Line of Sight: The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.
Angle of Elevation: The angle of elevation of an object viewed, is the angle formed by the line of sight with the horizontal when it is above the horizontal level.
Angle of Depression: The angle of depression of an object viewed, is the angle formed by the line of sight with the horizontal when it is below the horizontal level.
Snap Test
The angle of depression of the top and the bottom of a building 50 metres high as observed from the top of a tower are \[\mathbf{30}{}^\circ \] and \[\mathbf{60}{}^\circ \] respectively. Find the height of the tower and also the horizontal distance between the building and the tower.
(a) 73 m.43.3 m (b) 90.3 m, 75 m
(c) 75 m, 41.3 m (d) 75 m, 43.3 m
(e) None of these
Ans. (d)
Explanation: Let AB be the building and CD be the tower in the figure given below.
Let \[BD\text{ }=\text{ }x\text{ }metres\text{ }and~\,CD\,\,=\,\,h\,\,metres\].
Draw\[AE\parallel BD\].
Then, \[\angle ~CAE\text{ }=\,\,\angle ~FCA\text{ }=\text{ }30{}^\circ \]
And \[~\angle CBD\text{ }=~\angle FCB\text{ }=\text{ }60{}^\circ \]
In \[\Delta CAE\]
\[CE\text{ }=\text{ }h\text{ }-\text{ }50,\,\,\angle \,CAE\text{ }=\text{ }30{}^\circ \text{ }and\text{ }AE\text{ }=\text{ }x\]
\[\therefore \,\,\,\,\,\frac{CE}{AE}\,\,=\,\,\tan 30{}^\circ \,\,\,\,\Rightarrow \,\,\,\,\frac{h-50}{x}\,\,\,\,=\,\,\,\frac{1}{\sqrt{3}}\]
\[\Rightarrow \,\,\,h\,\,-\,\,50\,\,=\,\,\frac{x}{\sqrt{3}}\] ……. (i)
In \[\Delta CBD\]
\[CD\text{ }=\text{ }h,\text{ }BD\text{ }=\text{ }x\text{ }and\,\,\angle CBD\text{ }=\text{ }60{}^\circ \]
\[\therefore \,\,\,\frac{CD}{BD}\,\,\,=\,\,\tan \,60{}^\circ \,\,\Rightarrow \,\,=\,\,\frac{h}{x}\,\,=\,\,\sqrt{3}\,\,\Rightarrow \,\,h\,\,\,=\,\,\sqrt{3}\,x\]
Substituting \[h=~\sqrt{3}x\] from (ii) in (i), we get:
\[\sqrt{3}x-50\,\,=\,\,\frac{x}{\sqrt{3}}\,\,\Rightarrow \,\,\,3x\,-\,50\sqrt{3}\,\,=\,\,\,x\,\,\,\Rightarrow \,\,2x=50\sqrt{3}\,\,\,\Rightarrow \,\,x=25\sqrt{3}\,\,=\,\,75\].
Substituting \[x\text{ }=\text{ }25\sqrt{3}\] in (ii), we get: \[h=\sqrt{3}x\,\,=\sqrt{3}\,\,\times \,\,25\,\sqrt{3}=75.\]
Thus, height of the tower \[CD\text{ }=\text{ }h\text{ }=\text{ }75\] metres and the distance between the building and the tower \[=\text{ }BD\text{ }=\text{ }x\text{ }=~\,\,25\sqrt{3}\,\,metres\text{ }=\] \[\left( 25\text{ }\times \text{ 1}.732 \right)\text{ }m\text{ }=\text{ }43.3\text{ }m.\]
A man standing on the deck of a ship, which Is 10 m above the water level, observes the angle of elevation of the top of a hill as \[\mathbf{60}{}^\circ \] and the angel of depression of the base of the hill as\[\mathbf{30}{}^\circ \], Calculate the distance of the hill from the ship and the height of the hill.
Test for Collinearity of Three Points: To show that any three given points P, Q and R are collinear, we find the distance PQ, QR and PR and show that any one of these distance is equal to the sum of the other two.
