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AS Level Mathematics 9709 | Circular Measures 6 | Past Paper Questions | Raja Shafqat — Transcript

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0:08Assalam-o-Alaikum, child. Today, we are
0:10finally doing more practice on circular
0:12measurement from past papers. Look, in
0:15this entire topic, the new thing we
0:17have learned so far is called arc
0:20length. What was the formula for arc
0:22length? r*theta. And number two,
0:25another new thing we learned is what
0:26the formula for the area of a sector is
0:28. The area of a sector formula is 1/2
0:31rs or the area of a sector formula is
0:341/2 r² theta. That's all that's new.
0:38Otherwise, you have to use some common
0:40properties in this. Is that clear,
0:42child? Now, first, read this question
0:45properly to see what hints are given in
0:48it. Okay? What hint is given in this
0:50question? Then we will solve it. Number
0:53one, look, if you observe this diagram,
0:56I have deliberately copied this diagram
0:59below. Is that clear, child? I copied
1:01the diagram after enlarging it a bit
1:03more. Now, let's start slowly. Number
1:06one, what is required of us in this
1:08question? Let's start from here to see
1:11what is required. Number one. Let's go.
1:16First of all, if any student remembers
1:20the condition, this is the condition
1:23for tangents from the same external
1:26point. This is a circle property we
1:29studied. That is the property being
1:31applied here. Tangents from the same
1:33external points are equal in length.
1:34Look, this is that external point. A
1:36tangent was drawn from here. A tangent
1:38was drawn from here. And by the way, it
1:40is already given that these two lengths
1:41are equal. So, this means the condition
1:43is given to us directly. We don't need
1:45any extra effort. So, the first thing
1:48we learned was that when I join this to
1:50this, look what is formed? Two
1:53right-angled congruent triangles.
1:56Because, look, the radius is always
1:58perpendicular to the tangent line. The
2:01radius is always perpendicular to the
2:02tangent line. Is that clear, child? So,
2:04this angle is 90 °. Then, obviously,
2:06what will this angle be? This angle
2:08will also be 90 °. So, read it slowly
2:11first. Two conditions were used in this
2:12. The radius is always perpendicular to
2:14the tangent line. One more condition
2:16was used. So, this means that this
2:19length here is what? 8 cm, and we
2:21already know that these two angles are
2:24always the same in this. Meaning, if
2:26this angle is alpha, then we know this
2:28one is also alpha. And similarly, we
2:30also know that these two angles are
2:31also equal. Because these two triangles
2:34are congruent triangles. We have
2:37already studied that this triangle and
2:38this triangle. What are these two
2:41triangles? Two right-angled congruent
2:46triangles. Congruent triangles, meaning
2:50both are the same right-angled
2:52triangles. Number one, let's start
2:55slowly, what is our target in this
2:57whole question. Clear, kid? Now let's
2:59start even slower, our target is. Come
3:02on, first let's start from part one.
3:05What is required in part one? Number
3:07one. The examiner says, "Show that the
3:09whole angle AOB is 2.16." Meaning this
3:13entire angle, angle AOB, you need this
3:16whole angle. So if this angle is found
3:19out, then multiply it by two. So with
3:21respect to this angle, if you consider
3:23this small triangle, take this,
3:25consider this small triangle. Now you
3:28can work with respect to alpha or beta,
3:30but it is best to work with respect to
3:32beta. Here is beta. With respect to
3:34this, what is this side? This side is
3:3615 and what is this side? This side is
3:38eight. Clear, kid? You already know
3:41these sides. Okay? You already know
3:43this side. So now we applied it with
3:45respect to this angle. This is opposite
3:47, and with respect to this angle, this
3:49is adjacent. So you know the opposite
3:51and you also know the adjacent. Okay?
3:53You know the opposite and you also know
3:55the adjacent. So we applied tan. What
3:58will the answer for tan be? The answer
4:00for tan is opposite divided by adjacent
4:02. Opposite divided by adjacent. So here
4:04we took help from the calculator. tan
4:07inverse 15/8, so the answer for beta
4:09was found from here. The answer for
4:13beta came to 1.08, 1.080, or you can
4:17say 1.081, there it is. Now, what was
4:21your target, which angle is your target
4:23? Your target is angle AOB. So, what is
4:26angle AOB? Two times beta. This is the
4:29entire angle, angle AOB. Two times beta
4:31, meaning beta + beta. So, what will we
4:33do? 2*1.081, there it is, so we
4:36multiplied it by two and it came to
4:392.161. So the answer you need to show
4:42is to 3 s.f. So what did the answer
4:44come to? 2.1617, so we need to show the
4:47answer to 3 s.f. Here it is, 2.16, and
4:50the unit for the angle is also written
4:53as radians. And we already know that in
4:55this whole topic, the required unit for
4:57the angle is always radians. The
4:59required unit of the angle is always
5:01radians. So this is your radians. Clear
5:04, kid? Part one is solved. In the
5:06second part, the perimeter of the
5:07shaded region. Perimeter, the perimeter
5:10of the shaded region. Well, first of
5:13all, it's common sense. Perimeter means
5:16, one, this length here. Plus this one.
5:18Plus this one. Meaning, if you think
5:21technically, you have the diagram in
5:23this form. This is your target. I'm
5:25just giving you an idea. The situation
5:28is something like this, this is your
5:30target. Okay? This is your target. This
5:33kind of condition is given. Here it is.
5:36Right, son? I showed you a rough graph
5:38so you could get an idea. Here you go,
5:41so you get an idea. This is your target
5:43. By the way, what was this point? This
5:44point was A. This point was your T, and
5:47this point was your B. This is your
5:49shaded perimeter target, and you
5:50already know this length is 15 cm. This
5:53length is also already known as 15 cm;
5:55the question is, what is this? Now look
5:59, this whole thing is a sector. What is
6:01the angle of this sector? The angle of
6:03this entire sector was, you have a
6:06sector here. So, what is AB? AB is the
6:09arc length of that sector. So, if I ask
6:11you, what would the perimeter be?
6:13Because your target is the perimeter.
6:15So, first, what will you tell the
6:16examiner about the meaning of perimeter
6:18? Arc length AB + length BT + length AT
6:24, arc length AB, okay? Arc length AB. '
6:29s' means arc length. Now, how will this
6:31entire arc length come? The formula for
6:33arc length is r*theta. So, what is its
6:35radius? Eight. There it is, eight. Here
6:37it is. And what is this entire angle?
6:392.16. So, that means what will its
6:41perimeter be? Its perimeter will
6:45finally be r*theta, so 8*2.16, this
6:48gives you the arc length; BT and AT you
6:50already know are 15 and 15, here you go
6:53, BT and AT are already known as 15 and
6:5615, so if 8*2.16 + 30, then what is the
6:59answer? 47.28 47.28 I mean, we wrote
7:07three because we always need every
7:09answer to 3sf. Okay? Okay? In A-levels,
7:11we are required to provide answers to
7:13three significant figures. This is your
7:15final perimeter. This was your target.
7:18You needed the shaded perimeter. Now,
7:21read part three to see what the
7:22statement is in part three. In part
7:24three, the examiner requires the shaded
7:26area. Your target is the shaded area.
