Question

In ABC,Angle B is 90 degree and BD is the perpendicular bisector of AC the ratio of areas of triangle ADB and triangle ABC is ​

Answer 1

The ratio of areas of triangle ADB and triangle ABC is $[\frac{AD^2}{AB^2}] = [\frac{DB^2}{BC^2}] = [\frac{AB^2}{AC^2}]$.

Step-by-step explanation:

It is given that,

In ΔABC

∠B = 90°

BD is the perpendicular bisector of AC

 Since BD ⊥ AC ∴ ∠BDA = ∠BDC = 90°

Now, consider ΔADB and ΔABC, we have

∠A = ∠A ..... [common angle]

AB = AB ....... [common side]

∠ADB = ∠ABC = 90° ..... [given]

∴ By ASA similarity, ΔADB ~ ΔABC

We know that the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

∴  $\frac{Ar(Triangle ADB)}{Ar(Triangle ABC)} = [\frac{AD^2}{AB^2}] = [\frac{DB^2}{BC^2}] = [\frac{AB^2}{AC^2}]$

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Also View:

In a triangle ABC, angle BAC = 90 degree and AD is perpendicular to BC . Then which one is true and why?  

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In the figure , angle ABC = 90° and BD perpendicular AC . if BD = 8cm and AD =4cm , then find the value of CD .

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Answer 2

Answer:

1:2

Step-by-step explanation:

try solvig by using therom

area of triangle abc by area of triangle pqr =(ab)²/(pq)²

science the angle given is 90

substutute the value of carosponding sides and get the answer 1:2