10th Maths - Chapter 5.1 Book Back Questions and Answers

  

 

SSLC / 10th - Maths - Book Back Answers - Chapter 5 Unit Excercise 5.1 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 5 Unit Exercise 5.1: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 5 Unit Exercise 5.1 – Maths from the Tamil Nadu State Board 10th Standard textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 5 Unit Exercise 5.1 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 5 Coordinate Geometry Unit Ex 5.1

1. Find the area of the triangle formed by the points
(i) (1,-1), (-4, 6) and (-3, -5)
(ii) (-10, -4), (-8, -1) and (-3, -5)

Answer Key:

(i) Let the vertices A (1, -1), B (-4, 6) and C (-3, -5)

p

eq
= 12 [(6 + 20 + 3) – (4 – 18 – 5)] = 12 [29 – (-19)] = 12 [29 + 19]
= 12 × 48 = 24 sq. units.
Area of ∆ABC = 24 sq. units

(ii) Let the vertices be A(-10, -4), B(-8 -1) and C(-3, -5)

p

Area of ∆ABC = 12[(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]

= 12 [(50 + 3 + 32) – (12 + 40 + 10)]

eq

= 12 [85 – (62)] = 12 [23] = 11.5
Area of ∆ACB = 11.5 sq.units

2. Determine whether the sets of points are collinear?
(i) (-12,3)
(ii) (a, b + c), (b, c + a) and (c, a + b)

Answer Key:

(i) Let the points be A (-12,3), B (-5, 6) and C(-8, 8)
Area of ∆ABC = 12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₃ + x₁y₃)]
= 12 [(- 3 – 40 – 24) – (-15 – 48 – 4)]

eq

= 12 [-67 + 67] = 12 × 0 = 0
Area of a ∆ is 0.
∴ The three points are collinear.

(ii) Let the points be A (a, b + c), B (b, c + a) and C (c, a + b)

Area of the triangle = 12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]

eq

Since the area of a triangle is 0.
∴ The given points are collinear.

3. Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’

S.No	Vertices	Area (sq.units)
(i)	(0,0), (p,8), (6,2)	20
(ii)	(p,p), (5,6), (5,-2)	32

Answer Key:

(i) Let the vertices be A (0,0) B (p, 8), c (6, 2)
Area of a triangle = 20 sq. units
12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 20

eq

12 [(0 + 2p + 0) – (0 + 48 + 0)] = 20
12 [2p – 48] = 20
2p – 48 = 40
2p – 48 = 40 ⇒ 2p = 40 + 48
p = 
p = 882 = 44
The value of p = 44

(ii) Let the vertices be A (p, p), B (5, 6) and C (5, -2)

Area of a triangle = 32 sq. units

12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 32
eq

12 [6p – 10 + 5p) – (5p + 30 – 2p)] = 32
12 [11 p – 10 – 3p – 30] = 32
11p – 10 – 3p – 30 = 64
8p – 40 = 64
8p = 64 + 40 ⇒ 8p = 104
p = 1048 = 13
The value of p = 13

4. In each of the following, find the value of ‘a’ for which the given points are collinear.

(i) (2,3), (4, a) and (6, -3)
(ii) (a, 2 – 2a), (-a + 1, 2a) and (-4 -a, 6 – 2a).

Answer Key:

(i) Let the points be A (2, 3), B(4, a) and C(6, -3).
Since the given points are collinear.
Area of a triangle = 0
12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 0)

eq

12 [(2a – 12 + 18) – (12 + 6a – 6)] = 0

2a + 6 – (6 + 6a) = 0

2a + 6 – 6 – 6a = 0

-4a = 0 ⇒ a = 04 = 0

The value of a = 0

(ii) Let the points be A (a, 2 – 2a), B (-a + 1, 2a) C (-4 -a, 6 – 2a).
Since the given points are collinear.
Area of a ∆ = 0
eq

6a² – 2a – 2 – (-2a² – 6a + 2) = 0
6a² – 2a – 2 + 2a² + 6a – 2 = 0
8a² + 4a – 4 = 0 (Divided by 4)
2a² + a – 1 = 0
2a² + 2a – a – 1 = 0
2a (a + 1) – 1 (a + 1) = 0
p

