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RD Sharma Solutions for Class 10 Chapter 14 Coordinate Geometry Ex 14.5 Download PDF

Chapter 14: Coordinate Geometry Exercise – 14.5

Question: 1

Find the area of a triangles whose vertices are

(i) (6, 3), (-3, 5) and (4, – 2)

(ii) [(at₁², at₁),( at₂², 2at₂)( at₃², 2at₃)]

(iii) (a, c + a), (a, c) and (-a, c – a)

Solution:

(i) Area of a triangle is given by

1/2[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ + y₂)] 

Here, x₁ = 6,y₁ = 3, x₂ = -3, y₂ = 5, x₃ = 4,y₃ = -2]

Let A(6, 3), B(-3, 5) and C(4,-2) be the given points

Area of ∆ABC = 1/2 [6(5+2)+(-3)(- 2 -3)+ 4(3 – 5)]

=1/2 [6 × 7- 3 × ( – 5) + 4( – 2)]

= 1/2[42 +15 – 8]

= 49/2 sq. units

(ii) Let A = (x₁, y­₁) = (at₁², 2at₁)

B = (x₂,y₂) = (at₂², 2at₂)

= (x₃, y₃) = (at₃², 2at₂) be the given points.

The area of ∆ABC

 (iii) Let A = (x₁,y₁) = (a, c + a)

B = (x₂, y₂) = (a, c)

C = (x₃, y₃) = (- a, c – a) be the given points

The area of ∆ABC

= 1/2[a ( – {c – a}) + a(c – a – (c + a)) +( – a)(c + a – a)]

= 1/2 [a(c – c + a) + a(c – a – c – a) – a(c + a – c)]

= 1/2[a × a + ax( – 2a) – a × a]

= 1/2[a² – 2a² – a²]

= 1/2×(-2a)²

= – a²

Question: 2

Find the area of the quadrilaterals, the coordinates of whose vertices are

(i) (-3, 2), (5, 4), (7, -6) and (-5,- 4)

(ii) (1, 2), (6, 2), (5, 3) and (3, 4)

(iii) (-4, -2), (-3, -5), (3, -2), (2, 3)

Solution:

(i) Let A(-3, 2), B(5, 4), C(7,- 6) and D ( -5, – 4) be the given points.

Area of ∆ABC

= 1/2[-3(4 + 6) + 5(- 6 – 2) + 7(2 – 4)]

= 1/2[-3×1 + 5×(-8) + 7(-2)]

= 1/2[- 30 – 40 -14]

=  – 42

But area cannot be negative

∴  Area of ∆ADC = 42 square units

Area of ∆ADC

= 1/2[-3( – 6 + 4) + 7(- 4 – 2) + (- 5)(2 + 6)]

= 1/2[- 3( – 2) + 7(- 6) – 5 × 8]

= 1/2[6 – 42 – 40]

= 1/2 × – 76

= – 38

But area cannot be negative

∴ Area of ∆ADC = 38 square units

Now, area of quadrilateral ABCD

= Ar. of ABC+ Ar of ADC

= (42 + 38)

= 80 square. Units

(ii)  Let A(1, 2) , B (6, 2) , C (5, 3) and (3, 4) be the given points         

Area of ∆ABC

= 1/2[1(2 – 3) + 6(3 – 2) + 5(2 – 2)]             

= 1/2[ -1 + 6 × (1) + 0]

= 1/2[ – 1 + 6]

= 5/2

Area of ∆ADC

= 1/2[1(3 – 4) + 5(4 – 2) + 3(2 – 3)]

= 1/2[-1 × 5 × 2 + 3(-1)]

= 1/2[-1 + 10 – 3]

= 1/2[6]

= 3

Now, Area of quadrilateral ABCD

= Area of ABC + Area of ADC

(iii) Let A (- 4, 2), B( – 3, – 5), C (3,- 2) and D(2, 3) be the given points

Area of ∆ABC = 1/2|(- 4)(- 5 + 2) – 3(-2 + 2) + 3(- 2 + 5)|

= 1/2|(-4)(-3) – 3(0) + 3(3)|

=  21/2

Area of ∆ACD = 1/2|( – 4)(3 + 2) + 2( – 2 + 2) + 3( – 2 – 3)|

 = 1/2|- 4(5) + 2(0) + 3(- 5)|= (- 35)/2

But area can’t neative, hence area of ∆ADC = 35/2

Now, of quadrilateral (ABCD) = ar(∆ABC) + ar(∆ADC)

Area (quadrilateral ABCD) = 21/2 + 35/2

Area (quadrilateral ABCD) = 56/2

Area (quadrilateral ABCD) = 28 square. Units

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