We know that ,
The sum of the two sides of a triangle is greater than the third side.
∴ 5 + x > 9
⇒ x > 9 – 5
⇒ x > 4

Correct Option: D

We know that ,
The sum of the two sides of a triangle is greater than the third side.
∴ 5 + x > 9
⇒ x > 9 – 5
⇒ x > 4
∴ Required answer = 5

As per the given in question , we draw a figure triangle ABC

Let Angles of triangle = y°, 2y° and 3y°
∴ y + 2y + 3y = 180°
⇒ 6y = 180°
⇒ y = 30°
∴ Angles of triangle = 30°, 60° and 90°
∠ABC = 90°
∠BAC = 30°
∴ BC = 10 cm.

Correct Option: A

As per the given in question , we draw a figure triangle ABC

Let Angles of triangle = y°, 2y° and 3y°
∴ y + 2y + 3y = 180°
⇒ 6y = 180°
⇒ y = 30°
∴ Angles of triangle = 30°, 60° and 90°
∠ABC = 90°
∠BAC = 30°
∴ BC = 10 cm.

On the basis of question we draw a figure of triangle ABC ,

From ∆ ACD,
Given , AD = 8 cm.
CD = 6 cm.
∠CDA = 90°
∴ AC = √AD² + DC²
AC = √8² + 6²
AC = √64 + 36
AC = √100 = 10 cm
The straight line joining the midpoints of two sides of a triangle is parallel to the third side and half of third side.

Correct Option: D

On the basis of question we draw a figure of triangle ABC ,

From ∆ ACD,
Given , AD = 8 cm.
CD = 6 cm.
∠CDA = 90°
∴ AC = √AD² + DC²
AC = √8² + 6²
AC = √64 + 36
AC = √100 = 10 cm
The straight line joining the midpoints of two sides of a triangle is parallel to the third side and half of third side.

As we know that the sum of two sides of a triangle is greater than the third side.
Clearly,
(8 + 15) > 17
(15 + 17) > 8
(8 + 17) > 15

Correct Option: B

As we know that the sum of two sides of a triangle is greater than the third side.
Clearly,
(8 + 15) > 17
(15 + 17) > 8
(8 + 17) > 15
Therefore , Triangle possible segments are 8 cm, 15 cm and 17 cm .

According to question ,
The circumcentre of a triangle is equidistant from its vertices.

Correct Option: B

According to question ,
The point equidistant from the vertices of a triangle is called circumcentre . Or
The circumcentre of a triangle is equidistant from its vertices.