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If the parallelograms lie on the same base and between the same parallel lines, then their areas are equal.
A parallelogram and a rectangle are constructed on same base and between the same parallels, they have
Base = b, Distance between parallels = h
Then, Area of rectangle = area of parallelogram = bh
Point O is the point of intersection of AC and BD of parallelogram ABCD and, also, PQ ∥ AD, then,
∵ O lies in mid of the parallelogram.
∴ PQ divides parallelogram in two parts of equal areas.
∴ ar(PQDA) = 1/2 ar(ABCD)
P and Q are the points on AB and BC respectively of ∥gm ABCD, then
∵ ABCD and ∆PDC lie on same base, i.e., CD and between the same parallels, i.e., AB and CD.
∴
Similarly,
⇒ ar(∆PDC) = ar(∆AQD) = 1/2 ar(quad. ABCD)
In ∆ABC, P, Q, R are the midpoints of sides BC, CA, AB respectively, then ar(∆PQR) =
Using midpoint theorem,
BC = 2 QR, AB = 2 PQ, AC = 2 PR.
Let the area of ∆PQR be A, then, ABC is a ∆ resulted by doubling the length of every side of ∆PQR.
∴ ar (∆ABC) = 4 ar(∆PQR) [using Heron’s formula]
If P, Q, R and S are midpoints of AB, BC, CD and DA of parallelogram ABCD and ar(ABCD) = 26 m^{2}, then ar(PQRS) =
ar (parallelogram PQRS)
ABC is a right angled ∆ at A, BCED, ACFG and ABMN are squares on sides ABC, AC, AB respectively. AX ⊥ DE meets BC at Y then,
Let AB = x, AC = y, then,
BC = (Pythagoras’ theorem)
Area (ABMN) = x^{2}
Area (ACFG) = y^{2}
ar(BCED) =
∴ ar(BCED) = ar(ABMN) + ar(ACFG)
Ar(trap. DCYX) = K, Ar(XYBA), then K =
According to question
⇒ (40 + 50) = K (60 + 50)
⇒ k = 9/11
ABCD is a trapezium in which AB ∥ DC, then
∵ ∆ADC and ∆BDC lie on same base DC and between the same parallels, i.e., AB and DC.
∴ ar(∆ADC) = ar(∆BDC) ar(AOD) + ar(DOC) = ar(BOC) + ar(DOC) ⇒ ar(AOD) = ar(BOC).
ABCD is a rectangle with O as any point in its interior. If ar (∆AOD) = 3 cm^{2}, and ar(∆BOC) = 6cm^{2}, then ar(ABCD) =
Draw XY ∥ AD ∥ BC and PQ ∥ AB ∥ DC
Now ar(∆AOB) =
⇒ ar(∆AOB) + ar(∆DOC) =
⇒ ar(∆AOD) + ar(∆BOC) =
⇒ ar(ABCD) = 2 × (3 + 6) = 18 cm^{2}
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