NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.3) Exercise 7.3
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NCERT provides indepth knowledge of various subjects through the books they publish. However, books are not the only educational material provided by NCERT. NCERT, or the National Council of Educational Research and Teaching, is an organisation formed by the government to provide educational materials and teaching materials to schools. Many schools have been affiliated with NCERT in order to train new teachers and give educational resources to their students. The resources given by NCERT are widely used by students to prepare for many exams, like the board exams for Class 10 and Class 12, and various entrance exams like JEE, NEET, CUET, NDA, etc., for admission into different programmes and universities of theirdesire.
For some students, Mathematics is a particularly tricky subject. They might have trouble with Mathematics, and they might run into trouble with some problems. Mathematics has always piqued the curiosity of students. Geometry is new to all students and may scare them even if Algebra, Trigonometry, and Linear Equations in Two Variables are all included in Mathematics for Class 9 and Class 10. While some students struggle with it, others relish the challenges. Students may find it more difficult to finish the NCERT Exercises because they may find it challenging to comprehend the themes when they are studying them for the first time. Even though learning something new is never simple, after they have mastered it, they will feel incredibly satisfied when they are done with their studies. Understanding and studying the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 is important, and it takes a lot of practice. Students can better understand all of the chapter’s topics with the guidance of the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3.
The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 provide information about the topics in Class 10 Maths Chapter 7 Exercise 7.3. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 provides the solutions to the questions given in Class 10 Maths Ex 7.3. These NCERT Solutions are helpful for students. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 is a resourceful tool as it provides students with the resources to solve the questions given in the books published by NCERT. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 will help students understand the chapter and find out the correct answers and methods of answering each question.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.3) Exercise 7.3
To effectively manage their time during an exam, students must be able to quickly solve a question. Time management during the exam is crucial for students since it will enable them to review their answer sheets at the end and check for any major errors. Answering the questions requires thoroughly reviewing the materials provided to students and then figuring out the method of solution and formula, at the end, proper implementation is necessary to get the correct answers. This gives them time to remedy their errors while also assisting them in understanding how they should have answered the questions and where they must have gone wrong. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 can assist students in recognising and fixing their errors when practising, as well as becoming aware of several mistakes they are prone to make during the exam due to pressure and anxiety Hence, The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 is quite useful during taking exams. It will assist the students in time management during the exam, completion of their coursework, and preparation for the test.
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Access NCERT Solutions for Maths Chapter 7 – Coordinate Geometry
NCERT has been employed to teach the foundations of Mathematics and other topics ever since it began. NCERT is one of the best books to prepare for achieving higher marks on the CBSE board exams. NCERT books are a part of the curriculum for many students whose schools are affiliated with CBSE, or the Central Board of Secondary Education. CBSE has been recommending the NCERT books for study. The Mathematics NCERT book is one of the suggested textbooks for board exams. Most students prioritise solving NCERT problems before working on practise papers or taking a test. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 will be helpful to students as they progress through the NCERT. They can use the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 to quickly get the correct answers or faster solutions to these problems.
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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3
Students must practise Chapter 7 Coordinate Geometry in order to gain good marks in their exams. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 will students gain a thorough understanding of the chapter.The Class 10 Maths Exercise 7.3 Solutions by Extramarks is an epic tool for students to use in their preparation for getting good scores in exams. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 also gives an overview of the chapter to students to help them understand and draw rough sketches of the concepts that will be presented to them in the chapter. The concepts that are included in this chapter, as mentioned in the NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 are the use of the distance formula, the section formula, the midpoint formula and the area of a triangle. This helps students understand the application of these formulas of geometry in various different types of questions. This will help them solve more advanced problems. The NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 will help students understand the basic concepts from NCERT and help them solve advanced questions. In addition to these topics and exercises based on them, NCERT textbooks like the Exemplar also have optional exercises available to them for studying advanced questions that contain questions where more than one formula or concept might need to be applied. These exercises will help them solve that.
