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## AP Board Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.3 Book Answers AP Board Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.3 Book Answers

## Andhra Pradesh State Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.3 Books Solutions

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## AP Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.3 Textbooks Solutions with Answer PDF Download

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### 10th Class Maths 7th Lesson Coordinate Geometry Ex 7.3 Textbook Questions and Answers

Question 1.
Find the area of the triangle whose vertices are
i) (2, 3), (-1, 0), (2,-4)
Given: A (2, 3), B (- 1, 0) and C (2, – 4) are the vertices of a △ABC.
Area of the triangle ABC = 12∣∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣∣
= 12|2(0+4)−1(−4−3)+2(3−0)|
= 12|8+7+6|
= 212
= 1012 sq.units

ii) (-5, -1), (3, -5), (5, 2)
Given: A (- 5, – 1), B (3, – 5) and C (5, 2) are the vertices of △ABC.
Area of the △ABC
= 12∣∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣∣
= 12|−5(−5−2)+3(2+1)+5(−1+5)|
= 12|35+9+20|
= 642
= 32 sq.units

iii) (0, 0), (3, 0), (0, 2)
Given: O (0, 0), A (3, 0) and B (0, 2) are the vertices of a triangle, △AOB.
Area of the △AOB
= 12∣∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣∣
= 12|0(0−2)+3(2−0)+0(0−0)|
= 12|6|
= 3 sq.units

Or △AOB = 12 × OA × OB
= 12 × 3 × 2
= 3 sq.units

Question 2.
Find the value of ‘K’ for which the points are collinear.
i) (7, -2), (5, 1), (3, K)
Given: A (7, – 2), B (5, 1) and C (3, K) are collinear.
∴ Area of △ABC = 0
But area of triangle
12∣∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣∣
⇒ 12∣7(1−K)+5( K+2)+3(−2−1) = 0
⇒ |7−7𝐾+5𝐾+10−9| = 0
⇒ |−2 K+8| = 0
⇒ -2K + 8 = 0
⇒ -2K = -8
⇒ K = 82
i.e., K = 4

ii) (8, 1), (K,-4), (2,-5)
Given: A (8, 1), B(K, – 4) and C (2, – 5) are collinear.
∴ Area of △ABC = 0
⇒ 12∣∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣∣ = 0
⇒ 12|8(−4+5)+K(−5−1)+2(1+4)| = 0
⇒ |8−6 K+10| = 0
⇒ |18−6 K| = 0
⇒ 18 – 6K = 0
⇒ 6K = 18
⇒ K = 186
i.e., K = 3

iii) (K,K), (2, 3), and (4,-1)
A (K, K), B (2, 3) and C (4, – 1) are collinear.
∴ Area of △ABC = 0
⇒ 12∣∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣∣ = 0
⇒ 12| K(3+1)+2(−1−K)+4( K−3)| = 0
⇒ |4𝐾−2−2𝐾+4𝐾−12| = 0
⇒ |6 K−14| = 0
⇒ 6K – 14 = 0
⇒ 6K = 14
⇒ K = 146 = 73
∴ K = 73

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle. Given: A (0, – 1), B (2, 1) and C (0, 3) are the vertices of △ABC.
Let D, E and F be the midpoints of the sides AB⎯⎯⎯⎯⎯⎯⎯⎯, BC⎯⎯⎯⎯⎯⎯⎯⎯ and AC⎯⎯⎯⎯⎯⎯⎯⎯. Area of a triangle ABC =
12∣∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣∣
= 12|0(1−3)+2(3+1)+0(−1−1)|
= 12|8|
= 4 sq.units
Area of △DEF = 12|1(2−1)+1(1−0)+0(0−2)|
= 12|1+1|
= 22
= 1 sq.units
Ratio of areas = △ABC : △DEF = 4 : 1.
△ADF ≅ △BED ≅ △DEF ≅ △CEF
∴ △ABC : △DEF = 4 : 1

Question 4.
Find the area of the quadrilateral whose vertices taken inorder are (-4, -2), (-3, -5),(3, -2) and (2, 3). Given: A (- 4, – 2), B (- 3, – 5), C (3, – 2) and D (2, 3) are the vertices of the quadrilateral ▱ ABCD.
Area of ▱ ABCD = △ABC + △ACD.
Area of a triangle =
12∣∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣∣ Question 5.
Find the area of the triangle formed by the points by using Heron’s formula.
i) (1, 1), (1, 4) and (5, 1)
ii) (2, 3), (-1,3) and (2, -1)
i) (1, 1) (1, 4) and (5, 1)
let A (1, 1) B(l, 4) and C(5, 1) are the vertices then length of sides can be calculated using the formula
(𝑥2−𝑥1)2+(𝑦2−𝑦1)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
now now formula for area of triangle using Heron’s formula = △ = 𝑠(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
where s = 𝑎+𝑏+𝑐2
∴ s = 3+4+52 = 122 = 6
∴ △ = 6(6−5)(6−4)(6−3)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
= 6×1×2×3‾‾‾‾‾‾‾‾‾‾‾‾√
= 6×6‾‾‾‾‾√
= 6 sq. units
∴ area of given triangle = 6 sq units

ii) let the vertices of given triangle A (2, 3), B (-l, 3) and C (2, -1) ∴ a = 5, b = 4, c = 3 units
now from using Heron’s formula area of triangle
= △ = 𝑠(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
where s = 𝑎+𝑏+𝑐2
= 5+4+32
= 122 = 6
∴ △ = 6(6−5)(6−4)(6−3)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
= 6×1×2×3‾‾‾‾‾‾‾‾‾‾‾‾√
= 6×6‾‾‾‾‾√
= 6 sq. units
∴ area of given triangle = 6 sq units

## Andhra Pradesh Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.3 Textbooks for Exam Preparations

Andhra Pradesh Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.3 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.3 exam preparation. The AP Board STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.3 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 7 Coordinate Geometry Ex 7.3 Books State Board syllabus with maximum efficiency.

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