Mohan has two triangular plots connected to each other as shown in the figure. He thought to give the bigger isosceles triangular part with AC =BC to his daughter and son equally. (i)How a triangle can be divided into two parts of equal area?(ii) Also find the value of angleDCE.(iii) What values he is exibiting by doing so?
Answers
Answer:
Step-by-step explanation:
ABC is isosceles triangle with AC = BC
so to divide into two parts of equal area
we can draw a perpendicular bisector of AB
which will pass through point C as AC = BC
Let say perpendicular bisector of AB bisect at O
then Area of Δ AOC = Area of Δ BOC = (1/2) Area of Δ ABC
∠DCE = ∠ACB ( opposite angle)
∠BAC = ∠ABC ( as AC = BC)
=> ∠BAC = 48°
∠ABC + ∠BAC + ∠ACB = 180° ( Sum of Angles of a Triangle)
=> 48° + 48° + ∠ACB = 180°
=> ∠ACB = 84°
∠DCE = 84°
Answer:
Step-by-step explanation:
Draw perpendicular bisector of AB
∠DCE = 84°
Step-by-step explanation:
ABC is isosceles triangle with AC = BC
so to divide into two parts of equal area
we can draw a perpendicular bisector of AB
which will pass through point C as AC = BC
Let say perpendicular bisector of AB bisect at O
then Area of Δ AOC = Area of Δ BOC = (1/2) Area of Δ ABC
∠DCE = ∠ACB ( opposite angle)
∠BAC = ∠ABC ( as AC = BC)
=> ∠BAC = 48°
∠ABC + ∠BAC + ∠ACB = 180° ( Sum of Angles of a Triangle)
=> 48° + 48° + ∠ACB = 180°
=> ∠ACB = 84°
∠DCE = 84°