7:29Now, there is a little extra effort
7:30here. The shaded area is right here.
7:32Now, use a little common sense; this is
7:34a right-angled triangle, and this is
7:36also a right-angled triangle. So, if
7:39this sector is removed from these two
7:41right-angled triangles, the shaded area
7:42will be found. So, I can also say two
7:45right-angled triangles—two
7:46right-angled triangles means this whole
7:49shape is a quadrilateral, by the way.
7:51Two right-angled triangles minus—what
7:53needs to be subtracted from this? We
7:55need to subtract sector AOB from this.
7:58Subtracting sector AOB. Now, the
8:01triangle's height is 8. The triangle's
8:03base is 15. So, this entire
8:06quadrilateral, meaning the two
8:07triangles, represents the quadrilateral
8:09. So, what will be the area of the
8:11quadrilateral? 1/2*height*base, 1/2*8*
8:1315, this is the area of the two
8:16triangles, meaning the quadrilateral.
8:19What is the area of the sector? The
8:21area of the sector is 1/2*radius
8:23squared*theta. Here it is, 1/2*radius
8:25squared*theta. The radius is 8, the
8:27radius is 8, and who is this entire
8:29theta? You already know this theta is
8:322.16. So, 1/2*radius squared*theta.
8:36Clear, kid? Now, this two and two are
8:38canceled out. So, it will be 15*8, 15*8
8:41is what? 120, so this will be 64*2.16/2
8:46, this is finally 23.04 for you, there
8:50you go, 23.04. Now subtract this, 120-
8:5623.04, so the answer derived from here
8:59is 96.96. So it will be 96. Okay? 96,
9:06this is your final answer derived.
9:08Clear, kid? What did you finally find
9:11out here? You finally found the shaded
9:14area. This was the final target of your
9:17question. Okay, kid? So, basically, two
9:20properties of the circle were used in
9:22this question. Clear, kid? Two
9:23properties of the circle were used.
9:25Check it thoroughly. Clear? There
9:27wasn't anything very logical to it.
9:29Clear? So, what is the meaning of these
9:30two triangles? The two triangles meant
9:32the overall quadrilateral. Whether you
9:34call it two triangles or you call it
9:35quadrilateral AOBT, it's the same thing
9:39. Some students say two triangles, or
9:41you could alternatively say
9:42quadrilateral AOBT; both these things
9:45mean technically the same. Clear, kid?
9:47So we have solved all three parts of
9:49this one by one. The parts that were
9:51required. Clear, kid? We have solved
9:54all three parts that were required. And
9:56let's try to see what other varieties
9:58are in this. Come on, look at the
10:01question carefully first. This is a
10:03slightly technical question, just a
10:04little. Okay? So, I request everyone to
10:07please read this question thoroughly
10:09first. Let me zoom it in a little bit.
10:11I will enlarge this figure a bit
10:13because it requires a little more
10:15effort. All right? A little more effort
10:17. So, I will have to work on this.
10:19Meaning, I will have to check it. First
10:21, read the question properly. Well done
10:23. Then I will guide you. First, see
10:26what hints the examiner is giving in
10:28the question so that we can solve it
10:30accordingly. Clear, kid? We will solve
10:33the question according to that. Number
10:35one, it is clearly given that these two
10:37sides are parallel. Clearly, these two
10:40sides are parallel. Okay? I am not
10:43saying this; the hint is already given
10:45in the question. Okay? Let me bring it
10:47back to its actual size. This is your
10:50original question. Clear, kid? Now,
10:52first, take two minutes to read the
10:54question. Then I will guide you on what
10:57our main target is. Okay? Okay? What
11:02our main target is in this question.
11:04Okay? Look, I will copy the main
11:08diagram so that I can show you the
11:11workings on it. Okay, kid? I have
11:15copied the main diagram so I can show
11:17you all the work on it. Okay, kid? Take
11:21two minutes to read the question. Then
11:23I will finally guide you on what our
11:25main target is here. Okay, kid? What is
11:28our main target? Okay? Okay? I have
11:31copied the portion of the diagram that
11:33I am interested in. Okay? I will erase
11:36everything else. Okay? I will erase
11:39everything else to explain it to you so
11:42your focus remains only on the point
11:44that is our main question. Okay, kid?
11:47As much as our main question is. So, I
11:49have erased the extra portion of the
11:51diagram. Here you go. There, the rest
11:53of the extra portion is erased. Now you
11:55have the actual shape. Okay, kid. There
11:58, you have the actual shape. Now, if
12:01any student observes this properly,
12:02what is this shape? This shape is a
12:05trapezium. This shape is a trapezium.
12:07Now let's start slowly. Number one, if
12:10I drop a line from here. What did I do?
12:12I dropped a line from here. I dropped
12:15this line and this line. It’s common
12:17sense. Maths says that when two lines
12:19are parallel, what is the vertical
12:20distance between them always? It is the
12:22same. This is a basic rule of maths. Is
12:25that clear, son? So now, here is one
12:27right-angled triangle and here is
12:28another right-angled triangle. Both of
12:30these are right-angled triangles. So
12:32what did we do? The vertical distance
12:34between parallel lines is always the
12:36same. So, I have plotted the vertical
12:38distance for you here. Is that clear,
12:40son? Now we have two triangles. Now I
12:43will copy both of these triangles. Then
12:45, one by one, we will solve all its
12:47parts. Here is one triangle for you. I
12:50will copy this one by one. Here it is.
12:55First of all, here is one triangle. I
12:57will copy the name of this triangle.
12:58This one here is D. This one here is A,
13:01its hypotenuse is 6 cm, and let's name
13:03this point something. Come on, let me
13:06name it to explain. I will name this P.
13:09I will name this Q. Come on, just to
13:11guide you. Okay? Come on, here you go.
13:13We have labeled it. This is one
13:15triangle you have. A right-angled
13:17triangle. Clear? And this angle of this
13:21right-angled triangle is π/3. This
13:24angle is π/3. By now, we probably have
13:28an idea that π/3 means 180/3, which is
13:3160 °. Technically, if we think about
13:35it, is it clear, son? Come on, number
13:38one. Now, with respect to this angle,
13:40this is your opposite. This is your
13:42adjacent. We neither know the opposite
13:44nor the adjacent. We know neither the
13:45opposite nor the adjacent. So, I first
13:47applied sin. What is the sin formula?
13:50The sin formula is opposite divided by
13:52adjacent. Opposite...sorry, opposite
13:54over hypotenuse, here it is. So, who is
13:56the opposite? DP, here it is, divided
13:58by hypotenuse. What is the hypotenuse?
14:00Your hypotenuse is 6. Now, look at the
14:03answer for sin π/3 from the calculator
14:04. sin π/3 is √3/2. This is the
14:07calculator's job. There it is,
14:10multiplied by 6, so from here I found
14:12out what DP is. So, the answer for DP
14:16came to 3√3. This is your DP found.
