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Showing posts from April, 2015

### Geometry Sum | ABCD is a square. Prove in each case ∆APD ≅ ∆BPC | Find Angle DPC Angle PCD and Angle PCD |

ABCD is a square and APB is an equilateral triangle.
Prove in each case:
1) &#x2206APD ≅ &#x2206BPC
2) Find the angles of &#x2206DPC Solution :
First Part
Given &#x2206ABP = Equilateral triangle
R.T.P.
1) &#x2206APD ≅ &#x2206BPC
2) Find &#x2220DPC, &#x2220PDC and &#x2220PCD,
Proof: In &#x2206APD and &#x2206BPC

AP = BP | Same side of the equilateral triangle.
AD = BC | Same side of the Square.
and
&#x2220DAP = &#x2220 DAB - &#x2220PAB = 90°-60°=300
Similarly
&#x2220BPC = &#x2220 ABC - &#x2220ABP = 90°-60°=30°
∴ &#x2220DAP = &#x2220BPC
∴ &#x2206APD ≅ &#x2206BPC SAS_Proved
2nd Part
In &#x2206APD
AP=AD || As AP = AB (Equilateral Triangle).
We know that &#x2220DAP = 30 °
∴ &#x2220APD = (180°- 30°)/2 || &#x2206APD is an isoscless &#x2206 and &#x2220APD is one of the base angle.
or &#x2220APD = 150°/2
or …

### Sums of Speed Distance and Time Part 3 | Mathematics Class 8

Three sums of speed distance and time are discussed in the video. The sums are given here.     1. A train 150 m long passes a telegrah post in 15 seconds.
Find:
a) Speed of the train in km/hr.
b) Time taken by it to pass a plateform of 150 m long.
Watch Solution in the Video     2. 2. A person is walking parallel to a railway track, at a speed of 5 km/hr. A train , travelling in the same direction at speed of 50 km/hr, passes him in 16 seconds.
Find the length of the train.
Watch Solution in the Video     3. Two trains left Howra station for Chennai at different intervals. First trains left Howra station at 2 pm at a seed of 60 km/hr. Another train left Howra at 3:30 pm at a seed of 75km/hr.
At what time the second train will overtake the first train
and
at what distance from Howra it will overtake the first train.
Watch Solution in the Video

### Speed Distance and Time For Class 8 | Part 2 Mathematics | Relative Speed of Boat

In this video sums related to relative speed of boat in standstill water stream and in a water stream having speed in the same direction and also in the opposite direction have been discussed. Watch the video and comment below. If you have sum of boat and stream then post it in the comment box. Sum No-1: Speed of a boat in still water is 9 km per hr. If the boat goes 54 km downstream in 4 hours.; find the the speed of the stream.
Sum No-2: The speed of a boat in still water is 10 km per hour. If the boat goes 35 km upstream in 5 hrs. ; find the speed of the stream.

### Theorem: Tangent and Radius of a Circle | are Perpendicular to Each Other.

Theorem :Tangent and radius of a circle are perpendicular to each other.Solution :
Given:
A circle with centre O.
AB is the tangent to the circle at point B
and OB is the radius of the circle.
R.T.P. OB &#x22A5 AC
Proof: 1.  OB < OC || Since each point of the tangent other than point B is outside the circle.

2.  Similarly it can be shown that out of all the line segments which would be drawn from point O to the tangent line AC, OB is the shortest.

3.   OB &#x22A5 AC || The shortest line sgement drawn from a given point to a given line , is &#x22A5 to the line.
Proved

### Theorem : The Angle in a Semi-Circle is a Right Angle

Theorem: The angle in a semicircle is a right angle Given : A circle with centre 0 with  &#x2220A0B at centre and  &#x2220ACB at the circumference of the circle .
RT.P. : the angle in a semicircle is a right angle. ie : &#x2220 ACB= 90o
Statements: 1   &#x2220AOB = 2&#x2220ACB  ||  angle at the centre is twice the angle at the circumference  of a circle.
2 . &#x2220ACB = ½&#x2220AOB  ||  by statement 1
3 . &#x2220ACB =½ x 180o  || As  &#x2220 AOB = 180o
4. &there4 &#x2220ACB = 90o Proved

### Theorem : Opposite Angles of a Cyclic Quadrilateral are Supplementary | Class 8

Theorem : Opposite Angles of a Cyclic Quadrilateral are Supplementary.Solution :
Given: ABCD is a cyclic quadrilateral.
R.T.P.(Require To Prove)1. &#x2220BAD + &#x2220BCD = 1800
2. &#x2220ABC + &#x2220ADC = 1800Construction : Join OB and OD
Proof:1. &#x2220 BOD = 2 ∠BAD || Angle at the centre of a circle is twice the angle at the circumference.
2.  Reflex &#x2220 BOD + &#x2220 BOD = 2 (&#x2220BAD + &#x2220BCD) || Same as stated above.
or   2 (&#x2220BAD + &#x2220BCD) = 3600 || By statement 2.
or   (&#x2220BAD + &#x2220BCD) = 1800
3. Similarly by joining OA and OC it can be proved that
4.   ∴ &#x2220BAD + &#x2220BCD = 1800
&      &#x2220ABC + &#x2220ADC = 1800 Proved

### Theorem | the angle subtended by an arc of a circle | at the centre is double the angle subtended by it at | any point on the remaining part of the circle

Theorem : A circle  with centre O in which AB subtends  &#x2220AOB at centre and angle &#x2220ACB at any point  on the remaining part of the circle.
R.T.P. - &#x2220AOB =  2 x &#x2220ACB Solution :
Construction: Join CO and produce CO to point D . 1. In ∆AOC, OC = OA || Radii of same circle .
2. ∠OCA= ∠OAC || Base angles of isoscles &#x2206 in which OC = OA
3. Exterior &#x2220 AOD = ∠OCA+ ∠OAC || Exterior &#x2220 of a &#x2206 = sum of two opposite interior angles.
or &#x2220AOD = ∠OCA+ &#x2220OCA
or &#x2220AOD = 2 . &#x2220 OCA - - - - Eqn. 1
4. Similary ot can be proved that &#x2220BOD = 2. ∠OCB - - - - - - Eqn. 2
5. Adding Eqn 1 and Eqn. 2 we get
&#x2220AOD + &#x2220BOD = 2 . &#x2220OCA + 2. &#x2220OCB
&#x2220AOB = 2 (&#x2220OCA + &#x2220OCB)
∴ &#x2220AOB = 2 ∠ACB Proved