Let AB be the height of rod tip from the surface of water and BC is the horizontal distance between fly to tip of the rod. Then, the lenght of the string is AC. In ΔABC, by pythagoras theorem AC² = AB² + BC² ⇒ AB² = (1.8m)² + (2.4m)² ⇒ AB² = (3.24 + 5.76) m² ⇒ AB² = 9.00 m² ⇒ AB = √9 = 3 m Hence, thRead more
Let AB be the height of rod tip from the surface of water and BC is the horizontal distance between fly to tip of the rod.
Then, the lenght of the string is AC.
In ΔABC, by pythagoras theorem
AC² = AB² + BC²
⇒ AB² = (1.8m)² + (2.4m)²
⇒ AB² = (3.24 + 5.76) m²
⇒ AB² = 9.00 m²
⇒ AB = √9 = 3 m
Hence, the lenght of string, which is out, is 3 m.
If she pulls in the string, at the rate of 5 cm/s, then the distance travelled by fly in 12 seconds
= 12 × 5 = 60 cm = 0.6 m
Let, Dbe the position of string that is out after 12 seconds.
The lenght of the string pulls in by Nazima = AD = AC – 12
= (3.00 – 0.6) m
= 2.4 m
In ΔADB,
AB² + BD² = AD²
⇒ (1.8 m)² + BD² = (2.4 m)²
⇒ BD² = (5.76 – 3.24) m² = 2.52m²
⇒ BD = 1.587 m
Horizontal distance travelled by fly
= BD + 1.2 m
= (1.587 + 1.2) m
= 2.787 m
= 2.79 m.
A line RT is drawn parallel to SP, which intersects QP produced at T. Given that, SP bisects angle QPR, therefore ∠QPS = ∠SPR ...(1) By construction, ∠SPR = ∠PRT(As PS II TR) ...(2) ∠QPS = ∠QTR(As PS II TR) ...(3) From the above equations, we have ∠PRT = ∠QTR ∴ PT = PR By construction, PS II TR In ΔRead more
A line RT is drawn parallel to SP, which intersects QP produced at T.
Given that, SP bisects angle QPR, therefore
∠QPS = ∠SPR …(1)
By construction,
∠SPR = ∠PRT(As PS II TR) …(2)
∠QPS = ∠QTR(As PS II TR) …(3)
From the above equations, we have
∠PRT = ∠QTR
∴ PT = PR
By construction, PS II TR
In ΔQTR, by Thales theorem
QS/SR = QP/PT ⇒ QS/SR = PQ/PR [∵ PT = TR]
Join CB. (i) In ΔAPC and DPB, ∠APC = ∠DPB [Vertically Opposite Angles] ∠CAP = ∠BDP [Angles in the some segment] ΔAPC ∼ Δ DPB [ AA similarity] (ii) We have already proved that ΔAPC ∼ ΔDPB. We know that the corresponding sides of similar triangle are proportional. So, AP/DP = PC/PB = CA/BD ⇒ AP/DP = PRead more
Join CB.
(i) In ΔAPC and DPB,
∠APC = ∠DPB [Vertically Opposite Angles]
∠CAP = ∠BDP [Angles in the some segment]
ΔAPC ∼ Δ DPB [ AA similarity]
(ii) We have already proved that ΔAPC ∼ ΔDPB.
We know that the corresponding sides of similar triangle are proportional.
So, AP/DP = PC/PB = CA/BD ⇒ AP/DP = PC/PC ⇒ AP. PB = PC . DP
How many tangents can a circle have?
A Circle can have infinite number of tangents because a circle have infinite number of points on it and at every point a tangent can be drawn.
A Circle can have infinite number of tangents because a circle have infinite number of points on it and at every point a tangent can be drawn.
See lessNazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Let AB be the height of rod tip from the surface of water and BC is the horizontal distance between fly to tip of the rod. Then, the lenght of the string is AC. In ΔABC, by pythagoras theorem AC² = AB² + BC² ⇒ AB² = (1.8m)² + (2.4m)² ⇒ AB² = (3.24 + 5.76) m² ⇒ AB² = 9.00 m² ⇒ AB = √9 = 3 m Hence, thRead more
Let AB be the height of rod tip from the surface of water and BC is the horizontal distance between fly to tip of the rod.
