Question

CBSE CLASS X

MATHEMATICS

The vertices of ∆ ABC are A(4,6), B(1,5) and C(7,2). A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that AD/AB = AE/AC = 1/3. Calculate the area of ∆ADE and compare it with the area of ∆ABC.

Answer 1

Question :-

The vertices of ∆ ABC are A(4,6), B(1,5) and C(7,2). A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that AD/AB = AE/AC = 1/3. Calculate the area of ∆ ADE and compare it with the area of ∆ABC.

Answer :-

Given :

AD/BB = AE/AC 1/3

In Δ ADE and Δ ABC :

∠DAE = ∠BAC [ Common ]

AD/AB = AE/AC [ Given ]

Hence : Δ AED ≈ Δ ABC [ S.A.S criteria ]

Thus :

( Area of Δ AED ) / (  Area of Δ ABC ) = AD² / AB²

==> ( Area of Δ AED ) / (  Area of Δ ABC ) = 1² / 3²

==> ( Area of Δ AED ) / (  Area of Δ ABC ) = 1 / 9

Thus the ratio of ( Area of Δ AED ) : (  Area of Δ ABC ) = 1 : 9

Area of Δ ADE

Finding the coordinates of E and D

AD : AB = 1 : 3

AD / AB = 1/3

==> AD = AB/3

==> AB = 3 AD

AD + BD = AB

==> AD + BD = 3 AD

==> BD = 3 AD - AD

==> BD = 2 AD

==> BD / AD = 2 / 1

By invertendo we have AD / BD = 1 / 2

                                 ==> AD : BD = 1 : 2

Same way :- AE : EC = 1 : 2

Use section formula :

Finding the coordinates of D

A ( 4 , 6 ) ------------ D ( x , y ) --------------------------------B ( 1 , 5 )

                    1                                            2

Here we get :

x₁ = 4

x₂ = 1

y₁ = 6

y₂ = 5

m : n = 1 : 2

m = 1

n = 2

x = ( m x₂ + n x₁ ) / ( m + n )

==> x = ( 1 × 1 + 2 ×4 ) / ( 2 + 1 )

==> x = ( 1 + 8 ) / 3

==> x = 9 / 3

==> x = 3

y = ( m y₂ + n y₁ ) / ( m + n )

==> y = ( 1.5 + 2.6 ) / ( 2 + 1 )

==> y = ( 5 + 12 ) / 3

==> y = 17 / 3

Hence D = ( 3 , 17 / 3 )

Finding the coordinates of E

A ( 4 , 6 ) ------------- E ( x , y ) ------------------------------C ( 7 , 2 )

                   1                                      2

Here the values will be :

x₁ = 4

x₂ = 7

y₁ = 6

y₂ = 2

m : n = 1 : 2

m = 1

n = 2

x = ( m x₂ + n x₁ ) / ( m + n )

==> x = ( 1 × 7 + 2 × 4 ) / ( 1 + 2 )

==> x = ( 7 + 8  ) / 3

==> x = 15 / 3

==> x = 5

y = ( m y₂ + n y₁ ) / ( m + n )

==> y =  ( 1 × 2 + 2 × 6 ) / ( 1 + 2 )

==> y = ( 2 + 12 ) / 3

==> y = 14 / 3

So E = ( 5 , 14 / 3 )

Finding the area of Δ ADE

x₁ = 4

x₂ = 3

x₃ = 5

y₁ = 6

y₂ = 17 / 3

y₃ = 14 / 3

Area of ADE = 1/2 | x₁ ( y₂ - y₃ ) + x₂ ( y₃ - y₁ ) + x₃ ( y₁ - y₂ ) |

                     = 1/2 | 4 ( 17/3 - 14/3 ) + 3 ( 14/3 - 6 ) + 5 ( 6 - 17/3 ) |

                     = 1/2 | 4 ( 3/3 ) + 3 ( 14 - 18 )/3 + 5 ( 18 - 17 )/3 |

                     = 1/2 | 4 × 1 + 3 × ( -4/3 ) + 5 × 1/3 |

                     = 1/2 | 4 - 4 + 5 / 3 |

                     = 1/2 × | 0 + 5/3 |

                     = 1/2 × 5/3

                     = 5/6

$\textsf{\huge{\mathfrak{\underline{\underline{ANSWERS}}}}}$

(i) The area of Δ ADE is 5/6 units²

(ii) ( Area of Δ AED ) : (  Area of Δ ABC ) = 1 : 9

__________________________________________________________

Answer 2

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Here,  

Given  

➩ AD/AB = 1/3  

➩ AD/(AD + BD) = 1/3  

➩ 3AD = AD + BD  

➩ 3AD — AD = BD  

➩ 2AD = BD  

➩ 2AD/BD = 1  

➩ AD/BD = 1/2 = AE/EC  

Now,  

, We shall find out the  

coordinates of point 'D' using  

section formula  

Let the coordinates of point  

'D' be (x , y)  

Now,  

➩ x = (8 + 1)/3 ➩ 9/3 ➩ 3  

➩ y = (12 + 5)/3 ➩17/3  

So, coordinates of point 'D' are (3, 17/3)  

Now,  

, We shall find out the  

coordinates of point 'E' using  

section formula  

Let the coordinates of point  

'E' be (a , b)  
Now,  

➩ a = (8 + 7)/3 ➩ 15/3 ➩ 5  

➩ b = (12 + 2)/3 ➩14/3  

So, coordinates of point 'D' are (5, 14/3)  

Now, ar(∆ ADE)  

➩ 1 / 2 │ [ 4 (17/3 - 14/3) + 3  

(14/3 - 6) + 5 (6 - 17/3) ] │  

➩ 1/2 │ [ 4 - 4 + 5/3 ] │  

➩ 1/2 │ [ 5/3 ] │  

➩ 1/2 × 5/3 = 5/6 unit sq.  

Now, ar(∆ABC)  

➩ 1 / 2 │ [ 4 (5 -2 ) + 1 (2 - 6) + 7  

(6 - 5) ] │  

➩ 1 / 2 │ [ 4 (3) + 1 (-4) + 7 (1) ] │  

➩ 1 / 2 │ [ 12 - 4 + 7 ] │  

➩ 1 / 2 │ [ 12 + 3 ] │  

➩ 1 / 2 × 15 = 15/2 unit sq

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