14:20This is your DP found. Is that clear,
14:22son? Our main target was that we needed
14:24DP. So, from here, I found DP. Here is
14:27DP. Then I will guide you in more
14:29detail. Here you go. So, we considered
14:31right-angle triangles. First, we
14:32applied sine. DP was found, the height
14:35was found. Now, we applied cos. What is
14:38the formula for cos? The formula for
14:40cos is base over hypotenuse, here it is
14:41. So, with respect to this angle, this
14:43is the base. This is the hypotenuse. So
14:45, what will we say? We will say AP/6.
14:48Now, I already know cosπ/3 is half. By
14:53the way. cosπ/3 is 1/2. So, what did
14:56the answer turn out to be? It became
14:581/2*6, so from here you found the
15:00answer for AP. So, what is the answer
15:03for AP? The answer for AP came out to
15:05be 3 cm. There you go, your AP is found
15:07too. So, if I fill it in here, it means
15:10this DP that you got is 3√3, there
15:14you go. And the AP you got, this gap
15:17from A to P, this gap, this AP; so what
15:19did AP turn out to be? AP turned out to
15:22be 3 cm. There you go. We have only
15:24done two calculations so far. Not a
15:26very long or complicated task. There
15:28you go. This right-angle triangle. Now,
15:31you also have another right-angle
15:32triangle. Let's see, we looked at this
15:34right-angle triangle. There is another
15:35right-angle triangle. Let's copy that
15:37right-angle triangle too. This is
15:39right-angle triangle E. What is this
15:41point? Q, and this point is B, and with
15:43respect to this, what is this angle?
15:45This angle is theta. I copied this
15:48triangle here. Theta. You have this.
15:51And what is its hypotenuse? 10 cm. So
15:53with respect to this angle. This is
15:56opposite. This is adjacent. And
15:58opposite to 90, what is this? It's the
16:00hypotenuse. Clear, kid? Now, I applied
16:02sine and cos again, exactly the same
16:05way. Now, when I apply sine, what is
16:07the sine formula? The formula for sine
16:09is eq. /eq/10. There you go. So, from
16:15here I found out what the answer for eq
16:18is. The answer for eq came out to be 10
16:21*sinθ. 10*sin theta. This is eq, this
16:27is eq. Now I applied cos. What is the
16:31cos formula? The formula for cos is qb,
16:35by qb/10. Here it is. So, from here
16:39your qb was found. What is the answer
16:41for qb? 10*cos theta. So, we have found
16:44the height as well. We did the same
16:46work for the base in both triangles. So
16:47, let me just fill it in here. So what
16:49did I find out from here? I found out
16:51from here that this qb, what did this
16:53qb become? Q also became 10*cos theta
16:56and what did this height become? This
16:59height became 10*sin theta. So I
17:03considered both of these right-angle
17:05triangles. Now let’s move forward.
17:07Now let’s read the question. What
17:08needs to be done? By considering the
17:10perpendicular distances of D and E from
17:12AB. D and E from AB, these are the
17:15perpendicular distances. These are the
17:17perpendicular distances. So we already
17:18know that whenever two lines are
17:20parallel, what is the distance between
17:21them always? It is the same. So, by the
17:23way, there is a hint given in part one
17:25for you. It is even explaining to you
17:27what to do. If we aren't doing anything
17:29. It is giving you a hint itself. It is
17:31telling you that the perpendicular
17:33distance is the same. So that means it
17:35is saying that both dp and eq will be
17:38what? The same. I am not saying this,
17:40it is saying this because between
17:41parallel lines there is always what?
17:43The same vertical distance. So it means
17:45that there is always the same
17:47perpendicular distance between parallel
17:49lines. There is always the same
17:51perpendicular distance between parallel
17:53lines. It is even guiding you itself.
17:55It says equate them like this. So what
17:57was dp? The answer for dp was 3, what
18:00was eq? Its answer was 10*sin theta.
18:05Here it is. So now if you want theta,
18:08what will you do? You will say what is
18:10the answer for sin theta? The answer
18:13for sin theta came to be 3/10, so can I
18:17say that theta is equal to what? θ is
18:22equal to sin inverse of 3/10. 3/10.
18:30This is the final answer for theta you
18:34have. This is the final answer for
18:36theta you have. Clear, kid? This is
18:38your final answer for theta, that's
18:39what needs to be shown. And it was
18:41clearly written here, exact value. I
18:44have told you many times what the word
18:47exact value means, an answer without
18:50decimals. But it is interesting that
18:53the next part doesn't need an exact
18:54answer. So I will do one thing. I will
18:56also show you its answer in decimals.
18:59Because it will be needed in the next
19:02part, sin inverse of 3/10. So from the
19:04calculator, this came to 0.546. I
19:08deliberately wrote the answer below in
19:10radians, on purpose. Because this theta
19:13will help us in the second part. Next
19:15part. Okay? Alright, now read the
19:17question. Now, what does the examiner
19:20require in the next part? Read it.
19:23Perimeter of the shaded region. Okay,
19:26let's listen very carefully to what the
19:27perimeter of the shaded region means.
19:29The perimeter of the shaded region
19:31includes this arc length. This arc
19:33length is the second one. Because if
19:36you notice, this is a sector. If you
19:39notice, this is also a sector. The
19:41sector's length is this arc length. The
19:43sector's length is this arc length. And
19:45what is this part? DE, now we need to
19:47put in a little extra effort regarding
19:49DE. Just a little extra effort. Now, if
19:52we notice for a minute, just for a
19:54minute. Now notice one thing, if I ask
19:56you what is the total distance from A
19:58to B? A to B, you would say this is AX.
20:02AX is six. This is your 10. So, how
20:05much is this total distance? From A to
20:06X, this entire distance is 16. There
20:08you go. So that means what is the total
20:11distance of AX? The total distance is
20:1416 centimeters. 16 centimeters. Is that
20:18clear, dear? Okay, now look at one
20:20thing. This entire thing is 16. If we
20:24cut out this three and also cut out
20:26this length, then this small length
20:28remains, which is DE. So first, I will
20:31find out what DE is. First, we will
20:35find out what DE is. Clear, dear? So, I
20:37will repeat it again. First, I need DE.
20:40I know this entire length AB. That this
20:43entire length AB is how much? It's 16.
20:45Meaning, how much is 6 plus 10? 16.
20:48From this total length of 16, subtract
20:51AP and also subtract XB. Then the
20:54length remaining in the middle is PQ.
20:55Meaning this length is DE. Both these
20:57lengths are the same. So now, the first
20:59thing we did is, how will this length
21:01be found? Subtract AP from AB, and what
21:04else will you subtract? QB. Subtract QB
21:08. Clear, dear? Now tell me, what is AB?
21:12AB was 16 in total, AB was 16 in total.
21:15Sorry, this AB was the whole thing,
21:18this was your AB. By the way, how much
21:20was this AB of yours? It was 16.
21:23Actually, I wrote AX by mistake. By the
21:26way, this was AB. There you go, there
21:28you go. How much was AB? It was 16
21:30centimeters. Okay, it was 16
21:32centimeters. What is AB, AP? AP is 3A.
21:35Ignore this. Exclude this from here.
21:38And ignore this as well. What is this?
21:40tan cos theta. What is this thing?
