Class 9 Maths


Exercise 9.1

Question 1: Which of the following figures lie on the same base and between the same parallels? In such a case, write the common base and the two parallels.


Answer: (i) ABCD and PDC, (iii) PQRS and TQR (iv) ABCD and PQR. (v) ABCD and APCD, (vi) PQRS, PADS, ABCD and BQRC

Exercise 9.2

Question 1: In the given figure, ABCD is a parallelogram, AE perpendicular DC and CF perpendicular AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.


Answer: Given; AB = 16 cm, AE = 8 cm and CF = 10 cm
AB = DC = 16 cm (Because AD||BC)
Area (ABCD) `= He\ig\ht\ xx\ Ba\se`
`= AE xx DC = 8 xx 16 = 128  sq  cm`
Area (ABCD) `= CF xx AD`
Or, `10 xx AD = 128  sq  cm`
Or, `AD = 128/10 = 12.8  cm`

Question 2: If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that; ar (EFGH) = ½ ar(ABCD)

Answer: The figure shows a parallelogram ABCD in which E, F, G and H are the mid-points of sides. AM is the height of parallelogram ABCD. Altitude of ΔEFH is EO, and altitude of ΔFGH is GN.


EO = GN = ½ AM (because HF joins the mid-points of AD and BC)
Area (ABCD) `= He\ig\ht\ xx\ Ba\se = AM xx DC` ………………(1)
Area (EFGH) = ar(ΔEFH) + ar(ΔFGH)
`= ½ xx\ He\ig\ht\ xx\ Ba\se\ + ½ xx\ He\ig\ht\ xx\ Ba\se`
`= ½ xx EO xx HF + ½ xx GN xx HF`
`= ½ (EO xx HF + GN xx HF)`
`= ½ (EO xx HF + EO xx HF)` [Because EO = GN)
`= ½ xx 2 xx (EO xx HF) = EO xx HF` ………………(2)
As `EO = ½ AM` and `HF = DC`
Hence, from equations (1) and (2), it is clear that;
ar(EFGH) = ½ ar(ABCD) Proved

Question 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).

Answer: Let us assume that height for the base AB is h1 and height for the base BC is h2.
Area (ABCD) = h1 x AB = h2 x BC


Area(ΔAPB) `= ½ xx h_1 xx AB`
Area(ΔBQC) `= ½ xx h_2 xx BC`
From above equations, it is clear that; ar(ΔAPB) = ar(ΔBQC)

Question 4: In the given figure, P is a point in the interior of a parallelogram ABCD. Show that;

  1. ar(APB) + ar(PCD) = ½ ar(ABCD)
  2. ar(APD) + ar(PBC) = ar(APB) + ar(PCD)

(Hint: Through P, draw a line parallel to AB.)

Answer: Let us draw MN||AB passing through P. Let us assume that height of parallelogram ABNM is h1, that of MNCD is h2 and that of ABCD is h. The base for all parallelograms shall be equal to AB.


Ar(ABCD) = ar(ABNM) + ar(MNCD)
Ar(ΔAPB) = ½ ar(ABNM)
Ar(ΔPCD) = ½ ar(MNCD)
(Because area of a triangle is half the area of parallelogram on the same base and between the same altitude)
Hence, ar(ΔAPB) + ar(ΔPCD) = ½ ar(ABCD)
Similarly, ar(ΔAPD) + ar(ΔPBC) = ½ ar(ABCD) can be proven
This shows that; ar(APD) + ar(PBC) = ar(APB) + ar(PCD)

Question 5: In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

  1. ar(PQRS) = ar(ABRS)
  2. ar(AXS) = ½ ar(PQRS)

Answer:We know that area of two parallelograms on the same base and between same height is equal.
Hence, ar(PQRS) = ar(ABRS)

We also know that are of a triangle is half the area of parallelogram if they are on the same base and between same height.
Hence, ar(AXS) = ½ ar(PQRS)

Question 6: 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?

Answer: The following figure shows various parts in which the field has been divided.
The field is divided into three parts.
All parts are triangular in shape.
Area(ΔPQA) = ½ ar(PQRS)
This means that sum of areas of remaining triangles = ar(ΔPQA)


The farmer can sow wheat in triangle PQA and pulses in remaining triangles or vice-versa.