is a tangent to the circle with centre O to B. AB is a chord of length 24 cm at a distance of 5 cm from the centre. If the tangent is of length 20 cm, find the length of PO.Read more on Sarthaks.com – https://www.sarthaks.com/176103/pb-is-tangent-the-circle-with-centre-is-chord-of-length-24-cm-at-distance-of-cm-from-the-centre
OM=5~cm MSESHSEOM LAB and M is mid-point of AB, = 12 cm. In right-angled triangle A OMB, OB^{2}=OM^{2}+MB^{2} +OB^{2}=5^{2}+12^{2} OB^{2}=25+144 OB^{2}=169 \Rightarrow OB=169~cm OB=13~cm. As BP is tangent to circle at B, OB 1 BP. In right-angled triangle ▲ OBP, OP^{2}=OB^{2}+BP^{2} OP^{2}=13^{2}+20^{2} *OP=\sqrt{}569~cm Hence, the length of OP=\sqrt{}569~cm.
A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Length of the vertical pole = 6m (Given)Length of the shadow of the pole = 4 m (Given)Let Height of tower = h mLength of shadow of the tower = 28 m (Given)In ΔABC and ΔDEF,∠C=∠E(angular elevation)∠B=∠F=90∘∴ΔABC∼ΔDFE (By AAA similarity criterion)∴ABDF=BCEF (If two triangles are similar then their corresponding sides are proportional.)∴6h=428⇒h=6×284⇒h=6×7⇒h=42mHence, the height of the tower is
To prove that tangents PQ and PR are equal in length without any extra constructions, we can utilize the properties of tangents to a circle.
Tangents PQ and PR to a circle with center O. To prove: PQ = PR. Proof: OQ is common to both triangles:* OQ is the radius of the circle, and it is common to both triangles. OR is the radius of the circle, and it is common to both triangles. The common side OQ and