Question

In the given Fig. 4.47, if AB = 2, BC = 6, AE = 6, BF = 8, CE = 7, and CF = 7, compute the ratio of the area of quadrilateral ABDE to the area of ΔCDF.

Answer 1

Refer to the attached image.

Since triangle AEC have side measures as AE = 6cm , AC = 2+ 6 = 8cm and EC = 7 cm.

And triangle BFC have side measures as BC = 6cm, BF = 8cm and CF = 7cm

Since, both the triangles AEC and BFC have sides equal in measure.

therefore, the area of both the triangles are same.

area(triangle AEC) = area(triangle BFC)

Since triangle BDC is common in both the triangles.

So, let us subtract the area(triangle BDC) from the area of these two triangles.

Therefore, $ar(\Delta AEC) - ar(\Delta BDC) = ar(\Delta BFC)-ar(\Delta BDC)$

So, ar(ABED) = ar(CDF).

Now, we have to find the ratio of area of quadrilateral ABDE to the area of ΔCDF.

Since, ar(ABED) = ar(CDF).

Therefore, the ratio is 1:1.

Answer 2

Triangle BCD which is similar to ACE.

And BCF

To gets lengths BD and CD for triangle BCD we will use linear scale factor.

Here is the LSF

8/6 = 4/3

8(BF) /BD = 4/3

24 = 4BD

BD = 24/4 = 6

CE/CD = 8/6

7/(CD) = 8/6

42 = 8CD

CD = 42/8

= 5.25

The area of ACE :

Here is the formula:

A = √S(S - a) (S - b) (S - c)

S = (a + b + c) /2

S = (6 + 7 + 8)/2

= 10.5

A = √10.5(10.5 - 6)(10.5 - 7)(10.5 - 8) = √413.4375 = 20.33

Area of BCF is also equal to 20.33

Area of BCD :

S = (6 + 6+, 5.25) /2 = 8.625

A = √8.625(8.625 - 6)(8.625 - 6)(8.625 - 5.25) = √200.58 = 14.16

= 14.16

Area of CDF :

S = (5.25 + 7 + 2) / 2 = 7.125

A = √7.125(7.125 - 5.25) (7.125 - 7)(7.125 - 2) = √8.558 = 2.925

= 2.925

Area of ABDE.

20.33 - 14.16 = 6.17

The ratio of ABDE : CDF

6.17 : 2.925

Approximately :

6 : 3

= 2 : 1