Question

In the fiurgurein the figure ab is equal to 12 cm and area is equal to 4 cm and BD is equal to customer what is the ratio between the area of triangle ABC and triangle bdc

Answer 1

Answer:

(1). \frac{ar(ABD)}{ar(ADC)}=\frac{7}{13}

(2). \frac{ar(ABD)}{ar(ABC)}=\frac{7}{20}

(3). \frac{ar(ADC)}{ar(ABC)}=\frac{13}{20}

Step-by-step explanation:

From the given figure,

BDC is a straight line.

BD = 7 and BC = 20

So,

DC = BC - BD = 20 - 7 = 13

DC = 13

So,

Area of triangle ABD is,

Area(ABD)=\frac{1}{2}\times BD \times h

and,

Area of triangle ADC is,

Area(ADC)=\frac{1}{2}\times DC \times h

where, h is the height of the triangle which is same for all the triangles.

So,

\frac{ar(ABD)}{ar(ADC)}=\frac{\frac{1}{2}\times BD\times h}{\frac{1}{2}\times DC\times h}=\frac{7\times h}{13 \times h}\\\frac{ar(ABD)}{ar(ADC)}=\frac{7}{13}

Therefore, the ratio of area of triangles ABD and ADC is given by,

\frac{ar(ABD)}{ar(ADC)}=\frac{7}{13}

(2).

Area of triangle ABD is,

Area(ABD)=\frac{1}{2}\times BD \times h

and,

Area of triangle ABC is,

Area(ABC)=\frac{1}{2}\times BC \times h

So,

\frac{ar(ABD)}{ar(ABC)}=\frac{\frac{1}{2}\times BD\times h}{\frac{1}{2}\times BC\times h}=\frac{7\times h}{20 \times h}\\\frac{ar(ABD)}{ar(ABC)}=\frac{7}{20}

Therefore, the ratio of area of triangles ABD and ABC is given by,

\frac{ar(ABD)}{ar(ABC)}=\frac{7}{20}

(3).

Area of triangle ADC is,

Area(ADC)=\frac{1}{2}\times DC \times h

and,

Area of triangle ABC is,

Area(ABC)=\frac{1}{2}\times BC \times h

So,

\frac{ar(ADC)}{ar(ABC)}=\frac{\frac{1}{2}\times DC\times h}{\frac{1}{2}\times BC\times h}=\frac{13\times h}{20 \times h}\\\frac{ar(ADC)}{ar(ABC)}=\frac{13}{20}

Therefore, the ratio of area of triangles ADC and ABC is given by,

\frac{ar(ADC)}{ar(ABC)}=\frac{13}{20}