21:43There is tan cos theta. So what does
21:46this become? This becomes 13-10 cos
21:49theta. I already know theta above.
21:52Because now I need the answer in
21:54significant figures. So I needed de
21:56here. So what is the answer for de?
21:5913-10 times cos theta. What is theta?
22:020.546, here it is. So this comes to
22:054.454. 4.454 cm. So what did I find out
22:13from here? I found DE from here. Number
22:16one, look, here is your de. What is
22:19this thing? This is arc length. What is
22:21this thing? This is also arc length.
22:23Now, find the two arc lengths one by
22:25one. The first arc length is named arc
22:27length dx. The first arc length is
22:29named arc length dx. Here is arc length
22:31dx. Here is the radius. Here is the
22:33radius. Here is the angle. So what is
22:35its formula? r*theta. Its formula is r*
22:40θ, so 6*π/3, 6*π/3, so what is the
22:44answer? 2 π cm, this arc length is
22:49found. Okay, now this is also an arc
22:52length. Its radius is 10 and its angle
22:54is theta. So what is the name of the
22:56second arc length? The second arc
22:58length is named arc length ex, like
23:00this. So what is its radius? Its radius
23:02is 10 and its angle is theta. What was
23:04theta? 0.546, this was your theta.
23:07There you go, so 10*0.546. This came to
23:125.46, here it is, this is also found.
23:15And if I write this in decimals too,
23:192π, so 2π came to 6.283, there you go
23:22, 6.283. Now what was your target? You
23:25found both of these arc lengths too.
23:27There you go, these arc lengths are
23:29also found. Finally, your target was
23:31the perimeter. Shaded perimeter. Your
23:35target was the shaded perimeter. So
23:37what does shaded perimeter mean? That
23:39you have to add these three sides. What
23:41all do you have to add? Number one, de.
23:43Here is de in the shaded perimeter.
23:45Number two, arc length dx, number three
23:48, arc length ex, okay? So you have to
23:50add these in the perimeter. Arc length
23:53dx, arc length ex, now technically you
23:57have to add all three together. Add all
24:01three lengths. So this is 4.454 + 6.283
24:06+ 5.46, this equals 16. 197, 16.20, so
24:12the answer derived from here is 16.20.
24:18cm, this was your final target. Clear,
24:23kid? This was the final target of our
24:26question. Clear? So, I broke down
24:30everything step-by-step, piece-by-piece
24:33. Clear, kid? I showed how to break
24:36down every single thing. And then, I
24:39finally showed you how to solve it.
24:41Clear, kid? So, this topic is quite
24:43interesting. First, you need to know
24:46right angles, non-right angles, circle
24:48properties, and angle properties. Okay?
24:51Let's practice more questions of this
24:53type. Come on, here is another question
24:56. Take 2 minutes to look at this
24:57question. I will help you figure out
25:00what needs to be done. First, read the
25:02question properly to see what the
25:04examiner is asking. Read it. Well done.
25:08Come on, first look at what the
25:11examiner is saying. Let's start with
25:14number one. A circle is given. There
25:17are two sectors inside this circle. If
25:19you have noticed, your circle is
25:21divided into two sectors. Okay? What
25:24did he do? He formed two different
25:26sectors inside a circle. Now, let me
25:29guide you one by one on what the
25:31examiner's main target is. Okay, kid?
25:34What is the main target? First of all,
25:37let's start with part one. Let's start
25:41with part one. Look, we will begin with
25:45part one first. See, you are given a
25:48circle like this. Here, we will start
25:50with part one. We will work on it one
25:53by one. Now look, you have this
25:55condition. Here it is, and this given
25:58angle is theta. So, what did I do? I
26:01copied the diagram first. This is theta
26:04. This region is r2. This is your point
26:07A, this is your point B, and this angle
26:10is theta RAD, meaning radians, and the
26:12name of this region is R1. Come on,
26:15let's start slowly now. What do we need
26:17to do? Number one, perimeter of region
26:20R1 is equal to the length of the major
26:24arc. Now, we will deal with these two
26:26things one by one. One by one. First of
26:29all, he mentioned the perimeter of R1.
26:31So, let's read the first word slowly.
26:35He asked us for the perimeter of region
26:37R1. Okay? He asked for the perimeter of
26:42region R1, let's call it the perimeter
26:45of R1 or the perimeter of region R1. So
26:48here, perimeter of region R1, let's
26:51look at region R1 slowly. This is its
26:53radius. This is its radius. And what is
26:55this? Its arc length. So, what is the
26:58perimeter? The sum of all sides r + r +
27:01rθ, that is the perimeter of this
27:04region. Listen again. This r is the
27:06radius. This r is the radius. This is
27:08radius*theta. This is its perimeter. We
27:11have learned the first thing. Now we
27:13will learn the second thing: what does
27:14major arc mean? He mentioned another
27:16thing, didn't he? He used another word,
27:18major arc length. So now we will first
27:21learn what major arc length means. What
27:24does major arc AB mean? Look above,
27:27number one. We know that if this angle
27:28is theta, I have already told you in
27:30the introduction what the angle at the
27:32center is. It is always 360, and 360
27:35means 2π. So, if I ask you what this
27:37outer angle is. You will say this outer
27:40angle is 360-theta. Meaning, we will
27:43now say 2π-θ because we know 360
27:46means 2π. This length of yours, this
27:49is your major arc length. Major, we
27:52know, what is the formula for arc
27:54length? The formula for arc length is
27:57radius multiplied by the angle. So what
27:59did we do? We multiplied the radius by
28:01the angle so that the major arc length
28:03could be found. The examiner says these
28:06two are equal. So this means r + r + r
28:11theta is equal to 2π*r-rθ. And who do
28:16you have to make the subject? Theta.
28:18You have to make theta the subject.
28:20Shift r theta over there. This will be
28:222r theta. And take this 2r there. 2πr-
28:252r, here it is. Now you see, 2r theta
28:28is copied as is. If I take 2r common
28:32from here, it will be π-1, π-1. Now
28:35you note if something is canceling out.
28:39Look, 2r and 2r, what happened? They
28:41canceled out. So what answer for theta
28:43was derived from here? From here, the
28:45answer for theta was proven. What were
28:47we supposed to show? The answer for
28:48theta. So from here, I found out what
28:50the answer for theta is. The answer for
28:53theta is π-1 radians. This is the
28:56result you had to show. Clear, son? You
28:59had to show this same result. Meaning
29:01there is nothing extraordinary in this.
29:03We kept solving just as he guided us.
29:06Meaning we did not do any extraordinary
29:08work on our own. Just as the examiner
29:11guided us, I kept solving it
29:13accordingly. Okay, son? This result of
29:17yours is proven. There you go. Next.
29:20Now look at what is written in the next
29:22part. Given that area R1 = 30, find the
29:25area of the region R2; that is your
29:27question. Now, there are two options
29:29for this. Okay? I will copy this
29:32diagram and then, God willing, we will
29:34start. Look at number one. I have
29:36copied this same diagram here. The same
29:38. Okay? Here you go, I have copied the
29:40same diagram. Now let's start the
29:42second part. Okay? Now let's start the
29:44second part to see what the examiner is
29:46demanding from us. Second part. Come on
29:49, let's read it. Number one, the
29:51examiner is telling me that the area of
29:53R1 is 30. Find the area of R2. Come on,
29:57option number one. Option number one.