Section Formula: The coordinates of the point \[P\left( x,\text{ }y \right)\] which divides the line segment joining the points \[A({{x}_{1}}^{,}\,{{y}_{1}})\,\,and\,\,B({{x}_{2}},\,{{y}_{2}})\], internally, in the ratio \[{{m}_{1}}:\text{ }{{m}_{2}}\] are given by
Midpoint Formula: The midpoint of a line segment joining the points \[P\left( {{x}_{1}},\text{ }{{y}_{1}} \right)\] and \[Q({{x}_{2}},\,{{y}_{2\,\,}})\,is\,\,\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\,\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]
9. Area of a Triangle: The area of a \[\Delta \,PQR\] having vertices points \[P({{x}_{1}},\,\,{{y}_{1}}),\,Q({{x}_{2}},\,\,{{y}_{2}})\] and \[R({{x}_{3}},\,\,{{y}_{3}})\] is the numerical value of the expression \[\left| \frac{1}{2}\{{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})\} \right|\]
Snap Test
Find the coordinates of the point which divides the line segment joining the points \[\left( \mathbf{6},\text{ }\mathbf{3} \right)\text{ }\mathbf{and}\text{ }\left( -\mathbf{4},\text{ }\mathbf{5} \right)\] in the ratio \[\mathbf{3}\text{ }:\text{ }\mathbf{2}\] internally.
(a) \[\left( 0,\,\,\frac{21}{5} \right)\]
(b) \[\left( 1,\,\,\frac{21}{5} \right)\]
(c) \[\left( 1,\,\,\frac{21}{4} \right)\]
(d) \[\left( 2,\,\,\frac{21}{4} \right)\]
(e) None of these
Ans. (a)
Explanation: Let AB be the line segment with end points \[A\left( 6,\text{ }3 \right)\text{ }and\text{ }B\text{ }\left( -\text{ }4,\text{ }5 \right).\]
Then, \[\left( {{x}_{1}}=\text{ }6,\text{ }{{y}_{1}}=\text{ }3 \right)\] and \[\left( {{x}_{2}}=-\text{ }4,\text{ }{{y}_{2}}=\text{ }5 \right)\].
Also, \[{{m}_{1}}=\text{ }3\text{ }and\text{ }{{m}_{2}}=\text{ }2\].
Let the required point be \[P\text{ }\left( x,\text{ }y \right)\].
By section formula:
\[x=\frac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{1}}+{{m}_{2}}},\,\,y\,\,=\frac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{1}}+{{m}_{2}}}\]
\[x\,\,=\,\,\frac{3\times (-4)+2\times 6}{(3+2)},\,\,y\,\frac{(3\times 5+2\times 3)}{(3+2)}\,\,\,\Rightarrow \,x=0,\,\,y=\frac{21}{5}\]
Hence, the required point is \[P\left( 0,\,\,\frac{21}{5} more...
Congruent Figures: Two figures which are alike in all respects i.e., which have the same shape and size are known as congruent figures.
Similar Figures: Two figures having the same shape but not necessarily the same size are known as similar figures.
All the congruent figures are similar but converse is not true.
Equiangular Triangles: Two triangles are said to be equiangular if their corresponding angles are equal.
Similar Triangles: Two triangles are said to be similar if their corresponding angles are equal and their corresponding sides are proportional.
Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio.
If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangle are similar (AAA similarity).
If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar (AA similarity).
If in two triangles, corresponding sides are in the same ratio, then their corresponding angles are equal and hence the triangles are similar (SSS similarity).
If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are similar (SAS similarity).
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Snap Test
In the adjoining figure, if \[\mathbf{DE}\,~\parallel \mathbf{AB}\] and \[\mathbf{FE}~\,\,\parallel \,\,\mathbf{DB}\], then \[\mathbf{CF}.\mathbf{AC}\] =?
(a) \[F{{C}^{2}}\] (b) \[F{{E}^{2}}\]
(c) \[D{{C}^{2}}\] (d) \[D{{B}^{2}}\]
(e) None of these
Ans. (a)
Explanation: In \[\Delta \,ABC\], it is given that \[DE\,\,~\parallel \,\,AB\].
\[\therefore \,\,\,\,\,\frac{CD}{DA}=\frac{CE}{EB}\] …….. (i)
In \[\Delta \,CDB\], it is given that \[FE~\,\,\parallel \,\,DB\].
\[\therefore \,\,\,\,\,\frac{CF}{FD}=\frac{CE}{EB}\] …….. (ii)
From equations (i) and (ii), we get:
\[\frac{CD}{DA}=\frac{CF}{FD}\]
\[\Rightarrow \,\,\,\,\frac{DA}{CD}=\frac{FD}{CF}\Rightarrow \,\,\,\,\frac{DA}{CD}+1\,\,\,=\,\frac{FD}{CF}+1\,\,\Rightarrow \,\,\,\frac{DA+CD}{CD}=\frac{FD+CF}{CF}\]
\[\Rightarrow \,\,\,\,\,\frac{AC}{CD}=\frac{DC}{CF}\,\,\,\,\,\,\,\,\Rightarrow \,\,D{{C}^{2}}=CF\times AC\]
2. In the figure, AD is the bisector of \[\angle \,\mathbf{BAC}\]. more...
Secant: A line which intersects a circle in two distinct points.
