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If a triangle and a parallelogram are on the same base and between the same parallels, and area of triangle is A, then area of parallelogram is:
Area of ∆ = 1/2 x b x h = A
Area of parallelogram = bh = 2A.
ABCD is a parallelogram
AM = 7 cm, CN = 8 cm and AB = CD = 10 cm, Find AD.
Area of parallelogram
= AM × CD = AD × CN
⇒ 7 × 10 = AD × 8
⇒
P, Q, R and S are the midpoints of DC, BC, AB and AD respectively. area of parallelogram PQRS is
Joining QS, it can be clearly seen that,
QS ∥ DC ∥ AB,
[Area between QS and DC]
[Area between QS and AB]
Adding both,
ar (PQS) + ar (SRQ) = + ar (ASQB)]
ar (PQRS) = 1/2 ar (ABCD)
The median of a triangle divides is into two:
In ∆ABC, AD is the median
Let AE ⊥ BC, then,
ar(∆ABD) = 1/2 × AE × BD,
ar(∆ADC) = 1/2 × AE × DC = 1/2 × AE × BD
∴ ar(∆ABD) = ar(∆ADC)
Area of trapezium RQBC = ar(∆RBP) + ar(∆PQR) + ar(∆QPC)
= 3 × ar(∆PQR)
If AD is median of ∆ABC and P is a point on AC such that ar(∆ADP) : ar(∆ABD) = 2 : 3, then ar(∆PDC) : ar(∆ABC) is
∵ AD median of ∆ABC
∴ ar (ABD) = ar(ADC) = 1/2 ar(∆ABC) …(i)
∴ Required ratio = = 1 : 6
ABCD is a trapezium in which AB ∥ CD and CD = 40 cm, and AB = 60 cm. If X and Y are, respectively, the mid points of AD and BC, then XY =
∵ XY ∥ AB ∥ DC
[Let the length of XY be x cm]
∵ ar (DCYX) + ar (XYBA) = ar (ABCD)
∴
…(i)
∵ DN = NM =
∴ The equation (i) reduces to:
⇒ 100 + 2x = 200
⇒ 2x = 100
⇒ x = 50 cm.
Area of trapezium, PQRS in the given figure is:
Required area:
Find the area of quadrilateral ABCD
In ∆ADB,
ar(∆ADB) = 1/2 × 12 × 9 = 54 cm^{2}
ar(∆DBC) = 1/2 × 15 × 8 = 60 cm^{2}
∴ Total area = 54 + 60 = 114 cm^{2}
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area ∆DFB = 3 cm^{2}, area of parallelogram ABCD is:
ar(∆AFB) = 1/2 ar (parallelogram ABCD)
{areas between same parallels and same base}
⇒ ar (parallelogram ABCD) = 2 × ar(∆AFB) = 2 × ar(∆DCB) …(i)
In ∆ADF and ∆ECF
∠AFD = ∠EFC {vertically opposite angles}
∠ACF = ∠ADF {alternate angles}
BC = CE = AD
∴ ∆ADF ≅ ∆ECF {by AAS congruency}
∴ DF = CF
ar(parallelogram ABCD) = 2 × ar(∆DCB)
= 2 × 2 × (ar(∆DFB))
= 2 × 2 × 3
= 12 cm^{2}
ABCD and FECG are parallelograms equal in area. If ar(∆AQE) = 12cm^{2}, then ar(parallelogram FGBQ) =
ar(∆AQE) = 1/2ar (parallelogram AQED)
= 1/2 x 1/2 (ar(parallelogram ABCD))
= 1/4 × (ar(parallelogram FECG))
12 cm^{2} = 1/4 × ar(parallelogram FECG)
∴ ar(FECG) = 48 cm^{2}
⇒ ar(FGQB) = 1/2 ar(FECG) = 48/2 = 24 cm^{2}
ABCD is a parallelogram. P is the midpoint of AB. BD and CP intersect at Q such that QC : QP = 3 : 1. If the ar(∆PBQ) = 10 cm^{2}, then area of parallelogram ABCD is:
ar(∆PQB) + ar(∆DQC) = 1/2 ar(parallelogram ABCD)
⇒ ar(parallelogram ABCD)
= 2 [ar(∆PQB) + ar(∆DQC)]
= 2 [10 cm^{2} + ar(∆DQC)]
ar(∆DQC) = (3 + 3 + 1) × ar(∆PQB)
= 7 × 10 = 70 cm^{2}
∴ ar(parallelogram ABCD) = 2 × 80
= 160 cm^{2}
M is a point on base QR of ∆PQR, and N is the midpoint of QR. NX is drawn parallel to MP at X. If ar(∆PQR) = 12cm^{2}, then ar(∆XMR) =
Q ar(ΔXMR) = 1/2 ar(∆PQR)
∴ ar(∆XMR) = 1/2 × 12cm^{2} = 6 cm^{2}
ABCD is a parallelogram, P and Q are the midpoints of BC and CD respectively, then, ar(∆APQ) = K (ar(ABCD)), K =
Let AB = CD = x, and BC = AD = y, AM = h_{1}, BN = h_{2}
ar (∆ADQ) = 1/2 × AM × DQ =
ar (∆APB) =
ar (∆PQC) =
ar (parallelogram ABCD) = AM × DC = BN × AD = h_{1}x = h_{2}y
∴ ar(∆AQC) = h_{1}x –
∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. The area (in cm^{2}) of ∆AOB is:
OC = 6.5 cm,
∴ AC = BC = 6.5
[By similarly of ∆AOC and ∆BOC]
∴ AB = AC + BC = 2AC = 2 × 6.5 = 13 cm.
∴ ar(∆AOB) = 1/2 × 5 × 12 = 30 cm^{2}
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