Question

In figure 1.13 BC⊥AB,AD⊥AB,BC=4,AD=8, then find A( ΔABC)/A(ΔADB).

Answer 1

here is your answer.

Answer 2

Answer:

$\frac{Area(ABC)}{Area(ADB)}=\frac{1}{2}$

Step-by-step explanation:

In the given figure,

We have two triangles, ABC and ADB,

So, as we know that,

Area of triangle is given by,

$Area=\frac{1}{2}\times base\times height$

So,

Area of triangle, ABC, is given by,

$Area(ABC)=\frac{1}{2}\times BC \times AB\\Area(ABC)=\frac{1}{2}\times 4\times AB=2.AB\\Area(ABC)=2.AB$

Also,

Area of triangle, ADB, is given by,

$Area(ADB)=\frac{1}{2}\times AD \times AB\\Area(ADB)=\frac{1}{2}\times 8\times AB=4.AB\\Area(ADB)=4.AB$

So,

Ratio of the area of the triangles ABC and ADB is given by,

$\frac{Area(ABC)}{Area(ADB)}=\frac{2.AB}{4.AB}\\\frac{Area(ABC)}{Area(ADB)}=\frac{1}{2}$

Therefore, the ratio of the areas of the triangle ABC and ADB is 1:2.