Question

In ΔABC point D on side BC is such that DC = 6, BC = 15. Find A(Δ ABD) : A(Δ ABC)

1️⃣ 2:5
2️⃣ 5:2
3️⃣ 3:5
4️⃣ 2:3
​

Answer 1

Answer:

4 this is your answer and something it's rite

Answer 2

Given:

In ΔABC point D on side BC is such that DC = 6, BC = 15.

To find:

A(Δ ABD) : A(Δ ABC)

Solution:

Construction:- Draw a perpendicular line from vertex A on BC intersecting at point P.

We know,

$\boxed{\bold{Area \:of\:a\:triangle = \frac{1}{2}\times base \times height }}$

So,

$Area \:of\:\triangle ABD = \frac{1}{2}\times BD \times AP$ . . .  [Equation 1]

and

$Area \:of\:\triangle ABC = \frac{1}{2}\times BC \times AP$ . . . [Equation 2]

Now,

Area (Δ ABD) : Area (Δ ABC) is,

from equation 1 and equation 2, we get

= $\frac{\frac{1}{2}\times BD \times AP}{ \frac{1}{2}\times BC \times AP}$

= $\frac{ BD}{ BC }$

= $\frac{ BC - DC}{ BC }$

on substituting the given values of DC and BC, we get

= $\frac{ 15 - 6}{ 15 }$

= $\frac{9}{15}$

= $\bold{\frac{3}{5}}$ . . . option 3

Thus, A(Δ ABD) : A(Δ ABC) is → 3 : 5.

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