30:01Come on, let's start. First of all, he
30:04said what is area R1? He gave me area
30:07R1. Well, tell me one thing. What is R1
30:10? R1 is a sector. Area of the sector R1
30:14. What is it given equal to? It is
30:19given equal to 30 cm², here you go =
30:2330 cm² given. Now, do you know what?
30:25What is the area of a sector? 1/2 r²
30:28theta. The area of a sector is 1/2 r²
30:32theta. This is the area of a sector.
30:34Obviously, you just proved 2 minutes
30:37ago what theta is? Theta is π-1. So
30:42r²/2 copied the same. In place of
30:44theta, I put π-1, and what did this
30:46become? 30, so from here we found out
30:49what the answer is. 60/π-1. 60/π-1,
30:56this is r squared. The answer hasn't
31:00come yet, by the way. This is r squared
31:03. If I take its square root, then
31:04obviously the answer will come in
31:06decimals. Our target is area R2. Now
31:09what is your target? The target is area
31:11R2. Area of the sector R2. The area of
31:17the sector R2 is basically my target in
31:20this question. The area of the sector
31:23R2 is my target. Tell me one thing,
31:26this whole sector R2, this whole sector
31:28R2, what is the angle of the sector?
31:312π-theta. So tell me, what is the
31:33formula for the area of a sector? 1/2*
31:35radius squared*angle, and what is the
31:37angle? 2π-theta. Brother, simply, I
31:41already know what the angle of the
31:42major sector is. 2π-theta. Now let's
31:45make it a little easier. Now tell me,
31:47you have already found r². What is the
31:50answer for r²? The answer for r² is
31:5360/π-1, here you go, this is the
31:55answer for r². What does 2π-θ mean?
31:59Theta has already been found. There,
32:00the theta has already arrived. Take
32:02this, subtracted from 2π. When I
32:05simplified this further, it became 230,
32:07230, so it became 30. When simplified,
32:11it became π + 1, so it became 30 times
32:14π + 1 divided by what's at the bottom?
32:17At the bottom, it became π-1, that was
32:20your final target. But what did the
32:23examiner say? The examiner said they
32:25require the answer to 3sf, so just wait
32:271 minute. 30*π + 1 Bracket close by π
32:34-1, I put all this into the machine.
32:3758.02, answer to 3sf is 58.0. So what
32:43is the answer from here? The answer
32:45from here is 58.0, here you go, 58.0.
32:49This is your final derived answer.
32:52Clear, kid? This is your final derived
32:56answer. So this is option number one.
32:58You did the simple working. Your area
33:01is found. Clear, kid? Now look at the
33:04second option for this same question.
33:06Option number two for the same question
33:08. Let me try again. Option number two
33:11for the same question. Your target is
33:14area R2. So tell me, area R2 is your
33:16target, right? R2 area is the target.
33:19So area R2 can also be found like this.
33:21Subtract sector R1 from the circle area
33:24. Look, let me explain this in more
33:25detail. Your target is the area of
33:30sector R2. So one option is that you
33:35subtract the area of R1 from the total
33:38area of this circle. What is the area
33:41of a circle? πr² is the area of the
33:44circle. What was the area of R1? The
33:47area of R1 was 30. The area of R1 was
33:5030. Tell me, I have already found r².
33:5360/π-1 We have already found this,
33:59right? -30, here it is. If you take its
34:03LCM, the LCM will also be the same, π-
34:051. So 60π-30π, 30π + 30 is there, so
34:11if I take 30 common, then exactly the
34:14same answer will be derived, 30 times
34:17π + 1/π-1, so exactly the same answer
34:21will be derived. Ultimately, there is
34:24no difference between these two answers
34:27. Clear, kid? Ultimately, there is no
34:29difference between these two answers.
34:31It depends on you which approach you
34:34use. Clear, kid? It depends on you
34:36which approach you use. Clear? So, this
34:39is the area of region R2 found. So,
34:43Area of the whole circle minus R1 is
34:45also perfect. Finding the area of
34:47sector R2 directly is also perfect. By
34:50the way, what will be derived from both
34:52methods? It will be derived the same
34:53way. Okay, child? Let’s practice more
34:56of this. Here it is. Now look at the
34:59question. This is an interesting
35:01question. First, read the question
35:03properly. Okay? You read the question
35:07first. I am waiting for you. Read. Well
35:10done. By the way, all other questions
35:13were from A-Level Maths. I took this
35:15question from Add Maths. From Add Maths
35:17. The others were all A-Level Maths. I
35:19took this from Add Maths. Read it
35:21properly, child, so I can guide you.
35:24Okay? Read it properly. So that I can
35:27guide you through this question. Come,
35:29first of all, the interesting thing is
35:32you have to show something. Everything
35:34is clearly written about what to show.
35:35Look, first of all, the diagram shows a
35:38sector ABC, so who is this whole
35:40diagram? It is a sector. The name of
35:42the sector is ABC; what is its full
35:44radius? r + r = 2r, r + r = 2r, this is
35:48its full radius. What is its full angle
35:51? 3 theta. I am not saying this. It is
35:53given in the question. Well, what is
35:57the first given hint? The area of the
35:59shaded region is 5 cm squared. So,
36:02let's start with the given hint. Come,
36:05first thing, the examiner gave you the
36:09shaded area. The examiner gave you the
36:12shaded area. How will the shaded area
36:14come? The examiner gave you the shaded
36:15area. So you will say area of the whole
36:18sector. What is the name of this whole
36:19big sector? ACB From this whole big
36:24sector, this and this will both be
36:26subtracted. These two small sectors are
36:29identical. So what will we say? Minus
36:31two small sectors. Two small identical
36:36sectors. They are identical sectors.
36:39What is the shaded area given? The
36:41shaded area is given as 5 cm squared.
36:43It is written here. Shaded area is 5 cm
36:46squared. This whole big sector, what is
36:49the radius of this larger sector? 2r.
36:52What is the angle of this larger sector
36:54? 3 theta. So the formula for the area
36:58of a sector is 1/2*radius squared
37:01multiplied by angle. We already know,
37:04right? I will copy it here. We already
37:07know the formula for the area of a
37:09sector is 1/2 rs, or we already know
37:11the formula for the area of a sector is
37:131/2 r² theta. We already know this; we
37:16have done this in every question and
37:18topic. Now, what do we need to subtract
37:20from this? Two identical sectors,
37:22subtract exactly the same ones, exactly
37:24the same. Now tell me one thing, what
37:26is the radius of this sector? r. This
37:29is r, this is also r, this is r, and
37:31this is r too. If its radius is r and
37:33angle is theta, what is the area of
37:35this sector? 1/2*radius squared*angle.
37:401/2*radius squared multiplied by angle.
37:42Now make this a little easier. 42 2 6²
37:46theta, make it even easier. 2 and 2 cut
37:50r² theta. There you go. Now make this
37:53even easier. 5 = 5 r² theta, make it
37:57easier. r² theta = 1, make it easier.
