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Theoretical proof:
Given: The triangle ABC and the parallelogram BCDE are on the same base BC and between the same parallel lines BC and AD.
TO prove: Area ofΔABC = \(\frac{1}{2}\) BCDE
Construction: Draw AM⊥CB and EN⊥BC then AM is the height ofΔABC and EN be the height of BCDE.
S.N. | Statement | S.N. | Reasons |
1. | Area of BCDE = base *height = BC * EN | 1. | Area of BCDE = base * height |
2. | Area of ABC = \(\frac{1}{2}\) * BC * AM | 2. | Area ofΔ =\(\frac{1}{2}\) base * height |
3. | Area of ABC = \(\frac{1}{2}\) * BC * EN | 3. | Perpenicular distance between the same parallel lines are equal, i.e. AM = EN |
4. | Area of ABC = \(\frac{1}{2}\) of BCDE | 4. | From the statement (1) BC * EN = Area of BCDE |
Therefore, the area of the triangle is equal to half of the area of the parallelogram on the same base and between the same parallel lines.
.
Two circles with centre at O and radius more than 3 cm are drawn both angles are standing on the same arc PQ.
To verify:
\(\angle\)POQ = 2\(\angle\)PRQ
Experimental table:
Measure the angle by a protractor and write the value these anles on the following table:
Fig. No. | \(\angle\)POQ | \(\angle\)PRO | Results |
1 | 60° | 30° | \(\therefore\) \(\angle\)POQ = 2 \(\angle\)PRQ |
2 | 80° | 40° |
Conclusion:
\(\angle\)POQ = 2\(\angle\)PRQ so the centre angle is equal to the double of the angle at the circumference both stnding on the same arc of a circle.
Two circles with centre O and different radii are rawn.
The cyclic quadrilateral ABCD is drawn.
To verify:
\(\angle\)A + \(\angle\)C = 180° and \(\angle\)B + \(\angle\)D = 180°
Experimental table:
Measure the angles by protractor and write the values on the following table:
Fig. No. | \(\angle\)A | \(\angle\)C | \(\angle\)B | \(\angle\)D | Results |
1 | 130° | 50° | 20° | 60° | \(\therefore\)\(\angle\)A +\(\angle\)C = 180° \(\angle\)B +\(\angle\)D = 180° |
2 | 120° | 60° | 80° | 100° |
Conclusion:
\(\angle\)A + \(angle\)C = 180° and \(\angle\)B + \(\angle\)D = 180° so the opposite angles of a cyclic quadrilateral are supplementry.
Two circles with different radii is dran and the centre of each circle is named as O.
In whih \(\angle\)AOB = \(\angle\)COD
To verify:
\(\widehat{AB}\) = \(\widehat{CD}\)
Experimental table:
Measure the arcs AB and CD with the help of thread and ruler and tabulate the results:
Fig. No. | arc AB | arc CD | Results |
1 | 1.2 cm | 1.2 cm | arc AB = arc CD |
2 | 2.5 cm | 2.5 cm |
Conclusion:
\(\widehat{AB}\) = \(\widehat{CD}\) so two arcs of a circle subtend equal angles at the centre of the circle, the arcs are equal.
Draw two circles with radii of different measurement more than 3 cm and centre at O. Draw \(\angle\)QPR and \(\angle\)QSR t the circumference.
To verify:
\(\angle\)QPR = \(\angle\)QSR
Experimental table:
Fig. No. | \(\angle\)QPR | \(\angle\)QSR | Results |
1 | 95° | 95° | \(\angle\)QPR = \(\angle\)QSR |
2 | 60° | 60° |
Conclusion:
\(\angle\)QPR = \(\angle\)QSR so angles in the same segment of a circle are equal.
Draw two circles of different radii with centre O. Construct a cyclic quadrilateral PQRS and produced RS to T.
To verify:
\(\angle\)PST = \(\angle\)PQR
Experimental table:
Measure \(\angle\)PST and \(\angle\)PQR. The results are tabulated below:
Fig. No. | \(\angle\)PST | \(\angle\)PQR | Results |
1 | 120° | 120° | \(\angle\)PST = \(\angle\)PQR |
2 | 95° | 95° |
Conclusion:
\(\angle\)PQR = \(\angle\)PQR so one side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to thne opposite interior angle of the quadrilateral.
Two semi circle with centre at O and \(\angle\) ABC is an inscribed angle.
