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PQRS is a parallelogram whose area is \(180\ cm^2 \) and A is any point on the diagonal QS. The area of \(\triangle \)ASR is \(90\ cm^2 \). Is is true ?
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False
Area of parallelogram and is it diagonal which divides it into two triangles of equal area.
In the figure, PQRS and EFRS are two parallelogram. Is area of \(\triangle \)MFR equal to \(\dfrac{1}{2} \) area of  gm PQRS?
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and are on the same base and between the same parallels and .
So, …(1)
…(2)
BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendiculars from A and C respectively on BD. Show that ar \(\mathrm{(ABCD)=\dfrac{1}{2}BD(AM+CN)} \)
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Check whether the following statement is true. PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If \(PS = 5\ cm \), then \(ar\ (PAS) = 30\ cm^2 \).
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Since is any point on then
But the statement can be true of
Justification
Suppose
Suppose
In \(\triangle\ ABC,O\) is any point on its median \(AD\). Show that \(ar\ (\triangle ABO)=ar\ (\triangle ACO) \)
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In
is the median
…(1)
In
is the median
….(2)
(Proved)
In the figure, \(LM=\dfrac{3}{4}QR,LMQR \) and distance between \(LM \) and \(OR \) is \(3\ cm \). If length of \(QR=6\ cm \), find the area of \(LQRM \).
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Area
ABC and BDE are two equilateral triangle such that D is mid–point of BC. Show that ar (BDE) = \(\dfrac{1}{4} \) ar (ABC).
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Ratio of area of triangle is equal to the ratio of square of corresponding sides
area of (Proved)
Show that the segment joining the midpoints of a pair of opposite sides of a parallelgoram divides it into two equal parallelograms.
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and ( is a parallelogram)
and
So, is a parallelogram
and are two parallelogram on the base and between the same parallels and
In the figure, ABCD and EFGD are two parallegorams and G is the midpoint of CD. Check whether area of \(\triangle \)PDC is equal to half of area EFGD.
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The given statement is false
( and parallelogram are on the same base )
is the mid point of
In the figure, PQRS is square and T and U are respectively the midpoints of PS and QR. Find the area of \(\triangle \)OTS, if PQ = 8 cm.
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In is the mid point of and
In the figure, AD is a median of \(\triangle \)ABCE is any point on AD. Show that at (\(\triangle \)BED) = ar (\(\triangle \)CED)
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( is the median) …(1)
( is median) …(2)
Subtracting equation (2) from (1)
PQRS is a trapezium with PQSR. A line parallel to PR intersects PQ at L and QR at M. Prove that \(\mathrm{ar\ (\triangle PSL) = ar\ (\triangle PRM)} \).
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…(1)
(Two triangles having same base and between the same parallels)
…(2)
From (1) and (2)
(Proved)
P and Q are any two points lying on the sides DC and AD respectively of a paralleogram ABCD. Show that ar (\(\triangle \)APB = ar (\(\triangle \)BQC)
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…(1)
…(2)
From (1) and (2)
Diagonals AC and BD of a trapezium ABCD with ADCD intersect each other at O. Prove that ar (\(\triangle \)AOD) = ar (\(\triangle \)BOC)
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In and
(VOA)
(alt angles)
(alt )
(AAA)
In the figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar (BPC) = ar (DPQ).
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..(1)
( Triangles on the same base and between the same parallels)
Similarly …(2)
….(3)
From (1) and (3)
or (Proved)
D is the mid–point of side BC of a \(\triangle \)ABC and E is the mid point of BD. If O is the mid point of AE, then show that ar (BOE) = \(\dfrac{1}{8} \) ar (ABC).
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..(1)
( and are medians)
…(2)
…(3)
From (1), (2) and (3)
In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ=QR = RS and PA  QB  RC. Prove that ar (PQE) = ar (CFD)
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In Parallelogram ,
So is a parallelogram
…(1)
Similarly …(2) and ….(3)
Now …(4)
From equation (1), (2), (3) and (4)
In and
(alt. )
(ASA)
( congruent triangle have equal areas)
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
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Diagonal of a parallelogram but it each other
… (1) ( is median)
…(2) ( is median)
….(3) ( is median)
(Proved)
In the figure, ABCD is a square. E and F are respectively the mid–points of BC and CD. If R is the mid–point of EF, show that ar (AER) = ar (AFR)
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(Since is the mid point of so, )
(Proved)
In the figure, M, N are points on sides PQ and PR respectively of \(\triangle \)PQR , Such that ar (\(\triangle \)QRN) = ar (\(\triangle \)QRM) .Show that MNQR.
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and are on the same base and having equal areas
They lie between the same parallel lines
In the figure, O is any point on the diagonal PR of a parallelogram PQRS. Show that ar (PSO) = ar (PQO).
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is the mid point of ….(1)
….(2)
Adding equation (1) and (2)
(Proved)
In the figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE
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In and
(alt. int )
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. E and F are the midpoints of nonparallel sides. Show that ar (ABFE) : ar (EFCD) = (3a + b) : (a + 3b).
