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NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.2 - FREE PDF Download
In Class 10 Circles Exercise 10.2 Solutions, we're diving deeper into the world of circles and tangents that you explored in the previous chapter. Remember, a circle is the collection of all points at a constant distance from a central point. Tangents are special lines that touch the circle at exactly one spot, called the point of contact.
- 3.1Class 10 Exercise 10.2
This exercise will focus on applying your knowledge of circles and tangents to solve various problems. Get ready to:
Prove interesting theorems related to tangents.
Find lengths of tangents drawn from external points to circles.
Solve word problems that involve circles, tangents, and their properties.
By working through these problems, you will solidify your understanding of circles and tangents, and develop your problem-solving skills in geometry.
Glance on NCERT Solutions Maths Chapter 10 Exercise 10.2 Class 10 | Vedantu
This exercise revolves around understanding and applying the concept of tangents to circles. Remember, a tangent is a line that touches a circle at exactly one point.
The main focus is on finding the lengths of tangents drawn from a point outside the circle to the point of contact on the circle's circumference.
You will likely encounter a theorem stating that tangents drawn from an external point to a circle are equal in length.
There are no specific formulas directly related to tangents that you necessarily need to memorize for this exercise.
However, the distance formula (d = √((x2 - x1)² + (y2 - y1)²)) might be helpful in some problems where you need to find distances between points. Here, d represents distance, (x1, y1) and (x2, y2) represent the coordinates of two points.
NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.2 - Circles
ShareShareClass 10 Exercise 10.2
1. From a point \[Q\], the length of the tangent to a circle is \[24\ \text{cm}\] and the distance of \[Q\] from the centre is \[25\ \text{cm}\]. The radius of the circle is
(A) \[7\ \text{cm}\] (B) \[12\ \text{cm}\] (C) \[15\ \text{cm}\] (D) \[24.5\ \text{cm}\].
Ans:
From the figure given above, let us assume that \[O\] is the centre of the circle and given we have $PQ$ as the tangent to the circle of length $24\ \text{cm}$, and $OQ$ is of length $25\ \text{cm}$.
Now, by using the Pythagoras theorem we have –
$OP=\sqrt{O{{Q}^{2}}-P{{Q}^{2}}}$
$\Rightarrow OP=\sqrt{{{25}^{2}}-{{24}^{2}}}$
$\Rightarrow OP=\sqrt{625-576}$
$\Rightarrow OP=\sqrt{49}$
$\Rightarrow OP=7\ \text{cm}$.
Hence, option \[(A)\] is correct.
2. In the given figure, if \[TP\] and \[TQ\] are the two tangents to a circle with centre \[O\] so that \[\angle POQ=110{}^\circ \], then \[\angle PTQ\] is equal to
(A) \[60{}^\circ \] (B) \[70{}^\circ \] (C) \[80{}^\circ \] (D) \[90{}^\circ \].
Ans:
As, from the given figure, we can observe that there are two tangents perpendicular to the radius of the circle as $TP$ and $TQ$. Since, they are perpendicular to the radius hence, we have $\angle OPT=90{}^\circ $ and
$\angle OQT=90{}^\circ $.
Therefore, $\angle PTQ=360{}^\circ -\angle OPT-\angle OTQ-\angle POQ$
$\Rightarrow \angle PTQ=360{}^\circ -90{}^\circ -90{}^\circ -110{}^\circ $
$\Rightarrow \angle PTQ=360{}^\circ -290{}^\circ $
$\Rightarrow \angle PTQ=70{}^\circ $
Hence, option \[(B)\] is correct.
3. If tangents \[PA\] and \[PB\] from a point \[P\] to a circle with centre \[O\] are inclined to each other an angle of \[80{}^\circ \], then \[\angle POA\] is equal to
(A) \[50{}^\circ \] (B) \[60{}^\circ \] (C) \[70{}^\circ \] (D) \[80{}^\circ \].
Ans:
As, we have two tangents $PA$ and $PB$ which will be perpendicular to the radius of the circle. Hence, we have $\angle PAO=90{}^\circ $ and
$\angle PBO=90{}^\circ $. Since this is a quadrilateral, we will have the sum of all interior angles equal to $360{}^\circ $.