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Q.1
$\begin{array}{l}\text{Find the area of the triangle whose vertices are :}\\ \text{(i) (}2,\text{}3),\text{}(1,\text{}0),\text{}(2,4)\\ \text{(ii)}(5,\text{}1),\text{}(3,\text{}5),\text{}(5,\text{}2)\end{array}$
Ans
$\begin{array}{l}\text{Area of a triangle with vertives (}{\mathrm{x}}_{1}{\text{, y}}_{1}\text{), (}{\mathrm{x}}_{2}\text{,}{\mathrm{y}}_{2}\text{) and (}{\mathrm{x}}_{3}\text{,}{\mathrm{y}}_{3}\text{)}\\ \text{is given by the expression}\frac{1}{2}\left\{{\mathrm{x}}_{1}({\mathrm{y}}_{2}{\mathrm{y}}_{3})+{\mathrm{x}}_{2}({\mathrm{y}}_{3}{\mathrm{y}}_{1})+{\mathrm{x}}_{3}({\mathrm{y}}_{1}{\mathrm{y}}_{2})\right\}\text{.}\\ \text{Therefore,}\\ \text{(i) Area of the triangle with vertices (}2,\text{}3),\text{}(1,\text{}0\left)\text{and}\right(2,4)\\ \text{}=\frac{1}{2}\left\{2(0(4\left)\right)+(1)(43)+2(30)\right\}\\ \text{}=\frac{1}{2}(8+7+6)=\frac{21}{2}\text{square units}\\ \text{(ii) Area of the triangle with vertices}(5,\text{}1),\text{}(3,\text{}5)\text{and}(5,\text{}2)\\ \text{}=\frac{1}{2}\left\{5(52)+3(2(1\left)\right)+5(1(5)\right\}\\ \text{}=\frac{1}{2}(35+9+20)=32\text{square units}\end{array}$
Q.2 In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, –2), (5, 1), (3, k)
(ii) (8, 1), (k, – 4), (2, –5)
Ans
$\begin{array}{l}\text{Points (}{\mathrm{x}}_{1}{\text{, y}}_{1}\text{), (}{\mathrm{x}}_{2}\text{,}{\mathrm{y}}_{2}\text{) and (}{\mathrm{x}}_{3}\text{,}{\mathrm{y}}_{3}\text{) are collinear if}\\ \frac{1}{2}\left\{{\mathrm{x}}_{1}({\mathrm{y}}_{2}{\mathrm{y}}_{3})+{\mathrm{x}}_{2}({\mathrm{y}}_{3}{\mathrm{y}}_{1})+{\mathrm{x}}_{3}({\mathrm{y}}_{1}{\mathrm{y}}_{2})\right\}=0\\ \text{Therefore,}\\ \text{(i) Points (7,}\text{2)},\text{(5, 1) and (3, k) are collinear if}\\ \text{}\frac{1}{2}\left\{7(1\mathrm{k})+5(\mathrm{k}+2)+3(21)\right\}=0\\ \text{or 7}\text{7}\mathrm{k}+5\mathrm{k}+109=0\\ \text{or}\text{2k}+\text{8}=0\\ \text{or}\mathrm{k}=4\\ \text{(ii) Points (8, 1)},\text{(}\mathrm{k}\text{,}4\text{) and (2,}5\text{) are collinear if}\\ \text{}\frac{1}{2}\left[8\{4(5\left)\right\}+\mathrm{k}(51)+2\{1(4\left)\right\}\right]=0\\ \text{or 8}6\mathrm{k}+10=0\\ \text{or}\text{6k}=1\text{8}\\ \text{or}\mathrm{k}=3\text{}\end{array}$
Q.3
$\begin{array}{l}\text{Find the area of the triangle formed by joining the}\\ \text{midpoints of the sides of the triangle whose}\\ \text{vertices are}(0,\text{}1),\text{}(2,\text{}1)\text{and (0, 3). Find the}\\ \text{ratio of this area to the area of the given triangle.}\end{array}$
Ans