See lessThen, the lenght of the string is AC.
In ΔABC, by pythagoras theorem
AC² = AB² + BC²
⇒ AB² = (1.8m)² + (2.4m)²
⇒ AB² = (3.24 + 5.76) m²
⇒ AB² = 9.00 m²
⇒ AB = √9 = 3 m
Hence, the lenght of string, which is out, is 3 m.
If she pulls in the string, at the rate of 5 cm/s, then the distance travelled by fly in 12 seconds
= 12 × 5 = 60 cm = 0.6 m
Let, Dbe the position of string that is out after 12 seconds.
The lenght of the string pulls in by Nazima = AD = AC – 12
= (3.00 – 0.6) m
= 2.4 m
In ΔADB,
AB² + BD² = AD²
⇒ (1.8 m)² + BD² = (2.4 m)²
⇒ BD² = (5.76 – 3.24) m² = 2.52m²
⇒ BD = 1.587 m
Horizontal distance travelled by fly
= BD + 1.2 m
= (1.587 + 1.2) m
= 2.787 m
= 2.79 m.
PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/SR = PQ/PR.
A line RT is drawn parallel to SP, which intersects QP produced at T. Given that, SP bisects angle QPR, therefore ∠QPS = ∠SPR ...(1) By construction, ∠SPR = ∠PRT(As PS II TR) ...(2) ∠QPS = ∠QTR(As PS II TR) ...(3) From the above equations, we have ∠PRT = ∠QTR ∴ PT = PR By construction, PS II TR In ΔRead more
A line RT is drawn parallel to SP, which intersects QP produced at T.
See lessGiven that, SP bisects angle QPR, therefore
∠QPS = ∠SPR …(1)
By construction,
∠SPR = ∠PRT(As PS II TR) …(2)
∠QPS = ∠QTR(As PS II TR) …(3)
From the above equations, we have
∠PRT = ∠QTR
∴ PT = PR
By construction, PS II TR
In ΔQTR, by Thales theorem
QS/SR = QP/PT ⇒ QS/SR = PQ/PR [∵ PT = TR]
Two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (i) ∆ PAC ~ ∆ PDB (ii) PA . PB = PC . PD
Join CB. (i) In ΔAPC and DPB, ∠APC = ∠DPB [Vertically Opposite Angles] ∠CAP = ∠BDP [Angles in the some segment] ΔAPC ∼ Δ DPB [ AA similarity] (ii) We have already proved that ΔAPC ∼ ΔDPB. We know that the corresponding sides of similar triangle are proportional. So, AP/DP = PC/PB = CA/BD ⇒ AP/DP = PRead more
Join CB.
(i) In ΔAPC and DPB,
∠APC = ∠DPB [Vertically Opposite Angles]
∠CAP = ∠BDP [Angles in the some segment]
ΔAPC ∼ Δ DPB [ AA similarity]
(ii) We have already proved that ΔAPC ∼ ΔDPB.
See lessWe know that the corresponding sides of similar triangle are proportional.
So, AP/DP = PC/PB = CA/BD ⇒ AP/DP = PC/PC ⇒ AP. PB = PC . DP
ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² –2 BC . BD.
In ΔADB, by pythagoras theorem AD² + DB² = AB² ⇒ AD2 = AB2 - DB2 ...(1) In ΔADC, by pythagoras theorem, AD² + DC² = AC² ⇒ AB² - BD² + DC² = AC² [From the equation (1)] ⇒ AB² - BD² + (BC - BD)² = AC² ⇒ AC² = AB² - BD² + BC² + BD² - 2BC × BD = AB² + BC² - 2BC × BD
In ΔADB, by pythagoras theorem
See lessAD² + DB² = AB²
⇒ AD2 = AB2 – DB2 …(1)
In ΔADC, by pythagoras theorem, AD² + DC² = AC²
⇒ AB² – BD² + DC² = AC² [From the equation (1)]
⇒ AB² – BD² + (BC – BD)² = AC²
⇒ AC² = AB² – BD² + BC² + BD² – 2BC × BD = AB² + BC² – 2BC × BD