Advertisements
Advertisements
Question
In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Figure). Prove that ar (ABCD) = ar (APQD)
Advertisements
Solution
Given: In trapezium ABCD, AB || DC, DC produced in Q and L is the mid-point of BC.
∴ BL = CL
To prove: ar (ABCD) = ar (APQD)
Proof: Since, DC produced in Q and AB || DC.
So, DQ || AB
In ΔCLQ and ΔBLP,
CL = BL ...[Since, L is the mid-point of BC]
∠LCQ = ∠LBP ...[Alternate interior angles as BC is a transversal]
∠CQL = ∠LPB ...[Alternate interior angles as PQ is a transversal]
∴ ΔCLQ ≅ ΔBLP ...[By AAS congruence rule]
Then, ar (ΔCLQ) = ar (ΔBLP) [Since, congruent triangles have equal area] ...(i)
Now, ar (ABCD) = ar (APQD) – ar (ΔCQL) + ar (ΔBLP)
= ar (APQD) – ar (ΔBLP) + ar (ΔBLP) ...[From equation (i)]
⇒ ar (ABCD) = ar (APQD)
Hence proved.
APPEARS IN
RELATED QUESTIONS
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC
such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:
(1) ar ( ADEG) = ar (GBCD)
(2) ar (ΔEGB) = `1/6` ar (ABCD)
(3) ar (ΔEFC) = `1/2` ar (ΔEBF)
(4) ar (ΔEBG) = ar (ΔEFC)
(5)ΔFind what portion of the area of parallelogram is the area of EFG.
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)
In which of the following figures, you find two polygons on the same base and between the same parallels?
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 cm2.
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.
ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = `7/9` ar (XYBA)
In the following figure, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD). [Hint: Join PD].