Therefore,
$\angle PAO+\angle PBO+\angle APB+\angle AOB=360{}^\circ $
$\Rightarrow 90{}^\circ +90{}^\circ +80{}^\circ +\angle AOB=360{}^\circ $
$\Rightarrow \angle AOB=360{}^\circ -260{}^\circ $
$\Rightarrow \angle AOB=100{}^\circ $.
Now, we know that $\angle POA$ is half of the $\angle AOB$.
Therefore,
$\angle POA=\frac{\angle AOB}{2}$
$\Rightarrow \angle POA=\frac{100{}^\circ }{2}$
$\Rightarrow \angle POA=50{}^\circ $.
Hence, option \[(A)\] is correct.
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Ans:
As, from the figure given above, let us assume that \[O\] is the centre of the circle and \[AB\], \[CD\] are the two tangents at the ends of the diameter of the circle.
Now, we know that tangents are perpendicular to the radius of the circle.
Hence, \[\angle CQO=90{}^\circ \]
\[\angle DQO=90{}^\circ \]
\[\angle APO=90{}^\circ \]
\[\angle BPO=90{}^\circ \].
Therefore, we can say that \[\angle CPQ=\angle BQP\] because they are alternate angles. Similarly, \[\angle AQP=\angle QPD\].
Hence, if the interior alternate angles are equal then the lines \[AB\], \[CD\] should be parallel lines.
Hence, proved.
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Ans:
Let us assume that the line perpendicular at the point of contact to the tangent of a circle does not pass through the centre \[O\] but passes through a point \[Q\] as shown in the figure above.
Hence, we have the line \[PQ\bot AB\].
$\Rightarrow \angle QPB=90{}^\circ $
Also, we have $\angle OPB=90{}^\circ $.
After comparing both the equations we get –
$\Rightarrow \angle QPB=\angle OPB$
But, from the figure drawn above we can observe that this is not the case as $\angle QPB<\angle OPB$.
Therefore, we can conclude that $\angle QPB\ne \angle OPB$. They can only be equal when these two-line segments $QP$ and $OP$ will be equal. This implies that the line perpendicular at the point of contact to the tangent of a circle passes through the centre \[O\]. Hence, proved.
6. The length of a tangent from a point \[A\] at distance \[5\ \text{cm}\] from the centre of the circle is \[4\ \text{cm}\]. Find the radius of the circle.
Ans:
From the figure given above, let us assume that \[O\] is the centre of the circle and given we have $AB$ as the tangent to the circle of length $4\ \text{cm}$, and $OA$ is of length $5\ \text{cm}$.
Now, by using the Pythagoras theorem we have –
$OB=\sqrt{O{{A}^{2}}-A{{B}^{2}}}$
$\Rightarrow OB=\sqrt{{{5}^{2}}-{{4}^{2}}}$
$\Rightarrow OB=\sqrt{25-16}$
$\Rightarrow OB=\sqrt{9}$
$\Rightarrow OB=3\ \text{cm}$.
Therefore, the radius of the circle will be $3\ \text{cm}$.
7. Two concentric circles are of radii \[5\ \text{cm}\] and \[3\ \text{cm}\]. Find the length of the chord of the larger circle which touches the smaller circle.
Ans:
From the figure given above, we can observe that the line segment $PQ$ is the chord of the larger circle and a tangent to the smaller circle.
Therefore, \[OA\bot PQ\].
Now, we can observe that the \[\Delta OAP\] forms a right-angled triangle.
Hence, by applying Pythagoras theorem –
\[AP=\sqrt{O{{P}^{2}}-O{{A}^{2}}}\]
\[\Rightarrow AP=\sqrt{{{5}^{2}}-{{3}^{2}}}\]
\[\Rightarrow AP=\sqrt{25-9}\]
\[\Rightarrow AP=\sqrt{16}\]
\[\Rightarrow AP=4\ \text{cm}\].
Now, since the radius is perpendicular to the tangent therefore, we have \[AP=AQ\].
Hence, \[PQ=2AP\].
\[\Rightarrow PQ=2\times 4\]
\[\Rightarrow PQ=8\ \text{cm}\].
So, the length of the chord will be of \[8\ \text{cm}\].
