#### 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 PartGiven &#x2206ABP = Equilateral triangleR.T.P.1) &#x2206APD ≅ &#x2206BPC    2) Find &#x2220DPC, &#x2220PDC and &#x2220PCD, Proof: In &#x2206APD and &#x2206BPC AP = BP | Same side of

#### 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

#### 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.

#### 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

#### 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

#### 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 = 1800 Construction : Join OB and OD Proof: 1. &#x2220 BOD = 2 ∠BAD || Angle at the centre of a circle is

#### 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