To verify:
\(\angle\)ABC = 90°
Experiment table:
Measure the angle by a protractor and write the value of \(\angle\)ABC on the following table:
Fig. No. | \(\angle\)ABC | Results |
1 | 90° | \(\therefore\) \(\angle\)ABC = 90° |
2 | 90° |
Conclusion:
\(\angle\)ABC = 90° so the angle in a semi circle is a right angle.
Two circles with centre at O and radius more than 3 cm are drawn where \(\widehat {AB}\) = \(\widehat {MN}\).
To verify:
\(\angle\)AOB = \(\angle\)MON
Experimental table:
Measure the centre \(\angle\)AOB and \(\angle\)MON by a protractor and write the value on the followi table:
Fig. No. | \(\angle\)AOB | \(\angle\)MON | Results |
1 | 40° | 40° | \(angle\)AOB = \(\angle\)MON |
2 | 85° | 85° |
Conclusion:
\(\angle\)AOB = \(\angle\)MON so the angles substended by equal arcs of a circle at the centre are equal.
Draw two circles of different radii with centre O.
TAN is a tangent and OA is a radius of the circle are also drawn.
To verify:
\(\angle\)OAT + \(\angle\)OAN = 90°
Experimental table:
Measure the \(\angle\)OAT and \(\angle\)OAN by protractor and write the values on the following table:
Fig. No. | \(\angle\)OAT | \(\angle\)OAN | Results |
1 | 90° | 90° | \(\therefore\) \(\angle\)OAT = \(\angle\)OAN = 90° |
2 | 90° | 90° |
Conclusion:
\(\angle\)OAN = \(\angle\)OAT = 90° so tangent to a circle is perpendicular to the radius of the circle drawn at the point of contact.
TWo circles with different radii is drawn and the centre of each circle is named as O. Draw a tangent XAY in each circle. A chord AB is drawn from the point of contact A. \(\angle\)ACB and \(\angle\)ADB are drawn in the two alternate segments in each circle.
To verify:
\(\angle\)BAY = \(\angle\)BCA and \(\angle\)BAX = \(\angle\)BDA
Experimental table:
Measure the angle by a protractor of \(\angle\)BAY, \(\angle\)BCA, \(\angle\)BAX and \(\angle\)BDA the results are tabulated below:
Fig. No. | \(\angle\)BAY | \(\angle\)BCA | \(\angle\)BAX | \(\angle\)BDA | Results |
1 | 60° | 60° | 120° | 120° | \(\therefore\) \(\angle\)BAY = \(\angle\)BCA and \(\angle\)BAX = \(\\angle\)BDA |
2 | 55° | 55° | 125° | 125° |
Conclusion:
\(\angle\)BAY = \(\angle\)BCA and\(\angle\)BAX = \(\angle\)BDA so the angles formed by a tangent to a circle and a chord drawn from the point of contact are equal to respective angles in the alternative segment.
Two circles with different radii is drawn and the centre of each circle is O. Two tangents AMP and ANC are drawn.
To verify:
MA = NA
Experimental table:
Measure the side MA and NA. The result are tabulated below:
Fig. No. | MA | NA | Results |
1 | 4 cm | 4 cm | MA = NA |
2 | 2 cm | 2 cm |
Conclusion:
MA = NA so the length of two tangents to a circle at the point of contact from the same external point are equal.
Draw two triangles ABC and DEF both are standing on equal base BC and EF between parallel line AD and BF. In each triangle draw AP⊥ BC and DQ⊥ Ef.
To verify:
Area of \(\triangle\)ABC = Area of \(\triangle\)DEF
Experiment table:
Measure the BC and AP, EF and DQ. The results are tabulated below:
Fig. No. | BC | AP | EF | DQ | Area of \(\triangle\)ABC | Area of \(\triangle\)DEF | Results |
1 | 3 cm | 2.5 cm | 3 cm | 2.5 cm | 3.75 cm^{2} | 3.75 cm^{2} | Area of \(\triangle\)ABC = Area of \(\triangle\)DEF |
2 | 1 cm | 2.5 cm | 1 cm | 2.5 cm | 1.25 cm^{2} | 1.25 cm^{2} |
Conclusion:
Area of \(\triangle\)ABC = Area of \(\triangle\)DEF so triangles on equal bases and between the same parallels are equal in area.
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