$\begin{array}{l}\text{Area of a triangle with vertices (}{\mathrm{x}}_{1}{\text{, y}}_{1}\text{), (}{\mathrm{x}}_{2}\text{,}{\mathrm{y}}_{2}\text{) and (}{\mathrm{x}}_{3}\text{,}{\mathrm{y}}_{3}\text{)}\\ \text{is given by the expression}\frac{1}{2}\left\{{\mathrm{x}}_{1}({\mathrm{y}}_{2}{\mathrm{y}}_{3})+{\mathrm{x}}_{2}({\mathrm{y}}_{3}{\mathrm{y}}_{1})+{\mathrm{x}}_{3}({\mathrm{y}}_{1}{\mathrm{y}}_{2})\right\}\text{.}\\ \text{Therefore,}\\ \text{area of the triangle with vertices (0},\text{}1),\text{}(2,\text{1}\left)\text{and}\right(0,3)\\ \text{}=\frac{1}{2}\left\{0(13)+2(3+1)+0(11)\right\}\\ \text{}=\frac{1}{2}(0+8+0)=4\text{square units}\\ \text{Vertices of the triangle formed by joining the midpoints of the}\\ \text{sides of the triangle with vertices}(0,\text{}1),\text{}(2,\text{}1)\text{and (0, 3)}\\ \text{are}(\frac{0+2}{2},\text{}\frac{1+1}{2}),\text{}(\frac{2+0}{2},\text{}\frac{1+3}{2})\text{and}(\frac{0+0}{2},\text{}\frac{1+3}{2})\text{i.e.,}\\ (1,\text{}0),\text{}(1,\text{}2)\text{and}(0,\text{}1).\\ \text{Area of the triangle with vertices}(1,\text{}0),\text{}(1,\text{}2)\text{and}(0,\text{}1)\\ \text{}=\frac{1}{2}\left\{1(21)+1(10)+0(02)\right\}\\ \text{}=\frac{1}{2}(1+1+0)=1\text{square unit}\\ \text{Therefore, required ratio}=1:4\text{}\end{array}$
Q.4
$\begin{array}{l}\text{Find the area of the quadrilateral whose vertices, taken in order, are (}\text{4,}\text{2), (}\text{3,}\text{5), (}3,\text{}2\text{)}\\ \text{and (2, 3).}\end{array}$
Ans
\begin{array}{l}\text{Area of a triangle with vertices (}{x}_{1}{\text{, y}}_{1}\text{), (}{x}_{2}\text{,}{y}_{2}\text{) and (}{x}_{3}\text{,}{y}_{3}\text{)}\\ \text{is given by the expression}\frac{1}{2}\left\{{x}_{1}({y}_{2}{y}_{3})+{x}_{2}({y}_{3}{y}_{1})+{x}_{3}({y}_{1}{y}_{2})\right\}\text{.}\\ \text{Therefore,}\\ \text{area of}\Delta \text{ABC with vertices A(}4,\text{}2),\text{B(}3,\text{}5)\text{and C(3,}2)\\ \text{}=\frac{1}{2}\left\{4(5+2)+(3)(2+2)+3(2+5)\right\}\\ \text{}=\frac{1}{2}(12+0+9)=\frac{21}{2}\text{square units}\\ \text{area of}\Delta \text{ADC with vertices A(}4,\text{}2),\text{D(2},\text{3})\text{and C(3,}2)\\ \text{}=\frac{1}{2}\left\{4(3+2)+2(2+2)+3(23)\right\}\\ \text{}=\frac{1}{2}(20+015)=\frac{35}{2}\text{}\\ \text{But area is always a positive quantity}\text{.}\\ \therefore \text{area of}\Delta \text{ADC}=\frac{35}{2}\text{square units}\\ \text{Area of the given quadrilateral ABCD}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}=\text{area of}\Delta \text{ABC}+\text{area of}\Delta \text{ADC}\\ \text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}=\left(\frac{21}{2}+\frac{35}{2}\right)\text{square units}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}=\text{28 square units}\end{array}
Q.5
$\begin{array}{l}\mathrm{You}\text{\hspace{0.17em}}\mathrm{have}\text{\hspace{0.17em}}\mathrm{studied}\text{\hspace{0.17em}}\mathrm{in}\text{\hspace{0.17em}}\mathrm{Class}\text{\hspace{0.17em}}\mathrm{IX},\text{\hspace{0.17em}}\left(\mathrm{Chapter}\text{\hspace{0.17em}}9,\text{\hspace{0.17em}}\mathrm{Example}\text{\hspace{0.17em}}3\right),\\ \mathrm{that}\text{\hspace{0.17em}}\mathrm{a}\text{\hspace{0.17em}}\mathrm{median}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{a}\text{\hspace{0.17em}}\mathrm{triangle}\text{\hspace{0.17em}}\mathrm{divides}\text{\hspace{0.17em}}\mathrm{it}\text{\hspace{0.17em}}\mathrm{into}\text{\hspace{0.17em}}\mathrm{two}\text{\hspace{0.17em}}\mathrm{triangles}\\ \mathrm{of}\text{\hspace{0.17em}}\mathrm{equal}\text{\hspace{0.17em}}\mathrm{areas}.