8. A quadrilateral \[ABCD\] is drawn to circumscribe a circle. Prove that \[AB+CD=AD+BC\].
Ans:
From the figure given we can observe that the sides of the quadrilateral acts as tangents to the circle. As, the tangents drawn from any external point will have the same length. Therefore, we have –
\[DR=DS\],
\[CR=CQ\],
\[BP=BQ\], and
\[AP=AS\].
Now, we will add all the relations.
Hence, \[DR+CR+BP+AP=DS+CQ+BQ+AS\]
\[\Rightarrow \left( DR+CR \right)+\left( BP+AP \right)=\left( DS+AS \right)+\left( CQ+BQ \right)\]
\[\Rightarrow DC+AB=AD+BC\]
Hence, proved.
9. In the given figure, \[XY\] and \[X'Y'\] are two parallel tangents to a circle with centre \[O\] and another tangent \[AB\] with point of contact \[C\] intersecting \[XY\] at \[A\] and \[X'Y'\] at \[B\]. Prove that \[\angle AOB=90{}^\circ \].
Ans:
From the figure given we can observe that \[AB\], \[XY\] and \[X'Y'\] are tangents to the circle and will be perpendicular to its radius.
Now, let us consider two triangles \[\Delta OPA\] and \[\Delta OCA\], such that –
\[OP=OC\],
\[AP=AC\].
Hence, we have \[\Delta OPA\cong \Delta OCA\] by the SSS congruence rule.
\[\Rightarrow \angle POA=\angle COA\].
In the similar manner \[\Delta OQB\cong \Delta OCB\].
\[\Rightarrow \angle QOB=\angle COB\].
Now, we know that \[PQ\] is the diameter of the circle.
\[\Rightarrow \angle POA+\angle COA+\angle COB+\angle QOB=180{}^\circ \]
\[\Rightarrow 2\angle COA+2\angle COB=180{}^\circ \]
\[\Rightarrow \angle COA+\angle COB=90{}^\circ \]
\[\Rightarrow \angle AOB=90{}^\circ \].
Hence, proved.
10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Ans:
Let us assume a circle with centre at \[O\], which have two tangents \[PA\] and \[PB\] which are perpendicular to the radius of the circle as shown in the figure above.
Now, let us consider two triangles \[\Delta OAP\] and \[\Delta OBP\], such that –
\[PA=PB\], and
\[OA=OB\].
Hence, we have \[\Delta OAP\cong \Delta OBP\] by the SSS congruence criteria.
Therefore, \[\angle OPA=\angle OPB\] and
\[\angle AOP=\angle BOP\].
This implies that \[\angle APB=2\angle OPA\] and
\[\angle AOB=2\angle AOP\].
Hence, in the right-angled triangle \[\Delta OAP\], we have –
\[\angle AOP+\angle OPA=90{}^\circ \]
\[2\angle AOP+2\angle OPA=180{}^\circ \]
\[\Rightarrow \angle AOB=180{}^\circ -\angle APB\]
\[\Rightarrow \angle AOP+\angle OPA=180{}^\circ \].
Hence, proved.
11. Prove that the parallelogram circumscribing a circle is a rhombus.
Ans:
From the figure given above, we can observe that \[ABCD\] is a parallelogram whose sides are tangent to the circle. This implies that \[DR=DS\],
\[CR=CQ\],
\[BP=BQ\], and
\[AP=AS\].
Now, we will add all the relations.
Hence, \[DR+CR+BP+AP=DS+CQ+BQ+AS\]
\[\Rightarrow \left( DR+CR \right)+\left( BP+AP \right)=\left( DS+AS \right)+\left( CQ+BQ \right)\]
\[\Rightarrow DC+AB=AD+BC\]
As, the sides of a parallelogram are always parallel and equal in length.
Therefore, \[AB=DC\] and
\[AD=BC\].
\[\Rightarrow 2AB=2BC\]
\[\Rightarrow AB=BC\]
Hence, all the sides of parallelogram are equal.
Therefore, we can conclude that it is a rhombus. Hence, proved.
12. A triangle \[ABC\] is drawn to circumscribe a circle of radius \[4\ \text{cm}\] such that the Segments \[BD\] and \[DC\] into which \[BC\] is divided by the point of contact \[D\] are of lengths \[8\ \text{cm}\] and \[6\ \text{cm}\] respectively. Find the sides \[AB\] and \[AC\].
Ans:
From the figure given above, we can observe that the sides of a triangle \[ABC\] are the tangents to the circle and are perpendicular to the radius of the circle.
