Question

answer question number 14 and 15
also the question that which are the important questions from the figure above that can be asked in exam at least 8
the best one will be marked as brainlist

Answer 1

Question 14

AB║DE

and, D is the mid point of BC

E is the mid point of AC . Because we know that ,if a straight line is drawn parallel to BC through the mid point of AB , then it bisects the side AC.

Now, we can say that BE is the median.

Answer 2

(14)

In ΔABC,

⇒ AD is the median.

⇒ D is the mid-point of BC.

Given, DE || AB.

By the converse of mid point theorem,

E is the mid-point of AC.

⇒ BE is also a median.

(15).

Let the length of the diagonals be x and y.

We know that Diagonal of a rhombus bisect each other and perpendicular to each other.

Using Pythagoras theorem, we get

⇒ (x/2)^2 + (y/2)^2 = x^2

⇒ (y/2)^2 = (x)^2 - (x/2)^2

⇒ (y/2)^2 = x^2 - x^2/4

⇒ y^2/4 = 3x^2/4

⇒ y^2 = 3x^2

⇒ y = √3x.

Therefore, x/y = √3/1

                  x : y = √3  :1.

Hope it helps!