\text{\hspace{0.17em}}\mathrm{Verify}\text{\hspace{0.17em}}\mathrm{this}\text{\hspace{0.17em}}\mathrm{result}\text{\hspace{0.17em}}\mathrm{for}\text{\hspace{0.17em}}\mathrm{\Delta ABC}\text{\hspace{0.17em}}\mathrm{whose}\\ \mathrm{vertices}\text{\hspace{0.17em}}\mathrm{are}\text{\hspace{0.17em}}\mathrm{A}\left(4,6\right),\text{\hspace{0.17em}}\mathrm{B}\left(3,2\right)\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\mathrm{C}\left(5,2\right).\end{array}$
Ans
$\begin{array}{l}\text{Area of a triangle with vertives (}{\mathrm{x}}_{1}{\text{, y}}_{1}\text{), (}{\mathrm{x}}_{2}\text{,}{\mathrm{y}}_{2}\text{) and (}{\mathrm{x}}_{3}\text{,}{\mathrm{y}}_{3}\text{)}\\ \text{is given by the expression}\frac{1}{2}\left\{{\mathrm{x}}_{1}({\mathrm{y}}_{2}{\mathrm{y}}_{3})+{\mathrm{x}}_{2}({\mathrm{y}}_{3}{\mathrm{y}}_{1})+{\mathrm{x}}_{3}({\mathrm{y}}_{1}{\mathrm{y}}_{2})\right\}\text{.}\\ \text{Therefore,}\\ \text{area of}\mathrm{\Delta}\text{ABC with vertices A(}4,\text{}6),\text{B(}3,\text{}2)\text{and C(5,}2)\\ \text{}=\frac{1}{2}\left\{4(22)+3(2+6)+5(6+2)\right\}\\ \text{}=\frac{1}{2}(16+2420)=6\text{}\\ \text{But area is always a positive quantity.}\\ \therefore \text{area of}\mathrm{\Delta}\text{ABC}=6\text{square units}\\ \text{Now, D is the midpoint of BC. Therefore, coordinates of D are}\\ (\frac{3+5}{2},\text{}\frac{2+2}{2})\text{i.e., (4, 0).}\\ \text{area of}\mathrm{\Delta}\text{ADC with vertices A(}4,\text{}6),\text{D(4},\text{0})\text{and C(5,}2)\\ \text{}=\frac{1}{2}\left\{4(02)+4(2+6)+5(6+0)\right\}\\ \text{}=\frac{1}{2}(8+3230)=3\\ \text{But area is always a positive quantity.}\\ \therefore \text{area of}\mathrm{\Delta}\text{ADC}=3\text{square units}\\ \text{area of}\mathrm{\Delta}\text{ABD with vertices A(}4,\text{}6),\text{B(3,}2)\text{and D(4},\text{0})\\ \text{}=\frac{1}{2}\left\{4(20)+3(0+6)+4(6+2)\right\}\\ \text{}=\frac{1}{2}(8+1816)=3\\ \text{But area is always a positive quantity.}\\ \therefore \text{area of}\mathrm{\Delta}\text{ABD}=3\text{square units}\\ \text{Now, we observe that}\\ \text{area of}\mathrm{\Delta}\text{ADC}=\text{area of}\mathrm{\Delta}\text{ABD}=3\text{square units}\\ \text{Hence, we verified the result}\mathbf{that}\text{}\mathbf{a}\text{}\mathbf{median}\text{}\mathbf{of}\text{}\mathbf{a}\text{}\mathbf{triangle}\text{}\mathbf{divides}\\ \mathbf{it}\text{}\mathbf{into}\text{}\mathbf{two}\text{}\mathbf{triangles}\text{}\mathbf{of}\text{}\mathbf{equal}\text{}\mathbf{areas}.\end{array}$
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