Tangent: A line intersecting a circle only in one point is called a tangent to the circle at that point.
The tangent to a circle is perpendicular to the radius through the point of contract.
The point at which the tangent meets the circle is called the point of contact.
Length of Tangent: The length of the segment of the tangent from the given point P on the tangent to the point of contact with the circle is called the length of the tangent from the point P to the circle.
The lengths of the two tangents from an external point to a circle are equal.
Number of tangents to a circle:
(i) There is no tangent passing through a point lying inside the circle.
(ii) There is one and only one tangent passing through a point lying on the circle.
(iii) There are exactly two tangents through a point lying outside a circle.
A circle is inscribed in a \[\Delta \,\mathbf{ABC}\] having sides 8 cm, 10 cm and 12 cm, as shown in the adjoining figure. Find the length of CF.
(a) 7 cm (b) 3 cm
(c) 5 cm (d) 2 cm
(e) None of these
Ans. (b)
Explanation: We know that the lengths of tangents to a circle from an external point are equal.
Let \[AD\text{ }=\text{ }AF\text{ }=\text{ }x\text{ }cm,\text{ }BD\text{ }=\text{ }BE\text{ }=\text{ }y\text{ }cm\text{ }and\text{ }CE\text{ }=\text{ }CF\text{ }=\text{ }z\text{ }cm\]
Now, \[AB\text{ }=\text{ }12\,cm\] \[\Rightarrow \] \[AD\text{ }+\text{ }BD\text{ }=\text{ }12\text{ }cm\] \[\Rightarrow \] \[x\text{ }+\text{ }y\text{ }=\text{ }12\text{ }cm\] ..... (i)
\[BC\text{ }=\text{ }8\text{ }cm\] \[\Rightarrow \] \[BE\text{ }+\text{ }CE\text{ }=\text{ }8\text{ }cm\] \[\Rightarrow \] \[y\text{ }+\text{ }z\text{ }=\text{ }8\text{ }cm\] .... (ii)
and \[AC\text{ }=\text{ }10\text{ }cm\]\[\Rightarrow \] \[AF\text{ }+\text{ }CF\text{ }=\text{ }10\text{ }cm\] \[\Rightarrow \] \[x\text{ }+\text{ }z\text{ }=\text{ }10\text{ }cm\] .... (iii)
Adding (i), (ii) and (iii) we get:
\[2\left( x\text{ }+\text{ }y\text{ }+\text{ }z \right)=30\] \[\Rightarrow \] \[x\text{ }+\text{ }y\text{ }+\text{ }z\text{ }=\text{ }15\] more...
Circle: A circle is the locus of a point which moves in such a way that its distance from a fixed point always remains the same.
Chord: A line segment joining any two points on a circle is called a chord of the circle.
Arc: A continuous piece of a circle is called an arc of the circle. An arc AB is denoted by \[\overset\frown{AB}\].
Segment: A segment of a circle is the region bounded by an arc and a chord, including the arc and the chord.
Sector of a circle: The region enclosed by an arc of a circle and its two bounding radii is called a sector of the circle.
Circumference of a circle = \[2\pi r\]
Area of a circle = \[\pi {{r}^{2}}.\]
Length of an arc of a sector of a circle with radius r and angle with degree measure \[\theta \] is\[\frac{\theta }{360}\times 2\pi r\].
Area of a sector of a circle with radius rand angle with degrees measure \[\theta \] is \[\frac{\theta }{360}\times \pi {{r}^{2}}\]
10. Area of segment of a circle = Area of the corresponding sector - Area of the corresponding triangle.
Snap Test
A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.