(a + 1) (2a – 1) = 0
a + 1 = 0 (or) 2a – 1 = 0
a = -1 (or) 2a = 1 ⇒ a = 12
The value of a = -1 (or) 12

5. Find the area of the quadrilateral whose vertices are at

(i) (-9, -2), (-8, -4), (2, 2) and (1, -3)
(ii) (-9, 0), (-8,6), (-1, -2) and (-6, -3)

Answer Key:

(i) Let the vertices A (-9, -2), B(-8, -4), C(2, 2) and D(1, -3).
Plot the vertices in a graph.
p

[Note: Consider the points in counter clock wise order]

eq

Area of the Quadrilateral ABDC = 12 [36 + 24 + 2 – 4 – (16 – 4 – 6 – 18)]
= 12 [58 – (-12)] – 12[58 + 12]
= 12 × 70 = 35 sq. units 2
Area of the Quadrilateral = 35 sq. units

(ii) Let the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)
Plot the vertices in a graph and take them in counter – clock wise order.
Area of the Quadrilateral DCB

eq

= 12 [33 + 35] = 12 × 68 = 34 sq. units
Area of the Quadrilateral = 34 sq. units

p

6. Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (-4, -2), (-3, k), (3, -2) and (2, 3)

Answer Key:

Let the vertices A (-A, -2), B (-3, k), C (3, -2) and D (2, 3)
Area of the Quadrilateral = 28 sq. units
12 [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)] = 28
eq

-7k + 21 = 56
-7k = 56 – 21
-7k = 35 ⇒ 7k = – 35
k = – 357 = -5
The value of k = -5

7. If the points A(-3, 9), B(a, b) and C(4, -5) are collinear and if a + b = 1, then find a and b.

Answer Key:

Since the three points are collinear
Area of a ∆ = 0

eq

-3b – 5a + 36 – 9a – 4b – 15 = 0
-7b – 14a + 21 = 0
(÷ by 7) – b – 2a + 3 = 0
2a + b – 3 = 0

eq

Substitute the value of a = 2 in (2) ⇒ 2 + b = 1
b = 1 – 2 = -1
The value of a = 2 and b = -1

8. Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC . Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.

Answer Key:

Let the vertices of the ∆ABC be A(x₁,y₁), B(x₂,y₂), C(x₃,y₃)
p

eq

p

eq

9. In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.

p

Answer Key:

p
= 12 [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)]
= 12 [(16 + 80 + 36 + 80) – (-64 – 24 – 100 – 24)]
= 12 [212 – (-212)]

eq

= 12 [212 + 212] = 12 [424] = 212 sq. units

eq
= 12 [90 – (-90)]

eq

= 12 [90 + 90]

= 12 × 180 = 90 sq. units

Area of the patio = Area of the Quadrilateral ABCD – Area of the Quadrilateral EFGH

= (212 – 90) sq. units

    Area of the patio = 122 sq. units

10. A triangular shaped glass with vertices at A(-5, -4), B(l, 6) and C(7, -4) has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.

Answer Key:

Given the vertices of the triangular glass is A (-5, -4), B (1, 6), and C (7, -4)
p

eq

= 12 [(20 + 42 – 4) – (-28 – 4 – 30)]
= 12 [58 – (-62)]
= 12 [58 + 62]
= 12 × 120 = 60 sq. feet
Number of cans to paint 6 square feet = 1
∴ Number of cans = 606 = 10 ⇒ Number of cans = 10

11. In the figure, find the area of

(i) triangle AGF
(ii) triangle FED
(iii) quadrilateral BCEG.

p

Answer Key:

Area of a triangle = 12 [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
(i) Area of ∆AGF = 12 [(-2.5 – 13.5 – 6) – (-13.5 – 1 – 15)]
= 12 [-22 – (-29.5)]

eq

= 12 [-22 + 29.5]
= 12 × 7.5 = 3.75 sq.units

(ii) Area of ∆FED = 12 [(-2 + 4.5 + 3) – (4.5 + 1 – 6)]

= 12 [5.5 – (-0.5)]

eq

= 12 [5.5 + 0.5] = 12 × 6 = 3 sq.units

(iii)
eq

= 12 [(4 + 2 + 0.75 + 9) – (-4 -1.5 – 4.5 -2)]

= 12 [15.75 + 12]

= 12 [27.75] = 13.875

= 13.88 sq. units