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Ratio
Prove that a rectangle and a parallelogram on the same base and between the same parallels, the perimeter of the parallelogram is greater than the perimeter of the rectangle.
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Let be a parallelogram and be the rectangle
Perimeter of parallelogram
Perimeter of rectangle
(Proved)
If medians of a triangle \(ABC \) intersects at \(G \), show that \(ar\ \triangle AGB=ar(\triangle AGC)=ar\ (\triangle BGC)=\dfrac{1}{2}ar\ (\triangle ABC) \)
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( is median) ….(1)
….(2)
( is median)
….(3)
Similarly ….(4)
From equation (3) and (4)
…(5)
….(6)
(From 5 and 6)
(Proved)
In the figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
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…(1)
…(2)
Subtracting equation (1) from equation (2)
Since these two triangle have the same base . They must be lying between the same parallel lines
is a trapezium
Also
Hence is a trapezium
In the figure, ABCD is a quadrilateral. A line through D parallel to AC meets BC produced at E. Prove that ar (\(\triangle \)ABE) = ar qard (ABCD)
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(Triangles on the same base)
Adding on both the sides
In the figure, \(ABCD \) is a parallelogram. Points \(P \) and \(Q \) on \(BC \) trisect \(BC \). Show that \(ar\ (APQ) = ar\ (DPQ)=\dfrac{1}{6}\ ar\ (ABCD)\).
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and are on the same base and between the same parallels and
….(1)
….(2)
In \(\triangle \)ABC , if L and M are the points on AB and AC respectively, such that LMBC, prove that ar (LOB) = ar (MOC)
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and lie on the same base and between the same parallels and
(Proved)
In the figure, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD).
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In and
(Corresponding angles)
(opposite sides)
(corresponding angles)
(AAS)
( congruent triangles have equal area)
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed. Show that ar (ABCD) = ar (PBQR).
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( and are on the same base
…(1)
…(2)
….(3)
From (2) and (3)
In the figure, ABCD is a parallelograms in which BC is produced to E such that CE = BC. AE intersects CD at F. Show that \(\mathrm{(\triangle BDF)=\dfrac{1}{4}ar\ (ABCD)} \).
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…(1)
…(2)
In and
(VOA)
(Alt. )
(ASA)
(cpct)
The figure, ABCDE is a pentagon and a line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (ABCDE) = ar (AEDF)
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and lie on the same base and between the same parallels and
ABCD is a trapezium with ABDC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
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( They lie on the same base and between parallel lines and )
…(2)
(They lie on the same base and between parallel lines and )
From (1) and (2)
Diagonals PR and QS of quadrilateral PQRS intersect at T such that PT = TR. If PS = QR, show that ar (\(\triangle \)PTS) = ar (\(\triangle \)RTQ)
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In and
(given)
(VOA)
(AAS)
(cpct)
(cpct)
….(1)
In and
(each )
(Proved above)
(given)
(RHS)
…(2)
Adding (1) and (2)
Diagonal AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
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(given)
and lie on the same base and equal in area
In
So one pair of opposite side is parallel is a trapezium
PQRS and ABRS are parallelograms and X is any point on side BR. Show that :
(i) area PQRS = area ABRS
(ii) area AXS = \(\dfrac{1}{2} \) area PQRS.
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Since is a parallelogram and is also a parallelogram
So
So
and are two parallelograms with some base
area of
area of
In the given figure, APBQCR. Prove that \(ar\ (\triangle AQC)=ar\ (\triangle PBR) \)
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Since and lie on the same base and are between parallel lines and
…(1)
Also …(2)
Adding (1) and (2)
A point E is taken as the midpoint of the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that \(ar\ (\triangle ADF)=ar\ (ABFC) \)
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…(1)
Also ….(2)
Adding (1) and (2)
In the figure, M is a point in the interior of a parallelogram PQRS. Show that
(i) \(ar\ (\triangle PMQ)+ar\ (\triangle MRS)=\dfrac{1}{2}ar(gm\ PQRS) \)
(ii) \(ar\ (\triangle PMS)+ar\ (\triangle MQR)=ar(\triangle PMQ)+ar(\triangle MRS) \)
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(i) (by construction) …(1)
(opposite sides)
So …(2)
From (1) and (2)
and
is a parallelogram
…(3)
…(4)
Adding (3) and (4)
(ii) (by construction) …(1)
(opposite sides)
So
and
So is a parallelogram
(Proved)
In the figure, diagonals AC and BD of quadrilateral ABCD intersect at O, such that OB = OD. If AB = CD, show that
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = (ACB)
(iii) ABCD is a parallelogram.
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In and
(each 90)
(VOA)
(given)
(AAS)
(cpct) …(1)
Also, …(2)
In and
(each 90)
(given)
(From 1)
(RHS)
Also, …(3)
Adding (2) and (3) we get
(ii)
(iii)
….(4)
Adding (4) and (5) we get
we get and
So, is a parallelogram