(a) 70 cm (b) 50 cm
(c) 60 cm (d) 80 cm
(e) None of these
Ans. (a)
Explanation: Distance covered by the wheel in 1 revolution
\[=\,\,\,\frac{total\,dis\operatorname{tance}}{number\,of\,revolution}\,\,=\,\,\left( \frac{1100000}{5000} \right)\,cm\,=\,220\,cm\]
\[\therefore \] Circumference of the wheel \[=\text{ }220\text{ }cm\]
\[\Rightarrow ~~\,\,\,~2\pi r\,\,=\,\,220cm~~~\Rightarrow \text{ }\pi r\,\,=\,\,110\,cm\]
\[\Rightarrow ~~\,\,\,~r=\left( \frac{110\times 7}{22} \right)\,cm\,\,=\,\,35\,cm\]
Hence, the diameter of the wheel = 70 cm.
In the given figure, sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. Find the area of the shaded region.
(a) \[8.625\,\,c{{m}^{2}}\]
(b) \[9.1\,\,c{{m}^{2}}\]
(c) \[9.625\,\,c{{m}^{2}}\]
(d) \[7.625\,\,c{{m}^{2}}\]
(e) None of these
Ans. (c)
Explanation: It is given that outer radius \[R\text{ }=\text{ }7\text{ }cm\] and inner radius \[r\,\,=\,\,3.5\,cm\].
Area of shaded region
\[=\text{ }(area\text{ }of\text{ }the\text{ }sector\text{ }with\text{ }radius\text{ }=\text{ }R\text{ }=\text{ }7\text{ }cm\text{ }and\text{ }central\text{ }angle\text{ }=\,\,\theta \,\,=\text{ }30{}^\circ )\]
\[-\text{ }(area\text{ }of\text{ }the\text{ }sector\text{ }with\text{ }radius\text{ }=\text{ }r\text{ }=\text{ }3.5\text{ }cm\text{ }and\text{ }central\text{ }angle\text{ }=\,\,\theta \,\,~=\text{ }30{}^\circ )\]
= \[\frac{\pi {{R}^{2}}\theta }{360}-\frac{\pi {{r}^{2}}\theta more...
Probability: Probability is the quantitative measure of the degree of certainty of the occurrence of an event.
The probability of a sure event is 1
The probability of an impossible event is 0.
The probability of an event E is a number P (e) such that \[0\le P(E)\,\,\,\le \,\,1\]
A random experiment: An operation which produces some well-defined outcome such that all possible outcomes are known but the exact outcome is unpredictable is called a random experiment.
Event: The collection of all or some of the possible outcomes of a random experiment is called an event.
Probability of occurrence of an event: The probability of occurrence of a event E is given by:
Elementary Event: An event having only one outcome is called an elementary event.
The sum of the probabilities of all the elementary events of an experiment is 1
For any event E, \[\mathbf{P}\left( \mathbf{E} \right)\mathbf{+}\,\,\overline{\mathbf{E}}\,\,\mathbf{= 1}\], where stands for ‘not E’. E and \[\overline{E}\] are called complementary events.
Probability of an event cannot be negative and lies between 0 and 1.
In a pack of 52 cards we have:
(i) 4 suits - spades, hearts, clubs and diamonds having 13 cards each.
(ii) Each suit has one ace, one king, one queen, one jack and 9 other cards from 2 to 10.
(iii) King, queen and jack are called face cards or picture cards.
(iv) Hearts and diamonds are red coloured cards while spades and clubs are black coloured cards.
Snap Test
1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If Mahir has purchased one lottery ticket, what is the probability of winning a prize?
Chemical Reactions and Acids, Bases and Salts
Chemical Reactions
Its a process in which one or more substances/ the reactants are converted to one or more different substances, the products. During a chemical reaction, rearrangement of reacting substances takes place to form new substances, which have different properties than the original one. In a chemical reaction, substances are divided into reactants and products. Reactants are the substances that take part in a chemical reaction whereas products are the substances that are formed as a result of chemical reaction.
Endothermic and Exothermic Processes
In all the chemical reactions, transformation or change in energy is involved. On the basis of change in energy all the reactions are divided into two parts that are endothermic and exothermic reactions. The reaction in which heat is absorbed is called endothermic reaction. The reaction in which energy is given out in the surroundings is called exothermic reaction.
Photosynthesis is an example of an endothermic reaction. In the process of photosynthesis, plants by utilizing the energy of the sun convert carbon dioxide water and glucose and oxygen.
\[6C{{O}_{2}}+6{{H}_{2}}O\,\xrightarrow[Chlorophyll]{Light}\,{{C}_{6}}{{H}_{12}}{{O}_{6}}+6{{O}_{2}}\]
(Carbon dioxide) (Carbohydrate)
Sodium and chlorine are mixed together to yield table salt is an example of exothermic reaction. 411 kJ of energy is produced in this reaction.
\[Na+0.5C{{l}_{2}}(g)\to NaCl(s)+411\,\,KJ\]
Chemical Equation
A chemical equation is a way of writing or describing chemical reactions. It tells us that what happens when a chemical reaction takes place. It consists of information about reactants, products, the formulas of the reactants and products, the states of the reactants and products such as solid, liquid, gas and the amount of each substance.