38:02You can find theta from here. The
38:05answer for theta is 1/r squared. What
38:09did we see here? First, we used the
38:12given hint. I used the hint that was
38:16given to me. This is the hint they gave
38:19. I used the hint. Now the question is,
38:23what is our target? Our target is the
38:26shaded perimeter. Our target is the
38:29shaded perimeter. Now a little extra
38:31effort. Need to focus a little bit.
38:34Then the shaded perimeter. Tell me,
38:36what is this entire large arc length?
38:38This is the arc length of this entire
38:40larger sector. Its radius is 2r. What
38:43is its angle? 3 theta. So I will write
38:46one by one, what do we need first?
38:47First of all, we need the major arc AB.
38:50We need this largest arc. The largest
38:53arc. Here it is. That's one done.
38:55Number two. What is this? This is GB.
38:59This length is GB, done, this is also
39:02done, GB, what is this? This is arc
39:04length FG, arc length F G, then what is
39:08this? R, what is this? R + R + R, okay?
39:14This is also done. This is also done.
39:15What is this? Arc length DE, arc length
39:19D E plus this is also R. This is also
39:22R. What is this? AD plus what is this?
39:26Length AD, here it is. Now we have to
39:28add all these to get P for perimeter.
39:31Okay, let's start. Arc length AB, how
39:34will this arc length come? What is its
39:36total radius? 2R, what is the angle? 3
39:38theta. Radius multiplied by angle. So
39:41what will this be? 2R*3 theta. Done.
39:44This is done. Next is GB. Here is GB.
39:47Who is GB? r, here you go, GB is R.
39:51Next, GB is done. Who is this? This is
39:53FG. What is FG? Arc length. The radius
39:57is r, the angle is theta, so this
39:58becomes r theta. There you go, this is
40:00the arc length too. fg is done, this is
40:02r, this is r, so these are just r and r
40:04simply. There you go, r and r have been
40:06added. What is this? Length, arc length
40:09, arc length de, radius r, angle theta,
40:11so what does this become? r*theta. Here
40:15, plus what is this? ad yours is
40:18already r. There you go, now simplify
40:21it a bit more so your answer is proven.
40:24Now start, look R, It became 4R. You
40:30have 4R now. We added all these arcs
40:32together. These 4r have come. Done.
40:35Next 6R theta Rθ Ra 8r theta, this
40:40became 8R theta. The issue is that your
40:43answer doesn't involve theta. Your
40:46answer is without theta. Without theta.
40:50So we don't need theta. So what will we
40:52do? Copy 4r as it is, here it is; copy
40:558R as it is, here it is. What is theta?
40:581/r², 1/r². So from here I found out
41:03that my perimeter, what did that
41:05perimeter become? 4r, r and r cancelled
41:09out, 8/r. This was finally your final
41:14target. Clear, kid? This was your final
41:18target, so the question wasn't tough,
41:20it was interesting and different. Well,
41:23if we note, this is the smallest
41:25chapter in the entire AS Maths. Meaning
41:28the work is the least in this. In terms
41:30of work, if we talk about it, there is
41:32the least amount of work in this topic
41:34because there are only two formulas.
41:35But every question of it is different.
41:38Clear, kid? So you will have to work a
41:40bit attentively in this. Look at
41:42another example. Let's go. Now read
41:45this question carefully as well. I took
41:48this from Add Maths too. Clear, kid? I
41:50took this from Add Maths too. Look at
41:53the question for two minutes so that I
41:55can guide you. Okay? This is also a
41:57good question. It's a small question.
42:00Come on, now slowly I will guide you on
42:03what to do in this question. Clear, kid
42:06? What to do? Number one, I copied the
42:11diagram to explain. It was the same so
42:15that I could work on that diagram so
42:16when the student wants to read the
42:17question, they can see the question
42:18separately. So that if they want to
42:20read it again, they can read it again.
42:21Let's begin. Now, if you remember, the
42:24center of this circle is C. This R is
42:26the radius. So when I drop a line from
42:28here, this angle will also be 90. I
42:30will drop a line from the center
42:31because what is this line? Tangent to
42:33the circle. What is this line also?
42:34Tangent to the circle. So, it means
42:36this is an external point. From this, I
42:38drew this tangent. From this, I drew
42:39this tangent. So technically, which
42:41condition is here? Tangents from the
42:45same external point. Tangents from the
42:49same external point. This is a property
42:51of a circle. It was named Property
42:53Number Nine. So, what was the first
42:55thing I did? Look, here it is. First of
42:57all, I dropped a line from here. Let me
42:59change the color. I dropped a line from
43:01here. Here it is. So what will the
43:03radius form with this? Let me make the
43:06line a bit lighter. Here it is. Let's
43:08see, what will the radius form with
43:09this? What will the radius form with
43:11this? 90 °. Here you go. The radius
43:13will form 90 °. What will the radius
43:16form? The radius will form 90 ° with
43:18this. Now look, you already have this
43:21thing. Well, look at this entire dotted
43:23line, this dotted one, look, this was
43:25the entire dotted line. Tell me, what
43:27is this? This is r. What is the radius
43:31of this entire larger sector? 10, so
43:35this must also be 10. Because common
43:38sense says that any gap from the center
43:40of the circle to the circumference is
43:42the radius. So if this is an entire
43:45sector. If I tell you this is a sector.
43:47This is your A. This is O. This is B.
43:49If this is a radius. If this is a
43:50radius, then all these are radii. All
43:52of these will obviously be 10. So that
43:55means number one, how much is the
43:57entire OD? This means technically that
44:00your ODB is 10 cm. Understand this
44:04first. What is your ODB? ODB is 10 cm.
44:08Because what is ODB? ODB is your radius
44:12of the sector OAB. It is the radius of
44:17the sector OAB. Is it clear, kid? What
44:20is this whole thing? This is an entire
44:22sector. So what is OD too? OD is also
44:24its radius. What is CD? CD is R. So if
44:27I ask what is OC? OC. If I ask any kid
44:31what is OC? Then the kid will say, sir,
44:34subtract CD from OD. Clear? How much is
44:37this entire length? It is 10. Subtract
44:39this R from it. So this remaining part
44:41is OC, so from here we know what OC is?
44:43What was OD? OD was 10. What was OD? OD
44:48was 10. What is CD? CD was R, so this
44:53became 10-R, which means first of all,
44:56I confirmed this length, which I will
44:58highlight—let me highlight it. Here,
45:02let me make this length a bit darker.
45:04So from here, I found out what this
45:06entire length is? This entire length
45:09came out to be 10-r, 10-r, this entire
45:11length is 10-r. So, first of all, what
45:15did the examiner ask for? OC. So OC was
45:18a one-mark question, and I already
45:21found out what the answer to OC was.
45:24OD-CD, so the answer for OD was 10-R,
45:27the answer for OD was 10-R, so from
45:30here I found out what the answer to OC
45:33is. The answer for OC is 10-R; this is
45:37how I found OC. That is why you see,
45:40how many marks was this for? One mark.
