RD Sharma Class 10 Ex 14.5 Solutions Chapter 14 Coordinate Geometry

In this chapter, we provide RD Sharma Class 10 Ex 14.5 Solutions Chapter 14 Coordinate Geometry for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 14.5 Solutions Chapter 14 Coordinate Geometry pdf, Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 14
Chapter NameCoordinate Geometry
Exercise14.5
CategoryRD Sharma Solutions

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(y– y3) + x2(y– y1) + x3(y+ y2)] 

Here, x= 6,y= 3, x2 = -3, y= 5, x= 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[a– 2a– 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.

QuadrilateralsArea 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 quadrilateral 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 ∆ABCArea 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

All Chapter RD Sharma Solutions For Class10 Maths

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good

Leave a Comment

Your email address will not be published.