Advertisements
Advertisements
प्रश्न
ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF (Figure), prove that ar (AER) = ar (AFR)

Advertisements
उत्तर

Given: In square ABCD, E and F are the mid-points of BC and CD respectively. Also, R is the mid-point of EF.
To prove: ar (AER) = ar (AFR).
Construction: Draw AN ⊥ EF.
Proof: ar (ΔAER) = `1/2` × Base × Height
= `1/2 xx ER xx AN`
= `1/2 xx FR xx AN` ...[∵ R is the mid-point of EF, so ER = FR]
= ar (ΔAFR)
Hence proved.
APPEARS IN
संबंधित प्रश्न
In the given figure, ABCD is parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).

In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that
ar (BPC) = ar (DPQ).
[Hint: Join AC.]

In the given below fig. ABCD, ABFE and CDEF are parallelograms. Prove that ar (ΔADE)
= ar (ΔBCF)

In the below fig. ABCD and AEFD are two parallelograms. Prove that
(1) PE = FQ
(2) ar (Δ APE) : ar (ΔPFA) = ar Δ(QFD) : ar (Δ PFD)
(3) ar (ΔPEA) = ar (ΔQFD)
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is ______.
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Figure). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is ______.

In the following 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).

ABCD is a parallelogram in which BC is produced to E such that CE = BC (Figure). AE intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.
The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.