Balancing the Equation
To get the same number of atoms of every element on each side of the equation, apply the law of conservation of mass. Now balance an element that appears only once in reactant and product. After balancing one element, proceed further to balance the next and continue balancing until all the elements are balanced. Now you need to balance chemical formulas by placing coefficients in front of them. Do not add subscripts because this will change the formulas.
Types of Chemical Reactions
All the chemical reactions are divided into six categories. These six categories are as follows:
Combination Reaction
When two or more elements or compounds combine to form a compound, then combination reaction takes place. They are mostly exothermic.
The combination of iron and sulphur to form iron sulphide is an example of combination reaction:
\[8Fe+{{S}_{8}}\xrightarrow{{}}8FeS\]
(Iron) (Sulphur) (Iron sulphide)
Decomposition Reaction
When a complex compound breaks down to make simple molecule, decomposition reaction takes place. They are always endothermic.
The electrolysis of water to make oxygen and hydrogen gas is an example of decomposition reaction:
\[2{{H}_{2}}O\xrightarrow{{}}2{{H}_{2}}+{{O}_{2}}\]
(Water) (Hydrogen) (Oxygen)
Single Displacement Reaction
When in a reaction an atom or a group of atoms present in a molecule is displaced by another atom, is known as displacement reaction. The displacement of silver when more...
Let \[BD\text{ }=\text{ }x\text{ }metres\text{ }and~\,CD\,\,=\,\,h\,\,metres\].
Draw\[AE\parallel BD\].
Then, \[\angle ~CAE\text{ }=\,\,\angle ~FCA\text{ }=\text{ }30{}^\circ \]
And \[~\angle CBD\text{ }=~\angle FCB\text{ }=\text{ }60{}^\circ \]
In \[\Delta CAE\]
\[CE\text{ }=\text{ }h\text{ }-\text{ }50,\,\,\angle \,CAE\text{ }=\text{ }30{}^\circ \text{ }and\text{ }AE\text{ }=\text{ }x\]
\[\therefore \,\,\,\,\,\frac{CE}{AE}\,\,=\,\,\tan 30{}^\circ \,\,\,\,\Rightarrow \,\,\,\,\frac{h-50}{x}\,\,\,\,=\,\,\,\frac{1}{\sqrt{3}}\]
\[\Rightarrow \,\,\,h\,\,-\,\,50\,\,=\,\,\frac{x}{\sqrt{3}}\] ……. (i)
In \[\Delta CBD\]
\[CD\text{ }=\text{ }h,\text{ }BD\text{ }=\text{ }x\text{ }and\,\,\angle CBD\text{ }=\text{ }60{}^\circ \]
\[\therefore \,\,\,\frac{CD}{BD}\,\,\,=\,\,\tan \,60{}^\circ \,\,\Rightarrow \,\,=\,\,\frac{h}{x}\,\,=\,\,\sqrt{3}\,\,\Rightarrow \,\,h\,\,\,=\,\,\sqrt{3}\,x\]
Substituting \[h=~\sqrt{3}x\] from (ii) in (i), we get:
\[\sqrt{3}x-50\,\,=\,\,\frac{x}{\sqrt{3}}\,\,\Rightarrow \,\,\,3x\,-\,50\sqrt{3}\,\,=\,\,\,x\,\,\,\Rightarrow \,\,2x=50\sqrt{3}\,\,\,\Rightarrow \,\,x=25\sqrt{3}\,\,=\,\,75\].
