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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

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

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	10th
Subject	Maths
Chapters	Maths Chapter 7 Coordinate Geometry Ex 7.3
Provider	Hsslive

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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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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.
Answer:

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).
Answer:

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)
Answer:
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

AP Board Textbook Solutions PDF for Class 10th Maths

AP Board Class 10 Maths Chapter 1 Real Numbers Textbook Solutions PDF

AP Board Class 10 Maths Chapter 2 Sets Textbook Solutions PDF

AP Board Class 10 Maths Chapter 3 Polynomials Textbook Solutions PDF

AP Board Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variables Textbook Solutions PDF

AP Board Class 10 Maths Chapter 5 Quadratic Equations Textbook Solutions PDF

AP Board Class 10 Maths Chapter 6 Progressions Textbook Solutions PDF

AP Board Class 10 Maths Chapter 7 Coordinate Geometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 8 Similar Triangles Textbook Solutions PDF

AP Board Class 10 Maths Chapter 9 Tangents and Secants to a Circle Textbook Solutions PDF

AP Board Class 10 Maths Chapter 10 Mensuration Textbook Solutions PDF

AP Board Class 10 Maths Chapter 11 Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 12 Applications of Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 13 Probability Textbook Solutions PDF

AP Board Class 10 Maths Chapter 14 Statistics Textbook Solutions PDF

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Thursday, June 9, 2022