Access Answers for SAT Maths

Access Answers for SAT Maths Chapter 9 – Areas of Parallelograms and
Triangles
Exercise 9.1
1. Which of the following figures lie on the same base and in-between the same parallels? In such a
case, write the common base and the two parallels.




Solution:
(i) Trapezium ABCD and ΔPDC lie on the same DC and in-between the same parallel lines AB and
DC.
(ii) Parallelogram PQRS and trapezium SMNR lie on the same base SR but not in-between the same
parallel lines.
(iii) Parallelogram PQRS and ΔRTQ lie on the same base QR and in-between the same parallel lines
QR and PS.
(iv) Parallelogram ABCD and ΔPQR do not lie on the same base but in-between the same parallel
lines BC and AD.
(v) Quadrilateral ABQD and trapezium APCD lie on the same base AD and in-between the same
parallel lines AD and BQ.
(vi) Parallelogram PQRS and parallelogram ABCD do not lie on the same base SR but in-between the
same parallel lines SR and PQ.

,Exercise 9.2

1. In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF =
10 cm, find AD.




Solution:
Given,
AB = CD = 16 cm (Opposite sides of a parallelogram)
CF = 10 cm and AE = 8 cm
Now,
Area of parallelogram = Base × Altitude
= CD×AE = AD×CF
⇒ 16×8 = AD×10
⇒ AD = 128/10 cm
⇒ AD = 12.8 cm
2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that
ar (EFGH) = 1/2 ar(ABCD).
Solution:

,Given,
E, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.
To Prove,
ar (EFGH) = ½ ar(ABCD)
Construction,
H and F are joined.
Proof,
AD || BC and AD = BC (Opposite sides of a parallelogram)
⇒ ½ AD = ½ BC
Also,
AH || BF and DH || CF
⇒ AH = BF and DH = CF (H and F are mid points)
∴, ABFH and HFCD are parallelograms.
Now,
We know that, ΔEFH and parallelogram ABFH, both lie on the same FH the common base and in-
between the same parallel lines AB and HF.
∴ area of EFH = ½ area of ABFH — (i)
And, area of GHF = ½ area of HFCD — (ii)
Adding (i) and (ii),
area of ΔEFH + area of ΔGHF = ½ area of ABFH + ½ area of HFCD
⇒ area of EFGH = area of ABFH
∴ ar (EFGH) = ½ ar(ABCD)
3. 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).
Solution:

, ΔAPB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.
ar(ΔAPB) = ½ ar(parallelogram ABCD) — (i)
Similarly,
ar(ΔBQC) = ½ ar(parallelogram ABCD) — (ii)
From (i) and (ii), we have
ar(ΔAPB) = ar(ΔBQC)
4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB) + ar(PCD) = ½ ar(ABCD)
(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
[Hint : Through P, draw a line parallel to AB.]




Solution:




(i) A line GH is drawn parallel to AB passing through P.
In a parallelogram,