Math, asked by aulakhharsh064, 6 hours ago

step by step explanation ​

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Answered by tennetiraj86
2

Step-by-step explanation:

Solution :-

In ∆ ABC ,

AC = 52 cm

BC = 48 cm

Let AB = X cm

In ∆ ABC , there is another triangle ADB

In ∆ ADB , angle ADB = 90°

=> ∆ADB is a right angled triangle

AB is the hypotenuse = X cm

BD = 16 cm

AD = 12 cm

We know that

Pythagoras Theorem:-

In a right angled triangle The square of the hypotenuse is equal to the sum of the squares of the other two sides

AB² = AD² + BD²

=> X² = 12²+16²

=> X² = 144+256

=> X² = 400

=> X=±√400

=> X = ±20 cm

X acn not be negative since it is the length of the side

X = 20 cm

Therefore , AB = 20 cm

To find the area of the shaded region we have to find the area of ∆ABC and Subtract the area of ∆ ADB from it.

Area of the shaded region = Area of ∆ ABC - Area of ∆ ADB

Finding Area of ∆ ADB :-

Area of a right angled triangle = (1/2)ab sq.units

We have a = AD = 12 cm

and b = BD = 16 cm

Area of ∆ ADB = (1/2)×AD×BD

=> Area of ∆ADB = (1/2)×(12×16)

=> Area of ∆ ADB = 6×16

=> Area of ∆ ADB = 96 cm² -----(1)

Finding Area of ∆ ABC:-

We have

AB = 20 cm , BC = 48 cm , CA = 52 cm

Area of triangle by Heron's formula

∆ =√[S(S-a)(S-b)(S-c)] sq.units

Where , S = (a+b+c)/2 units

Let a = 20 cm , b= 48 cm , c = 52 cm

S = (20+48+52)/2

=> S = 120/2

=> S = 60 cm

Now

Area of ∆ ABC

=> ∆=√[60(60-20)(60-48)(60-52)]

=> ∆ = √(60×40×12×8)

=> ∆ =√(3×20×20×2×3×4×4×2)

=> ∆ = √[(20×20)×(3×3)×(2×2)×(4×4)]

=> ∆ = 20×3×2×4 cm²

=> ∆ = 480 cm²

Area of ∆ ABC = 480 cm²-------(2)

Area of the shaded region

=> Area of ∆ABC - Area of ∆ ADB

=> (2)-(1)

=>480-96

=> 384 cm²

Answer:-

Area of the shaded region for the given problem is 384 cm²

Used formulae:-

Pythagoras Theorem:-

"In a right angled triangle The square of the hypotenuse is equal to the sum of the squares of the other two sides".

  • The opposite side of the right angle is the hypotenuse.

  • Area of a right angled triangle = (1/2)ab sq.units

  • Area of triangle by Heron's formula
  • ∆ =√[S(S-a)(S-b)(S-c)] sq.units

  • S = (a+b+c)/2

  • a ,b,c are the three sides of the triangle
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