i)∠A = ∠B
(ii)∠C = ∠D
(iii)∆ ABC ≅ ∆ BAD
(iv)Diagonal AC = Diagonal = BD
Construction :- Draw DA || CE
Solution :-
(i) Since it's given ABCD is a trapezium
AB || CD
DA || CE ( By construction)
Therefore, ADCE is a parallelogram
So, DA = CE &
DC = AE ( Opposite side of parallelogram are equal )
But, AD = BC
Therefore, BC = CE ( Given )
∠CEB = ∠CBE ( In ∆ CBE angles opposite to equal sides are equal )
180° - ∠DAB = 180° - ∠ABC
[ ADCE is a parallelogram and ∠A + ∠E = 180° ∠B & ∠CBE form a linear pair ]
∠A = ∠B ( Cancelling 180° from both sides)
(ii) Co interior angles on the same side of a transversal are supplementary
∠A + ∠D = 180° & ∠B + ∠C = 180°
∠A + ∠D = ∠B + ∠C
∠B + ∠D = ∠B + ∠C ( ∠A = ∠B proved above)
∠D = ∠C
(iii) In ∆ ABC & ∆ BAD
AB = BA
∠B = ∠A ( proved above )
BC = BD ( Given )
∆ ABC ≅ ∆ BAD ( By SAS criteria)
(iv) AC = BD ( CPCT )