Hence, in \[\Delta ABC\]
\[CF=CD=6\ \text{cm}\]
Similarly, \[BE=BD=8\ \text{cm}\] and
\[AE=AF=x\ \text{cm}\].
In \[\Delta ABE\],
\[AB=AE+BE\]
\[\Rightarrow AB=x+8\]
Similarly, \[BC=8+6=14\] and
\[CA=6+x\].
Therefore, we have \[2s=AB+BC+CA\]
\[\Rightarrow 2s=x+8+14+6+x\]
\[\Rightarrow 2s=2x+28\]
\[\Rightarrow s=x+14\].
Now, we know that the area of a triangle can be calculated by using the formula \[Area=\sqrt{s(s-a)(s-b)(s-c)}\].
Therefore,
\[Area\ \text{of}\ \Delta \text{ABC}=\sqrt{(14+x)(14+x-6-x)(14+x-8-x)}\]
\[\Rightarrow \sqrt{x(14+x)(8)(6)}\]
\[\Rightarrow 4\sqrt{3(14x+{{x}^{2}})}\].
Hence, \[Area\ \text{of}\ \Delta \text{OBC = }\frac{1}{2}\times OD\times BC\]
\[Area\ \text{of}\ \Delta \text{OBC =}28\]
\[\Rightarrow Area\ \text{of}\ \Delta \text{OCA = }\frac{1}{2}\times OF\times AC\]
\[Area\ \text{of}\ \Delta \text{OCA = 12+2x}\]
\[\Rightarrow Area\ \text{of}\ \Delta \text{OAB = }\frac{1}{2}\times OE\times AB\]
\[Area\ \text{of}\ \Delta \text{OAB = 16+2x}\]
Therefore, the total area of the triangle \[ABC\] will be –
\[=Area\ \Delta \text{OBC}+Area\ \Delta \text{OCA+}Area\ \Delta \text{OAB}\]
\[\Rightarrow 4\sqrt{3(14x+{{x}^{2}})}=28+12+2x+16x+2x\]
\[\Rightarrow \sqrt{3(14x+{{x}^{2}})}=14+x\]
\[\Rightarrow 3(14x+{{x}^{2}})={{(14+x)}^{2}}\]
Hence, after further solving we have –
\[(x+14)(x-7)=0\]
\[\Rightarrow x=-14\] or
\[\Rightarrow x=7\]
As, the side cannot be negative in nature, therefore, \[x=7\ \text{cm}\].
Hence, \[AB=7+8\]
\[\Rightarrow AB=15\ \text{cm}\] and
\[CA=6+7\]
\[\Rightarrow CA=13\ \text{cm}\].
13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend Supplementary angles at the centre of the circle.
Ans:
From the figure given we can observe that the sides of the quadrilateral acts as tangents to the circle. Therefore, we have to prove that–
\[\angle AOB+\angle COD=180{}^\circ \] and
\[\angle BOC+\angle DOA=180{}^\circ \]
Now, let us consider two triangles \[\Delta OAP\] and \[\Delta OAS\],
Hence, we have \[AP=AS\], and
\[OP=OS\]
\[\Rightarrow \Delta OAP\cong \Delta OAS\] by the SSS congruence criteria.
Thus, \[\angle POA=\angle AOS\]
\[\angle 7=\angle 8\]
Also,
\[\angle 1=\angle 2\],
\[\angle 3=\angle 4\],
\[\angle 5=\angle 6\]
Therefore, \[\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360{}^\circ \]
\[\Rightarrow \left( \angle 1+\angle 2 \right)+\left( \angle 4+\angle 3 \right)+\left( \angle 6+\angle 5 \right)+\left( \angle 8+\angle 7 \right)=360{}^\circ \]
\[\Rightarrow 2\angle 1+2\angle 4+2\angle 5+2\angle 8=360{}^\circ \]
\[\Rightarrow \left( \angle 1+\angle 8 \right)+\left( \angle 4+\angle 5 \right)=180{}^\circ \]
\[\Rightarrow \angle AOB+\angle COD=180{}^\circ \]
In the similar manner we can prove that \[\angle BOC+\angle DOA=180{}^\circ \].