Substituting \[x\text{ }=\text{ }25\sqrt{3}\] in (ii), we get: \[h=\sqrt{3}x\,\,=\sqrt{3}\,\,\times \,\,25\,\sqrt{3}=75.\]
Thus, height of the tower \[CD\text{ }=\text{ }h\text{ }=\text{ }75\] metres and the distance between the building and the tower \[=\text{ }BD\text{ }=\text{ }x\text{ }=~\,\,25\sqrt{3}\,\,metres\text{ }=\] \[\left( 25\text{ }\times \text{ 1}.732 \right)\text{ }m\text{ }=\text{ }43.3\text{ }m.\]
Snap Test
(a) \[F{{C}^{2}}\] (b) \[F{{E}^{2}}\]
(c) \[D{{C}^{2}}\] (d) \[D{{B}^{2}}\]
(e) None of these
Ans. (a)
Explanation: In \[\Delta \,ABC\], it is given that \[DE\,\,~\parallel \,\,AB\].
\[\therefore \,\,\,\,\,\frac{CD}{DA}=\frac{CE}{EB}\] …….. (i)
In \[\Delta \,CDB\], it is given that \[FE~\,\,\parallel \,\,DB\].
\[\therefore \,\,\,\,\,\frac{CF}{FD}=\frac{CE}{EB}\] …….. (ii)
From equations (i) and (ii), we get:
\[\frac{CD}{DA}=\frac{CF}{FD}\]
\[\Rightarrow \,\,\,\,\frac{DA}{CD}=\frac{FD}{CF}\Rightarrow \,\,\,\,\frac{DA}{CD}+1\,\,\,=\,\frac{FD}{CF}+1\,\,\Rightarrow \,\,\,\frac{DA+CD}{CD}=\frac{FD+CF}{CF}\]
\[\Rightarrow \,\,\,\,\,\frac{AC}{CD}=\frac{DC}{CF}\,\,\,\,\,\,\,\,\Rightarrow \,\,D{{C}^{2}}=CF\times AC\]
2. In the figure, AD is the bisector of \[\angle \,\mathbf{BAC}\]. more... 
(a) 7 cm (b) 3 cm
(c) 5 cm (d) 2 cm
(e) None of these
Ans. (b)
Explanation: We know that the lengths of tangents to a circle from an external point are equal.
Let \[AD\text{ }=\text{ }AF\text{ }=\text{ }x\text{ }cm,\text{ }BD\text{ }=\text{ }BE\text{ }=\text{ }y\text{ }cm\text{ }and\text{ }CE\text{ }=\text{ }CF\text{ }=\text{ }z\text{ }cm\]
Now, \[AB\text{ }=\text{ }12\,cm\] \[\Rightarrow \] \[AD\text{ }+\text{ }BD\text{ }=\text{ }12\text{ }cm\] \[\Rightarrow \] \[x\text{ }+\text{ }y\text{ }=\text{ }12\text{ }cm\] ..... (i)
\[BC\text{ }=\text{ }8\text{ }cm\] \[\Rightarrow \] \[BE\text{ }+\text{ }CE\text{ }=\text{ }8\text{ }cm\] \[\Rightarrow \] \[y\text{ }+\text{ }z\text{ }=\text{ }8\text{ }cm\] .... (ii)
and \[AC\text{ }=\text{ }10\text{ }cm\]\[\Rightarrow \] \[AF\text{ }+\text{ }CF\text{ }=\text{ }10\text{ }cm\] \[\Rightarrow \] \[x\text{ }+\text{ }z\text{ }=\text{ }10\text{ }cm\] .... (iii)
Adding (i), (ii) and (iii) we get:
\[2\left( x\text{ }+\text{ }y\text{ }+\text{ }z \right)=30\] \[\Rightarrow \] \[x\text{ }+\text{ }y\text{ }+\text{ }z\text{ }=\text{ }15\] more... 
(a) \[8.625\,\,c{{m}^{2}}\]
(b) \[9.1\,\,c{{m}^{2}}\]
(c) \[9.625\,\,c{{m}^{2}}\]
(d) \[7.625\,\,c{{m}^{2}}\]
(e) None of these
Ans. (c)
Explanation: It is given that outer radius \[R\text{ }=\text{ }7\text{ }cm\] and inner radius \[r\,\,=\,\,3.5\,cm\].
Area of shaded region
\[=\text{ }(area\text{ }of\text{ }the\text{ }sector\text{ }with\text{ }radius\text{ }=\text{ }R\text{ }=\text{ }7\text{ }cm\text{ }and\text{ }central\text{ }angle\text{ }=\,\,\theta \,\,=\text{ }30{}^\circ )\]
\[-\text{ }(area\text{ }of\text{ }the\text{ }sector\text{ }with\text{ }radius\text{ }=\text{ }r\text{ }=\text{ }3.5\text{ }cm\text{ }and\text{ }central\text{ }angle\text{ }=\,\,\theta \,\,~=\text{ }30{}^\circ )\]
= \[\frac{\pi {{R}^{2}}\theta }{360}-\frac{\pi {{r}^{2}}\theta more... 
