Areas of Parallelograms and Triangles (Exercise 9.2)

  • by

Areas of Parallelograms and Triangles


Chapter 9


Exercise 9.2


EX 9.2 QUESTION 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:

AE ⊥ DC and AB = 16 cm

∵ AB = CD [Opposite sides of parallelogram]

∴ CD = 16 cm

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


EX 9.2 QUESTION 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 AB = CD

⇒ ½AB =  ½CD

⇒ BE = CG  ——-(i) [E and G are the midpoints of sides AB and CD]

BE || CG  ———-(ii) [AB || CD]

From equation (i) and (ii) BEGC is a parallelogram

Hence, ar(GEF) = ½ar(BEGC) ———-(3)

[If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram]

Similarly, AB = CD     [Given]

⇒ ½AB =  ½CD

⇒ AE = DG  ——-(4)  [E and G  are the midpoints of sides AB and CD]

and AE || DG  ——(5)  [AB || CD]

From equation (4) and (5), BEGC is a parallelogram.

Hence, ar(GEH)= ½ar(ADGE) ——(6)

Adding equation (3) and (6),

ar(GEH) + ar(GEH) =½ar(BEGC) + ½ar(ADGE)

ar(EFGH) = ½ar(ABCD)


EX 9.2 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).

Solution:

ΔAPB and parallelogram ABCD lie on the same base AB

and between same parallel, AB || DC.

ar(ΔAPB) = ½ ar(parallelogram ABCD) — (i)  [If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram]

Similarly,

ΔBQC and parallelogram ABCD lie on the same base AB

and between same parallel, AB || DC.

ar(ΔBQC) = ½ ar(parallelogram ABCD) — (ii)

From (i) and (ii), we have

ar(ΔAPB) = ar(ΔBQC)


EX 9.2 QUESTION 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,

AB || GH (by construction) — (i)

∴, AD || BC ⇒ AG || BH — (ii)

From equations (i) and (ii),

ABHG is a parallelogram.

Now, ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel AB || GH.

∴ ar(ΔAPB) = ½ ar(ABHG) — (iii)

also, ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel CD || GH.

∴ ar(ΔPCD) = ½ ar(CDGH) — (iv)

Adding equations (iii) and (iv),

ar(ΔAPB) + ar(ΔPCD) = ½ [ar(ABHG)+ar(CDGH)]

⇒ ar(APB)+ ar(PCD) = ½ ar(ABCD)

(ii) A line EF is drawn parallel to AD passing through P.

In the parallelogram,

AD || EF (by construction) — (i)

∴ AB || CD ⇒ AE || DF — (ii)

From equations (i) and (ii),

AEDF is a parallelogram.

Now, ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD || EF.

∴ar(ΔAPD) = ½ ar(AEFD) — (iii)

also, ΔPBC and parallelogram BCFE is lying on the same base BC and between the same parallel lines BC || EF.

∴ar(ΔPBC) = ½ ar(BCFE) — (iv)

Adding equations (iii) and (iv),

ar(ΔAPD)+ ar(ΔPBC) = ½ {ar(AEFD)+ar(BCFE)}

⇒ar(APD)+ar(PBC) = ar(APB)+ar(PCD)


EX 9.2 QUESTION 5.


In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = ½ ar (PQRS) 

Solution:

(i) Parallelogram PQRS and ABRS lie on the same base SR and between the same parallel SR || PB.

∴ ar(PQRS) = ar(ABRS) — (i) [Parallerogram on the same base and between same parallel line are equal in area.]

(ii) ΔAXS and parallelogram ABRS are lying on the same base AS and between the same parallel AS || BR.

∴ ar(ΔAXS) = ½ ar(ABRS) — (ii) [If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram]

From (i) and (ii),

ar(ΔAXS) = ½ ar(PQRS)


EX 9.2 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 fields 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?

Solution:

The field is divided into three parts each in a triangular shape.

Let, ΔAPS, ΔAPQ and ΔARQ be the triangles.

ΔAPQ Parallelogram PQRS lie on the same base PQ and between the same parallel PQ || SR.

Area of ΔAPQ = ½ area of PQRS

[If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram]

The farmer must sow wheat or pulses in ΔAPQ or either in both ΔASP and ΔARQ.