Therefore, we have proved that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Conclusion
The purpose of Vedantu's NCERT Solutions for Class 10 Maths CH 10 Ex 10.2 is to assist students in understanding key ideas about circles, including tangents, chords, and the characteristics of angles created by tangents. Comprehending the relationship between tangents and circles, as well as the associated theorems, is essential for acing board exams. In recent exams, questions from Class 10 Maths Chapter 10 Exercise 10.2 frequently appeared, highlighting the importance of mastering these concepts. Specifically, the problems often involve finding the lengths and angles associated with tangents and chords, proving certain geometric properties, and solving equations derived from circle theorems.
Class 10 Maths Chapter 10: Exercises Breakdown
CBSE Class 10 Maths Chapter 10 Other Study Materials
Chapter-Specific NCERT Solutions for Class 10 Maths
Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.
Study Resources for Class 10 Maths
For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.
FAQs on NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.2 - Circles
1. How many examples and questions are there in Exercise 10.2 of Class 10 Maths?
There are 5 examples involving 10.2 and its subsequent subtopics. Apart from that at the end of section 10.2, you would find Exercise 1 which will consist of a total of 14 questions arranged well in a hierarchy of difficulty. Practice all of these to get a good idea of the concepts mentioned in section 10.2.
2. Is Exercise 10.2 of Class 10 Maths easy?
Maths is enjoyable if you understand the methods, but if you are unfamiliar with the themes and formulas, things may get difficult and perplexing because there are various ways to answer a question or problem. Mathematics is a big topic, and it's easy to become lost in all of the formulae and derivations. Concentrate on the basics, and with enough practise, you'll be fine.
3. Where can I find the solution to Class 10 Maths Chapter 10?
To get the solutions of NCERT Class 10 Maths Chapter 10, you can go and visit the Vedantu site and search for the solutions of Class 10 Maths Chapter 10. Apart from this you can avail multiple modules and help to crack good scores in math exams. The link for the solutions to the exercise is NCERT Class 10 Maths Chapter 10. Click on that to get a free of cost PDF version of the solutions.
4. Is it compulsory to solve all examples and exercises of NCERT Class 10 Maths Chapter 10?
You definitely must practice each one of them as you never know what concept you might be facing difficulty in unless you test your knowledge by solving different types of questions and problems. Hence it is of utmost importance that you solve all examples and exercises of NCERT Class 10 Maths Chapter 10. Revise all the concepts and practice as many problems as you can to grasp the concepts better. Also, visit the Vedantu site for more problems regarding these topics.
5. How many Chapters are there in the NCERT Solutions Class 10 Maths apart from Chapter 10?
The NCERT Class 10 Maths textbook has 16 chapters as well as some selected problems for each topic. Each chapter must be given equal weight in order to have a comprehensive grasp of all of the topics provided for Class 10 students and to perform well in your examinations. If you are interested in obtaining various modules and example papers relating to these areas, you can simply get them through the Vedantu website as well as the Vedantu mobile app.
6. What is theorem 10.1 and 10.2 Class 10?
The Theorems 10.1 and 10.2 in Class 10 Ex 10.2 are:
Theorem 10.1: Tangent and Radius
The tangent drawn to a circle at any point is perpendicular (exactly at a 90-degree angle) to the radius drawn to the same point of contact.
When a line touches a circle at just one point, a right angle (90 degrees) is formed between the line and another line drawn from the circle's center to the touch point.
Theorem 10.2: Equal Lengths of Tangents from an External Point
If two tangents are drawn from an external point to a circle, they will have equal lengths.
When a point is outside a circle, and two lines are drawn from that point to touch the circle at separate points, these two tangent lines will have the same length, despite their different positions on the circle, according to this theorem.
7. Which theorem is important for class 10 Circles?
Both Theorem 10.1 (Tangent and Radius) and Theorem 10.2 (Equal Lengths of Tangents from an External Point) are important for the Circles Chapter 10.
Theorem 10.1:This lays the groundwork for understanding the relationship between the tangent line, the radius drawn to the point of contact, and right angles within circles, helping to solve related problems.
Theorem 10.2: Understanding how to find the lengths of tangents drawn from an external point to a circle is important. It's a commonly used theorem in questions about tangents. Although both theorems are important, Theorem 10.2 may be slightly emphasized more because it directly deals with calculating tangent lengths, which is a common type of problem in exercises.
