Ncert Solutions Maths class 11th

2. Let ABC be the equilateral triangle of side 2a and 0 be the origin. Then

AB = BC = AC = 2a

O is the mid-point of AB we have AO = a

BO = a

We know that A and B lies on y-axis so they have co-ordinate of the form (0, y).

Hence, co-ordinate of A is (0, a) and that of B is (0, –a)

Since OC, bisects AB at right angle, by Pythagoras theorem,

AC²= OA² + OC²

$\Rightarrow {\left(2a\right)}^2={\left(a\right)}^2+OC{}^2$

$\Rightarrow OC{}^2=4a^2-a^2$

And as C we on x-axis it has co-ordinate of the form (x, 0)

Exercise 9.1

1. Let the given points be A(–4, 5), B(0, 7), C(5, –5) and D(–4, –2).

Then quadrilated ABCD can be plotted on the graph by joining the points A, B, C and D.

We connect diagonal AC such that

area (ABCD) = (ΔABC) + (ΔADC)

Now,

$\left(\Delta ABC\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

$=\frac{1}{2}\left|-4\left[7-\left(-5\right)\right]+0\left[-5-5\right]+5\left[5-7\right]\right|\text{\hspace{0.17em}unit}{}^2$

$=\frac{1}{2}\left|-4\left(7+5\right)+0+5\left(5-7\right)\right|\text{\hspace{0.17em}unit}{}^2$

$=\frac{1}{2}|\left(-4\right)*12+5*\left(-2\right)|unit{}^2$

$=\frac{1}{2}|-48-10|unit{}^2$

$=\frac{1}{2}\left|-58\right|\text{\hspace{0.17em}unit}{}^2$

$=\frac{58}{2}\text{unit\hspace{0.17em}}{}^2=29\text{\hspace{0.17em}unit}{}^2$

Similarly, $\left(\Delta ACD\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

$=\frac{1}{2}\left|-4\left[-5-\left(2\right)\right]+5\left[-2-5\right]+\left(-4\right)\left[5-\left(-5\right)\right]\right|\text{\hspace{0.17em}unit}{}^2$

$=\frac{1}{2}|-4\left(-5+2\right)+5\left(-2-5\right)-4\left(5+5\right)|unit{}^2$

$=\frac{1}{2}|\left(-4\right)*\left(-3\right)+5*\left(-7\right)-4*\left(10\right)|unit{}^2$

$=\frac{1}{2}|12-35-40|unit{}^2$

$=\frac{1}{2}\left|-63\right|\text{\hspace{0.17em}unit}{}^2$

$=\frac{63}{2}\text{\hspace{0.17em}unit}{}^2$

Hence, area (ABCD) = $29+\frac{63}{2}\text{\hspace{0.17em}unit}{}^2$

$=\frac{58+63}{2}\text{\hspace{0.17em}unit}{}^2=\frac{121}{2}\text{\hspace{0.17em}unit}{}^2$

The Class 12 Relations and Functions explores various types of Functions, Students can check main types dicussed in this chapter below;

• One-One Function (Injective)

• Onto Function (Surjective)

• One-One and Onto Function (Bijective)

• Identity Function

• Constant Function

• Inverse of a Function

• Composite Functions

NCERT Solutions for 3D Geometry serves as a mathematical foundation developer for all the students. All competition exam aspirant must use NCERT Textbooks and its solutions for better understanding for fundamental concepts in 3D Geometry. Having a proper understanding of foundation knowledge of chapter is important for competitive exams like JEE and NEET. 

Students can download the free Class 11 Limits and Derivatives NCERT Solutions PDF from our website Shiksha, which offers chapter-wise and exercise-wise solutions. These PDFs are available for free and students can access  and download it to use it offline. Studernts can access the PDF directly on the Limits and Derivatives NCERT Solutions page provided by Shiksha as a part of class 11 chapter-wise matgh solutions.

The NCERT Solutions for Class 11 Limits and Derivatives provided by shiksha, cover all important concepts, including the definition of a limit, techniques to evaluate limits, left-hand and right-hand limits, and derivatives.  Our NCERT Solutions provide step-by-step explanations to questions . Our Solutions also include basic derivative formulas, rules, and their application to simple algebraic and trigonometric functions in solutions to make students aware of their usage.

Students must cover the foundational concepts of statistics such as deviation, variance, and standard deviation in order to get 100% marks. We recommend students practice every example and exercise of Statistics from the NCERT textbook for scoring full marks. We have provided NCERT Solutions for Statistics to help students in their preparation journey.

Students can visit our website shiksha.co to download the NCERT Statistics Solutions PDF. To download the PDF follow these steps;

• visit the Class 11 Math Chapter 13 Statistics Page on our website.
• go to the Class 11 Math Statistics NCERT Solution PDF: Download Free PDF
• Click on the "Download Free PDF"
• Save the PDF in your Device

No, Our NCERT Solutions include solutions of all the exercises of the statistics chapter. Students can check the solutions of the solved examples directly on the NCERT Textbooks. Students can take help of our NCERT Solution to practice the exercise, revise, and clear their doubts.

Statistics carries a weightage of 8-10 marks in the class 11 annual exams. For sure it proves that this chapter is a high scoring chapter, questions aremostly based on data and straight forword formulas. We have provided NCERT Solutions for Statistics, which can be really helpful in coving all the topics of the chapter and equip you with enough tools to get good marks in class 11 annaul exams.