Step 1
Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5 cm radius respectively.
Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.
Step 2
Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
Step 3
Locate 7 points, A₁, A₂, A₃, A₄, A₅, A₆, A₇, (as 7 is greater between 5 and 7), on line AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.
Step 4
Join BA₅ and draw a line through A₇ parallel to BA₅ to intersect extended line segment AB at point B’.
Step 5
Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’ ΔAB’C’ is the required triangle.
Justification
The construction can be justified by proving that
AB’ = 7/5 AB, B’C’ = 7/5 BC, AC’ = 7/5 AC
In ΔABC and Δ AB’C’,
∠ABC = ∠AB’C’ (Corresponding angles)
∠BAC = ∠B’AC’ (common)
∴ ΔABC ∼ ΔAB’C’ (AA similarity Criterion)
⇒ AB’/AB = B’C’/BC = AC’/AC …(1)
In ΔAA₅B/AA₇B’,
∠A₅AB = ∠A₇AB’ (common)
∠AA₅B = ∠AA₇B'(Corresponding angles)
∴ ΔAA₅B ∼ ΔAA₇B’ (AA similarity criterion)
⇒ AB’/AB = AA₅/AA₇ ⇒ AB’/AB = 5/7 …(2)
On comparing equations (1) and (2), we obtain.
AB’/AB = B’C’/BC = AC’/AC = 7/5
⇒ AB’ = 7/5 AB, B’C’ = 7/5 BC, AC’ = 7/5 AC
This justifies the construction.

Let O be the centre of the circle.
Given: OQ = 25 cm and PQ = 24 CM
We Know that the radius is perpendicular to tangent. Therefore, OP ⊥PQ
In ΔOPQ, by pythagoras theorem,
OP² + 24² = 25² ⇒ OP² = 625 – 576 ⇒ OP² = 49 ⇒OP = 7
Therefore, the radius of circle is 7 cm. Hence, the option (A) is correct.

Given: PA and PB are two tangents of the circle.
We know that the radius is perpendicular to tangent. Therefore, OA ⊥PA and OB PB.
⇒ ∠OBP = 90° and ∠OAP = 90°
In quadrilaterral AOBP, ∠OAP + ∠APB + ∠PBO + ∠BOA = 360°
⇒ 90° + 80° + 90° + ∠BOA = 360°
⇒ ∠BOA = 360° – 260° = 100°
In ΔOPB and ΔOPA,
AP = BP [Tangents drawn from same external point]
OA = OB [Radii]
OP = OP [Common]
Therefore, ΔOPB ≅ ΔOPA [SSS Congruency rule]
Hence, ∠POB = ∠POA
∠POA = 1/2 ∠AOB = 1/2(100°) = 50°. Hence, the option (A) is correct.

Let, O be the centre of circle and AB is tangent at P.
We have to prove that the perpendicular at P to AB, passes through O.
Let the Perpendicular drawn at P point of AB does not pass throught O.
It passes through O’. Join and O’P.
Tangent drawn at P passes throught O’ therefore,
∠O’PB = 90° … (1)
We know that the radius is perpendicular to tangent.
Therefore, ∠OPB = 90° … (2)
Comparing equation (1) and (2), we have
∠O’PB = ∠OPB … (3)
From figure, it is clean that,
∠O’PB < ∠OPB … (4)
Therefore, ∠O'PB = ∠OPB is not possible. It is possible only when OP and O'P coincident lines.
Therefore, the perpendicular drawn at P passes through the centre O.