45:42There was nothing very interesting in
45:44this. Now you have to show R. Now, note
45:47this for two minutes. Look, if I
45:50highlight this triangle and show it to
45:52you. I am intentionally highlighting
45:54this triangle to show that I am picking
45:56this triangle. Then I justify my point.
45:59I explain. I picked this triangle. I
46:02intentionally highlighted it. Clear,
46:04beta? I intentionally highlighted it. I
46:07am interested in this triangle. Okay? I
46:11will mark this triangle. Okay? What is
46:14the name of this triangle? The name of
46:16the triangle is OCE, here is the
46:18triangle. OC, OCE, this is the triangle
46:22. Okay? So I will copy this triangle.
46:25Okay? I will copy this triangle. The
46:27name of the triangle is OCE, here it is
46:29. I have copied this triangle. Okay? I
46:33will copy this triangle. Here it is.
46:37Okay? What was the angle in this
46:38triangle? It was 90 °. I will copy
46:40this triangle from here. This point was
46:42your O. This point was your C. This
46:44point was your E. Done. And okay, let's
46:46see. Number one. What was OC? OC was 10
46:49-R. Clear, beta? And what was CE? CE is
46:53your R. This is R. This is also R. This
46:55is also R. This is also R. This is your
46:57R. And this small angle here, what was
46:59this small angle? This small angle was
47:02theta. This small angle right here was
47:06theta. Clear, kid? So, what did I do? I
47:09copied this right-angled triangle over
47:13here. Clear, kid? Now, I can see what
47:17is required in my answer. R is the
47:20subject and it's in terms of sin. So, I
47:22got the idea that I need sin. By the
47:25way. With respect to this angle, who is
47:28this? Opposite, and opposite to 90, who
47:30is this? Hypotenuse. So, I applied sin.
47:35Okay? I applied sin. So, what is the
47:38formula for sin? The formula for sin is
47:40opposite divided by hypotenuse.
47:42Opposite divided by hypotenuse. Your
47:44target is to make r the subject. So,
47:46you cross-multiply. tan sin theta-r sin
47:50theta equals what? It became r. Your
47:53target is to make r the subject. Read
47:55it. So now, what did we do? Shift this
47:58r sin theta over there. Shift this r
48:01sin theta over there. So, when it
48:03shifted there, what happened here? Here
48:05it became tan sin theta, and what came
48:08here? Here we got r + r sin theta. Now,
48:14there is something common in this. r is
48:15common in this. When r was taken common
48:18, what was left here? When r was taken
48:21common, what was left here? 1 + sin
48:23theta was left here, see. And what was
48:25left here? 10 times sin theta remained
48:28copied as it is. What is your target?
48:30Your target is R. By the way. Your
48:33target is R. You have to make R the
48:35subject. So, what answer for R was
48:37found from here? What answer for R was
48:40found from here? 10 sin theta/1 + sin
48:47theta. Your target has been achieved.
48:51In this question, what did you finally
48:52have to do? Your target was to show the
48:56radius. Clear, kid? Your target was to
49:00show the radius. 10 sin theta/1 + sin
49:03theta, this was your final target.
49:06Clear, kid? This was your final target.
49:10So, you should know the properties very
49:12well. No concept should be weak. Clear,
49:14kid? The topic is the smallest. But it
49:17is interesting. Interesting topic. Come
49:20on, let's try another question. You
49:22look at the question first. Well done.
49:24I am waiting for you for 2 minutes.
49:25Read the question first. Are you okay,
49:27son? First, please read the question so
49:30I can guide you on what to do. Look,
49:32well done. The interesting thing here
49:35is that every question is different. So
49:36, you have to stay a bit focused. Clear
49:39, son? Your focus should be quite good.
49:42Okay? Come, I'm showing you another
49:43question. Please read the question
49:45first. Then I'll guide you. Okay?
49:49Please read this question properly
49:51first. It's a slightly technical
49:53question. Okay? Now look for a minute.
49:59I'll copy the diagram. Then I'll get
50:02you started. Come, this is your
50:06original diagram. Okay? Original
50:08diagram. First of all, what is the
50:11examiner saying? Show the radius of the
50:14larger circle is r 2. Look, this is
50:17your entire larger circle. The bigger
50:19one. This is its center. This is its
50:21radius. This is its radius. Simple. It
50:25is written here. A larger circle at
50:27center B passes through C and D. This
50:29is a whole larger circle. This is its
50:31center. So, BC is its radius and BD is
50:34also its radius. Isn't that right, son?
50:36If we have read it properly and
50:38carefully. Clear? Read it properly and
50:40carefully. Look, the first thing is
50:43that I'll copy this diagram here. Here
50:47you go. I copied this diagram so I can
50:52show you some work on it. Okay? Come,
50:54part one. First, let's read what our
50:58target is for part one. Part one is a
51:01small question. What is our target in
51:03part one? Our target is the radius of
51:05the larger circle. So come, let me
51:07start slowly. Number one. What is this?
51:10Radius of the larger circle. What is
51:13this? Radius of the larger circle. Now,
51:15the interesting thing is this whole
51:16thing is a smaller circle. This is the
51:18center of the smaller circle. Its
51:21center is A, so this became its
51:23diameter, right? So that means this
51:25became a semicircle, a semi. We had
51:27learned that when both points of a
51:29diameter join in a semicircle, what
51:31angle is formed? 90 °. So, a property
51:34is being applied here. The name of the
51:36property is the semicircle property. So
51:38, a property is being used in this
51:41question. Which property is that? It
51:44was a circle property, and the
51:46property's name was the semicircle
51:49property. Meaning we learned that the
51:52angle subtended by a diameter is always
51:5490 °. Listen again. This is its radius
51:57. This is its diameter. So, this is a
51:59semi-circle. So, wherever the two
52:01points of the diameter join, what do
52:02they form? 90 °. So, first of all, in
52:05this one-mark question, we applied this
52:08property. Applied the property. Now,
52:11tell me one thing. Look, first of all,
52:15this is your r. This is also your r.
52:18Whose radius is this r? r is the radius
52:20of the smaller one. Now, what is the
52:22radius of the larger one? I don't know.
52:24Consider the radius of the larger one
52:26as R. Let's name the larger radius R.
52:28That is, this larger radius, this
52:30larger radius, whose radius is this?
52:32This is the radius of the larger circle
52:34. Take its radius as R. Is this capital
52:37R clear, beta? We have taken its radius
52:40as R. Let's see, there is one more
52:43option. Let's make it even easier. Okay
52:46? There are two options. I can explain
52:48to you using both. Let's do it this way
52:51. Let me make it even easier. Let's
52:53make it even easier. Look, if this is r
52:55, then this is also r. Let me make it
52:57even easier. Here you go. Look, this is
53:00the radius of the smaller one. This is
53:02also the radius of the smaller one.
53:03This is also a radius. This is also a
53:04radius. This is also r. This is also r.
53:06So, if you look at a small triangle, a
53:09small one. Look at this smaller
53:10triangle, the small one. Here is a
53:12smaller triangle. Okay? If you look at
53:15this, a smaller triangle. Here is the
53:17smaller triangle. Okay? Let's copy the
53:19name of the smaller triangle. This was
53:21C. This was your A. This was your B.
