On the basis of given in question , we draw a figure of a circle with centre O in which T is an external point where OT = 13 cm and OT intersects the circle at point E and AB is a tangent at point E

∠OPT = 90°
In ∆OPT,
OT² = OP² + PT²
⇒ 13² = 5² + PT²
⇒ PT² = 169 – 25 = 144
⇒ PT = √144 = 12 cm.
Tangents drawn from an external point on a circle are equal.
∴ Let AP = AE = y
⇒ AT = PT – AP = (12 – y) cm
OE ⊥ AB
⇒ ∠OEA = 90°
⇒ ∠ AET = 90°
∴ AT² = AE² + ET²
⇒ (12 – y)² = y² + (13 – 5)²
⇒ 144 – 24y + y² = y² + 64
⇒ 24y = 80

Correct Option: B

On the basis of given in question , we draw a figure of a circle with centre O in which T is an external point where OT = 13 cm and OT intersects the circle at point E and AB is a tangent at point E

∠OPT = 90°
In ∆OPT,
OT² = OP² + PT²
⇒ 13² = 5² + PT²
⇒ PT² = 169 – 25 = 144
⇒ PT = √144 = 12 cm.
Tangents drawn from an external point on a circle are equal.
∴ Let AP = AE = y
⇒ AT = PT – AP = (12 – y) cm
OE ⊥ AB
⇒ ∠OEA = 90°
⇒ ∠ AET = 90°
∴ AT² = AE² + ET²
⇒ (12 – y)² = y² + (13 – 5)²
⇒ 144 – 24y + y² = y² + 64
⇒ 24y = 80

As per the given figure in question ,

In ∆ OCD,
OC = OD = CD
∴ ∆ OCD is an equilateral triangle.
As we know that each angle of an equilateral triangle is 60° .
∴ ∠COD = 60°
Now, ∠CBD = ( 1/2 )∠COD
⇒ ∠CBD = 30°
∠ACB is angle of semi-circle.
∴ ∠ACB = 90°
⇒ ∠ BCE = 180° – ∠ACB = 180° – 90° = 90°
In ∆ BCE,

Correct Option: B

As per the given figure in question ,

In ∆ OCD,
OC = OD = CD
∴ ∆ OCD is an equilateral triangle.
As we know that each angle of an equilateral triangle is 60° .
∴ ∠COD = 60°
Now, ∠CBD = ( 1/2 )∠COD
⇒ ∠CBD = 30°
∠ACB is angle of semi-circle.
∴ ∠ACB = 90°
⇒ ∠ BCE = 180° – ∠ACB = 180° – 90° = 90°
In ∆ BCE,
∠BCE = 90°, ∠CBE = ∠CBD = 30°
∴ ∠BCE + ∠CBE +∠CEB = 180°
⇒ 90° + 30° + ∠CEB = 180°
⇒ ∠CEB = 180° – 120° = 60°
⇒ ∠ AEB = 60°

On the basis of given figure in question ,

Given that , ∠ ABC = 69°,∠ ACB = 31°
In ∆ ABC,
As we know that ,
∠BAC + ∠ABC + ∠ACB = 180°
⇒ ∠BAC + 69° + 31° = 180°

Correct Option: A

On the basis of given figure in question ,

Given that , ∠ ABC = 69°,∠ ACB = 31°
In ∆ ABC,
As we know that ,
∠BAC + ∠ABC + ∠ACB = 180°
⇒ ∠BAC + 69° + 31° = 180°
⇒ ∠ BAC = 180° – 100° = 80°
Since angles in the same segment are equal.
∴ ∠BDC = 80°

According to question , we draw a figure of two circles whose radii are 10 cm and 8 cm, intersect each other and their common chord is 12 cm long

OP = 10 cm, O’P = 8 cm and PQ = 12 cm ;

Correct Option: D

According to question , we draw a figure of two circles whose radii are 10 cm and 8 cm, intersect each other and their common chord is 12 cm long

OP = 10 cm, O’P = 8 cm and PQ = 12 cm ;

On the basis of given in question , we draw a figure of triangle ABC whose AD, BE and CF are medians

We know that the sum of any two sides of a triangle is greater than twice the median bisecting the third side.
AD is the median on BC.
∴ AB + AC > 2AD .....( 1 )
Similarly,
AB + BC > 2BE .....( 2 )
BC + AC > 2CF .....( 3 )
On adding all three equations, we get

Correct Option: B

On the basis of given in question , we draw a figure of triangle ABC whose AD, BE and CF are medians

We know that the sum of any two sides of a triangle is greater than twice the median bisecting the third side.
AD is the median on BC.
∴ AB + AC > 2AD .....( 1 )
Similarly,
AB + BC > 2BE .....( 2 )
BC + AC > 2CF .....( 3 )
On adding all three equations, we get
AB + AC + AB + BC + BC + AC > 2AD + 2BE + 2CF
⇒ 2 (AB + BC + AC) > 2 (AD + BE + CF)
⇒ AB+BC+AC > AD + BE + CF