Question

If triangle abc ,d is a point on bc, such that bd : bc = 2 : 5, what is the ratio area(\triangle abc): area(\triangle abd)?

Answer 1

Answer:

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Answer 2

The ratio of $area(\triangle ABC): area(\triangle ABD)$ is 5:2.

Step-by-step explanation:

Given information: In triangle ABC, D is a point on BC, such that BD : BC = 2 : 5.

Draw a perpendicular on BC from A of length h.

Area of a triangle is

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

BD : BC = 2 : 5, Let BD and BC are 2x and 5x respectively.

Area of triangle ABC is

$Area=\dfrac{1}{2}\times (5x)\times h=2.5xh$

Area of triangle ABD is

$Area=\dfrac{1}{2}\times (2x)\times h=xh$

The ratio of $area(\triangle ABC): area(\triangle ABD)$ is

$\dfrac{\triangle ABC}{\triangle ABD}=\dfrac{2.5xh}{xh}=2.5=\dfrac{5}{2}$

Therefore, the ratio of $area(\triangle ABC): area(\triangle ABD)$ is 5:2.

#Learn more

If A(5,6) B(3,-2) c(5,2) are the vertices of triangle ABC, If D is the mid point of BC then triangle ABD : triangle ACD is ​

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