53:23This was the radius of the smaller
53:25circle. This is also 90 °. This is
53:27clearly visible. This is your radius of
53:29the larger circle. Radius of the larger
53:32circle. Understand it like this, I have
53:34copied this triangle. Here it is, I
53:35copied this triangle. What is this
53:37triangle? It's a right-angled one. I am
53:39not saying it, it is already r. This is
53:41r. This 90 is already given. So, I
53:43applied Pythagoras. So, according to
53:45Pythagoras, what will be the answer?
53:46r² + r², Pythagoras applied. This
53:50became 2 r², here it is. So, for the
53:52radius, you will have to take the
53:54square root. The square root of two
53:55will obviously be root two. Because you
53:57require the root in the answer too.
53:59Here it is, and the radius of r² will
54:02be r. So, you wanted the answer in surd
54:05form. You wanted the answer in surd
54:08form. So, what was your target in part
54:10one? What was your target, what was
54:12required? The radius of the larger
54:15circle was required, sorry. The radius
54:18of the larger circle was required. That
54:21was your final target. Alright, that's
54:23one thing. Now the examiner says find
54:26the area of the shaded region in terms
54:28of R. We need to find the shaded area
54:30in terms of R. This is our final target
54:33. The shaded area in terms of R, this
54:35is our final target. We need to find
54:37this. Is that clear, dear? We need to
54:39find this. Okay? Now, let's see what we
54:42will do. If I copy this diagram. I'll
54:45show you the diagram again to explain
54:47it so that it becomes a bit more
54:49obvious from the diagram. Here you go.
54:52Okay? Okay? Now I need to solve the
54:53second part. Look, I copied this
54:55diagram from the previous question for
54:58the second part. Now I will solve the
55:00second part. Okay, dear? My target is
55:02the second part. Take 2 minutes to
55:04check the question. Here it is. So what
55:07did I do? I copied this diagram from
55:09the second part. I request you to look
55:11at the question for 2 minutes, dear. I
55:14will erase the extra portion. Okay,
55:18dear? I will erase the extra portion.
55:20Now you see what your target is. Okay,
55:23dear? Here you go. I erased the extra
55:25portion so that you don't get confused
55:28by it. Here, I erased the extra portion
55:31. Now, I have shown only the portion we
55:33need to work with. Clear, dear? I have
55:36shown only that portion. I have erased
55:38all the rest of the extra work. Okay,
55:40dear? I erased the extra portion. Now,
55:42shall we begin? What is your target?
55:45Your target is the shaded area. Now
55:48tell me one thing. What is this thing?
55:50If you note one thing, what is this? If
55:54this angle is 90 °, then what is this
55:55entire shape? The name of this shape
55:57should be a quadrant. Also, one thing
56:00is obvious, this radius has been found,
56:02right? This radius has been found. What
56:05was its answer? 2*r, I just found this
56:08in the previous part. So what was its
56:11answer? 2*r 2*r, I have already found
56:17this. 2*R, clear, dear? Let's go.
56:21Alright, now what is my target? I need
56:23the shaded area. So, tell me, what is
56:26this entire shape? This whole shape is
56:28a quadrant. Here it is. Remove this
56:30triangle and this triangle from it.
56:32Then this area will be obtained. This
56:33small one here. So, I will do one thing
56:36. First, I will highlight this area for
56:39you. I will highlight this area for you
56:42. Okay? I have intentionally
56:44highlighted this area. Okay? I have
56:48intentionally highlighted this area.
56:50Here you go. Okay? Now, what is my
56:53target? Obviously, what will I call
56:56this portion? A segment. Brother,
56:59technically this forms a whole sector.
57:01Call it a sector or a quadrant. From
57:03that, this triangle and this triangle
57:04will be removed. Then this area will be
57:06obtained. This area will be obtained.
57:08What is this area? This area is segment
57:11AD. Meaning, you can call it the entire
57:14segment CAD. Name this whole thing
57:17segment CAD. So, first, I will find out
57:20the segment area. Segment area. The
57:24area of which segment? What is the name
57:26of this segment? Segment CAD. Segment
57:29CAD, here it is. So, how will this come
57:31about? What is this whole shape? This
57:33whole shape is. A quadrant. A quadrant.
57:36It is 1/4 of a circle. So, what is the
57:38formula for 1/4 of a circle? 1/4*π*r²
57:41, here it is. You have a full larger
57:45sector, and this was 1/4 of it. Meaning
57:48, you know this was a large sector, so
57:50this was 1/4 of it. Meaning, it would
57:54be like this, one 4th of it. One will
57:56be this, one will be this. Out of that,
57:57this was one 1/4. Here, from that 1/4,
58:00what needs to be subtracted? This
58:02triangle, what is the height of this
58:04triangle? r, what is the base? r, so
58:06two such triangles need to be
58:08subtracted. So what will I say? Two
58:10times 1/2*height*base. Two times 1/2*
58:13height*base. Now what happened? 2 and
58:15this canceled out. So what is the
58:17answer here? πr²/2, here it is. And
58:20what is here?-r², so the segment area
58:24is found. Segment area. What is the
58:27segment's name? The segment's name is
58:29segment CAD. The segment's name is
58:32segment CAD. Clear, kid? So, half the
58:35work is done. Segment CAD is obtained.
58:37Okay? We have done a bit of work.
58:39Segment CAD has been found. Here you go
58:41. But your actual target is right here.
58:44This is your original target called the
58:46shaded area. So you already know that
58:48this whole shape is a semi-circle.
58:50Clear, kid? So what is your final
58:52target? Your target is the shaded area.
58:55Your target is the shaded area. The
58:58shaded area. Here it is. This is your
58:59original target. So how will you get
59:02this original target, the shaded area?
59:04From this entire semi-circle. The
59:06semi-circle is named DEC. That this
59:08entire thing is a semi-circle. So what
59:10do we need to do? Semi-circle DEC. What
59:19do we need to subtract from it? We need
59:22to subtract the segment you just found
59:24out. What was the name of the segment?
59:26The name of the segment was C. What is
59:29the area of the semi-circle? πr²,
59:30look here, brother. Its radius is r.
59:33The radius is r. This whole thing is a
59:35semi-circle. Subtract this from it. The
59:37shaded area is obtained. So we found
59:39the area of the segment 2 minutes ago.
59:42I found the area of the segment 2
59:44minutes ago. What was the area of the
59:46segment? πr²/2-r² By the way, what
59:53is the area of the semi-circle? πr²/2
59:55, this is a whole semi-circle. What is
59:58the area of this entire semi? πr²/2,
1:00:00minus this area from it. So now when
1:00:04you simplify this, what happens? This
1:00:06and this get cancelled out. So finally,
1:00:08what answer for your question was
1:00:10derived? Finally, this answer to your
1:00:13question was derived. So what was the
1:00:16final answer derived? The final answer
1:00:19is r². So this was a good and slightly
1:00:23different question. Clear, kid? So
1:00:25today, we finally practiced a bit more
1:00:27on circular measurement. Alright, kid?
1:00:30We will practice this topic for one or
1:00:32two more classes to gain a bit more
1:00:34confidence. Alright, kid?

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