NCERT Solutions Class 9 Mathematics Areas of Parallelograms and Triangles Exercise 9.4

Answer :

(i) Let G and H be the mid-points of side AB and AC respectively.

Line segment GH is joining the mid-points. Therefore, it will be parallel to third side BC and

also its length will be half of the length of BC (mid-point theorem).

GH = BC/2 and GH || BD

GH = BD = DC and GH || BD                      [D is the mid-point of BC]

Consider quadrilateral GHDB,

GH ||BD and GH = BD

Two line segments joining two parallel line segments of equal length will also be equal and

parallel to each other.

Therefore, BG = DH and BG || DH

Hence, quadrilateral GHDB is a parallelogram.

We know that in a parallelogram, the diagonal bisects it into two triangles of equal area.

Hence, Area (∆BDG) = Area (∆HGD)

Similarly, it can be proved that quadrilaterals DCHG, GDHA, and BEDG are parallelograms and

their respective diagonals are dividing them into two triangles of equal area.

ar(∆GDH) = ar(∆CHD)               [For parallelogram DCHG]

ar(∆GDH) = ar(∆HAG)               [For parallelogram GDHA]

ar(∆BDE) = ar(∆DBG)                [For parallelogram BEDG]

ar(∆ABC) = ar(∆BDG) + ar(∆GDH) + ar(∆DCH) + ar(∆AGH)

ar(∆ABC) = 4 * ar(∆BDE)

Hence, ar(∆BDE) = ar(∆ABC)/4

(ii)Area (∆BDE) = Area (∆AED)                     [Common base DE and DE||AB]

Area (∆BDE) − Area (∆FED) = Area (∆AED) − Area (∆FED)

Area (∆BEF) = Area (∆AFD) ………….(1)

Area (∆ABD) = Area (∆ABF) + Area (∆AFD)

Area (∆ABD) = Area (∆ABF) + Area (∆BEF)              [From equation (1)]

Area (∆ABD) = Area (∆ABE) …………(2)

AD is the median in ∆ABC.

ar(∆ABD) = ar(∆ABC)/2

                  = 4 * ar(∆BDE)/2                                [As proved earlier]

                  = 2 * ar(∆BDE)

⇒ ar(∆ABD) = 2 * ar(∆BDE) ………..(3)

From equation (2) and (3), we get

2 * ar(∆BDE) = ar(∆ABE)

⇒ ar(∆BDE) = ar(∆ABE)/2

(iii) ar(∆ABE) = ar(∆BEC)                           [Common base BE and BE||AC]

      ar(∆ABF) + ar(∆BEF) = ar(∆BEC)

Using equation (1), we obtain

ar(∆ABF) + ar(∆AFD) = ar(∆BEC)

⇒ ar(∆ABD) = ar(∆BEC)

⇒ ar(∆ABC)/2 = ar(∆BEC)

⇒ ar(∆ABC) = 2 * ar(∆BEC)

(iv) It is seen that ∆BDE and ar ∆AED lie on the same base (DE) and between the parallels DE

and AB.

ar (∆BDE) = ar (∆AED)

ar (∆BDE) − ar (∆FED) = ar (∆AED) − ar (∆FED)

ar (∆BFE) = ar (∆AFD)

(v) Let h be the height of vertex E, corresponding to the side BD in ∆BDE.

Let H be the height of vertex A, corresponding to the side BC in ∆ABC.

In (i), it was shown that ar(∆BDE) = ar(∆ABC)/4

⇒ 1/2 * BD * h = (1/2 * BC * H)/4

⇒ BD * h = (2BD * H)/4

⇒ h = H/2

In (iv), it was shown that ar (∆BFE) = ar (∆AFD).

ar (∆BFE) = ar (∆AFD)

                 = 1/2 * FD * H

                 = 1/2 * FD * 2h

                 = 2(1/2 * FD * h)

                 = 2 * ar (∆FED)       

Hence, ar (∆BFE) = 2 * ar (∆FED)

(vi) Area (AFC) = area (AFD) + area (ADC)   

[in (iv), ar(∆BFE) = ar(∆AFD), AD is the median of ΔABC]

                              = ar (∆BFE) + ar (∆ABC)/2

                              = ar (∆BFE) + 4 * ar (∆BDE)/2

                             = ar (∆BFE) + 2 * ar (∆BDE)    …………..(5)

Now, by (v), ar (∆BFE) = 2 * ar (∆FED)    …………..(6)

ar (∆BDE) = ar (∆BFE) + ar (∆FED) = 2 * ar (∆FED) + ar (∆FED) = 3 * ar (∆FED)   ………7

From equation (5), (6) and (7), we get

ar (∆AFC) = 2 * ar (∆FED) + 2 * 3 * ar (∆FED)

                  = 2 * ar (∆FED) + 6 * ar (∆FED)

                  = 8 * ar (∆FED)

Hence, ar (∆FED) = ar (∆AFC)/8

Question : 6: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB) * ar(CPD) = ar(APD) * ar(BPC).                                                                                               [Hint: From A and C, draw perpendiculars to BD.]

Answer :