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If ABCD is a parallelogram, AE \(\bot \) DC and CF \(\bot \) AD. If AB = 10 cm, AE = 6 cm and CF = 5 cm, then AD is equal to:
AB = CD = 10 cm (Opposite sides of a parallelogram)
CF = 5 cm and AE = 6 cm
Now,
Area of parallelogram = Base Ã— Altitude
If E, F, G and H are the midpoints of the sides of a parallelogram ABCD, respectively, then ar (EFGH) is equal to:
and
and
and (H and F are midpoints)
and are parallelograms.
and llgmABFH, both lie on a common base, FH.
area of EFH area of ABFH â€” 1
area of GHF area of HFCD â€” 2
Adding eq. 1 and 2 we get;
area of + area of (area of ABFH + area of HFCD)
ar (EFGH) ar(ABCD)
If P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD, then:
and parallelogram ABCD lie on the same base AB and inbetween same parallel AB and DC.
ar(APB) = ar(parallelogram ABCD) â€” 1
ar(BQC) = ar(parallelogram ABCD) â€” 2
From eq. 1 and 2:
If ABCD and EFGH are two parallelograms between same parallel lines and on the same base, then:
A median of a triangle divides it into two
Suppose, ABC is a triangle and AD is the median.
AD is the median of ABC.
It will divide ABC into two triangles of equal area.
ar(ABD) = ar(ACD) â€” (i)
also,
ED is the median of ABC.
ar(EBD) = ar(ECD) â€” (ii)
Subtracting (ii) from (i),
ar(ABD) â€“ ar(EBD) = ar(ACD) â€“ ar(ECD)
ar(ABE) = ar(ACE)
In a triangle ABC, E is the midpoint of median AD. Then:
ar(BED) = BD.DE
AE = DE (E is the midpoint)
BD = DC (AD is the median on side BC)
DE = AD â€” 1
BD = BC â€” 2
From eq. 1 and 2, we get;
ar(BED) = BC AD
ar(BED) = ar(ABC)
ar(BED) = ar (ABC)
If D and E are points on sides AB and AC respectively of \(\triangle \)ABC such that ar(DBC) = ar(EBC). Then:
DBC and EBC are on the same base BC and also having equal areas.
They will lie between the same parallel lines.
DE  BC.
If Diagonals AC and BD of a trapezium ABCD with AB  DC intersect each other at O. Then,
DAC and DBC lie on the same base DC and between the same parallels AB and CD.
ar(DAC) = ar(DBC)
ar(DAC) âˆ’ ar(DOC) = ar(DBC) âˆ’ ar(DOC)
ar(AOD) = ar(BOC)
If Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(\(\triangle \)AOD) = ar(\(\triangle \)BOC). Then ABCD is a:
: ar(AOD) = ar(BOC)
ar(AOD) = ar(BOC)
â‡’ ar(AOD) + ar(AOB) = ar(BOC) + ar(AOB)
â‡’ ar(ADB) = ar(ACB)
Areas of ADB and ACB are equal.
Therefore, they must lie between the same parallel lines.
Therefore, AB âˆ¥ CD
Hence, ABCD is a trapezium.
If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of the parallelogram will be:
For parallelogram ABCD
Area â€¦.(1)
For
Area
….(2)
(Distance between the parallel lines are equal)
So, PN = DM
From of (1) and (2)
Area of PAB ar ABCD
The area of trapezium ABCD in the given figure is
ABCD is a trapezium, AB  CD, AE  CD
Area of trapezium ABCD
In the given figure, ABCD is a parallelogram in which AB = CD = 5 cm and BD \(\bot \) DC such that BD = 6.8 cm. The area of parallelogram ABCD is
Area of parallelogram
In the given figure, ABCD is a parallelogram in which diagonals AC and BD intersect at O. If ar (\(\mathrm{^{gm}} \) ABCD) is \(52\ cm^2 \) then the ar(\( \triangle\)OAB) =
Area of OAB = ar (ABD)
=
ABCD is a parallelogram in which DL \(\bot \) AB, If AB = 10 cm and DL = 4 cm then the ar (\(\mathrm{^{gm}} \) ABCD is
Area of
The midpoint of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to
Area of parallelogram
The length of the diagonals of a rhombus are 12 cm and 16 cm the area of the rhombus is
Area of rhombus Product of its diagonals
In a quadrilateral ABCD it is given that BD = 16 cm. If AL \(\bot \) BD and CM \(\bot \) BD such that AL = 9 cm and CM = 7 cm then ar(quad ABCD) is
Area of quadrilateral ABCD
ABCD is a rhombus in which \(\angle C=60^o \). Then AC : BD = ?
ABCD is a rhombus
In
ABCD and ABFE are parallelogram such that ar(quad EABC) \( =17\ cm^2\) and ar(\(^{gm} \)ABCD) \( =25\ cm^2\). Then ar (\(\triangle \)BCF = ?
\(\triangle \)ABC and \(\triangle \)BDE are two equilateral triangles such that D is the midpoint of BC. Then \(ar\ (\triangle BDE):ar(\triangle ABC)=? \)
Area of an equilateral
One of the diagonals of a quadrilateral is 16 cm. The perpendiculars drawn to it from its opposite vertices are 2.6 cm and 1.4 cm. Find its area
Area
The ratio of the bases of two triangles is a : b. If the ratio of their corresponding alttudes is c : d. The ratio of their areas is
Ratio of areas
The area of the given quadrilateral ABCD is
Area of quadrilateral ABCD
= Area of ABC + area of ACD
In the given figure, AB  DC Identify the triangle that have equal areas
AXD and AXC are on the same base and between the same parallels
Which of the following is not a property of a rhombus?