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(1) Prove that a diagonal of a parallelogram divides it into two triangles of equal area.
Given:Â A parallelogram ABCD in which BD is one of the diagonals.
To prove: Â Â
Proof: Since two congruent geometrical figures have equal area. Therefore, in order to prove that it is sufficient to show that
In ABD and CDB, we have
And,
So, by SSS criterion of congruence, we have
Hence,
(2) Prove that parallelograms on the same base and between the same parallels are equal in area.
Given: Two parallelograms ABCD and ABEF, which have the same base AB and which are between the same parallel lines AB and FC.
To prove: Proof: In ADF and BCE, we have
And, Â Â Â Â Â Â Â Â [ â¸ª and ]
So, by SAS criterion of congruence, we have
Â Â â€¦..(i)
Now,
Â Â [Using(i)]
Hence,
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(3) Prove that the area of a parallelogram is the product of its base and the corresponding altitude.
Given: A parallelogram ABCD in which AB is the base and AL the corresponding altitude.
To prove:
Construction: Complete the rectangle ALMB by drawing .Proof: Since and rectangle ALMB are on the same base and between the same parallels.
Â Â Â [By rect. Area axiom area of a rectangle = Base X Height]
Hence,
(4) Prove that parallelograms on equal bases and between the same parallels are equal in area.
Given: Two parallelograms ABCD and PQRS with equal bases AB and PQ and between the same parallels AQ and DR.
To prove:
Construction: Draw and Proof: Since , and
Now,
Â [ and ]
(5) Prove that triangles on the same bases and between the same parallels are equal in area.
Proof: We have,
And,
is a parallelogram.Similarly, is a parallelogram.
Now, parallelograms ECQP and BCAD are on the same base BC, and between the same parallels.
Â Â Â â€¦.(i)
We know that the diagonals of a parallelogram divides it into two triangles of equal area.
Â â€¦..(ii)
And, Â ....(iii)
Now, [ From (i)]
Â Â Â [From (ii) and (iii)]
Hence,
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(6) Prove that the area of a triangle is half the product of any of its sides and the corresponding altitude.
Given: A in which AL is the altitude to the side BC.
To prove:
Construction: Through C and A draw and respectively, intersecting each other at D.Proof: We have,
And,
BCDA is a parallelogram.
Since AC is a diagonal of parallelogram BCDA.
[BC is the base and AL is the corresponding altitude of parallelogram BCDA]
(7) Prove that if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to the half of the parallelogram.
Given: A and a parallelogram BCDE on the same base BC and between the same parallel BC and AD.
To prove:
Construction: Draw and , meeting BC produced in M.Proof: Since A, E and D are collinear and
Â Â Â â€¦..(i)
Now,
Â Â [ (from (i)]
(8) Prove that the area of a trapezium is half the product of its height and the sum of parallel sides.
Given: A trapezium ABCD in which ; , and , where and .
To prove:
Construction: Join ACProof: We have,
Â [ and ]
(9) Prove that triangles having equal areas and having one side of one of the triangles, equal to one side of the other, have their corresponding altitudes equal.
Given: Two triangles ABC and PQR such that:
CN and RT are the altitudes corresponding to AB and PQ respectively of the two triangles.
To prove: Proof: In , CN is the altitude corresponding to side AB.
Â Â Â â€¦.(i)
Similarly, we have,
Â Â â€¦..(ii)
Now,
Â Â [ (Given)]
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(10) Prove that if each diagonal of a quadrilateral separates it into two triangles of equal area, then the quadrilateral is a parallelogram.
Given: A quadrilateral ABCD such that its diagonals AC and BD are such that
and
To prove: Quadrilateral ABCD is a parallelogram.Proof: Since diagonal AC of the quadrilateral ABCD separates it into two triangles of equal area. Therefore,
Â Â Â â€¦..(i)
But,
Â Â Â [Using (i)]
â€¦.(ii)
Since diagonal BD of the quadrilateral ABCD separates it into triangles of equal area.
Â Â â€¦.(iii)
But,
Â Â Â [Using(iii)]
Â Â â€¦..(iv)
From (ii) and (iv), we get
Since ABC and ABD are on the same base AB. Therefore they must have equal corresponding altitudes.
i.e. Altitude from C of = Altitude from D of
Similarly,
Hence, quadrilateral ABCD is a parallelogram.
(11) Prove that the area of a rhombus is half the product of the lengths of its diagonals.
Given: A rhombus ABCD whose diagonals AC and BD intersect at O.
To prove: ar(rhombus ABCD) Proof: Since the diagonals of a rhombus intersect at right angles. Therefore,
and
ar(rhombus)
ar(rhombus)
ar(rhombus)
ar(rhombus)
(12) Prove that diagonals of a parallelogram divide it into four triangles of equal area.
Given: A parallelogram ABCD. The diagonals AC and BD intersect at O.
To prove: Proof: Since the diagonals of a parallelogram bisect each other at the point of intersection.
and
Also, the median of a triangle divides it into two equal parts.
Now, in , BO is the median.
Â â€¦.(i)
In , CO is the median
Â Â Â â€¦..(ii)
In , DO is the median
Â Â â€¦.(iii)
From (i), (ii) and (iii), we get
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(13) Prove that if the diagonals AC and BD of a quadrilateral ABCD, intersect at O and separate the quadrilateral into four triangles of equal area, then the quadrilateral ABCD is parallelogram.
Given: A quadrilateral ABCD such that its diagonals AC and BD intersect at O and separate it into four parts such that
To prove: Quadrilateral ABCD is a parallelogram.Proof: We have,
Thus, ABD and ABC have the same base AB and have equal areas. So, their corresponding altitudes must be equal.
Altitude from Altitude from C of
Similarly, we have, .
Hence, quadrilateral ABCD is a parallelogram.
(14) Prove that a median of a triangle divides it into two triangles of equal area.
Given: A in which AD is the median.
To prove:
Construction: Draw .Proof: Since AD is the median of . Therefore, D is the mid point of BC.
Â [ Multiplying both sides by AL]
ALITER Since ABD and ADC have equal bases and the same altitude AL.
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It helped me in my exam preparations....Thanks
very helpful and good for studies
aur Nani hai