In an acute-angled triangle, express a median in terms of its sides.
Question: In an acute-angled triangle, express a median in terms of its sides. Solution: LetΔABC be acute angled triangle where AD is its median with respect side BC. It is known that in any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side. $\therefore A B^{2}+A C^{2}=2 A D^{2}+2\left(\frac{1}{2} B C\right)^{2}$ $A B^{2}+A C^{2}=2 A D^{2}+\frac{1}{2} B C^{2}$ $A D^{2}...
Read More →Each side of a rhombus is 10 cm. If one its diagonals is 16 cm find the length of the other diagonal.
Question: Each side of a rhombus is 10 cm. If one its diagonals is 16 cm find the length of the other diagonal. Solution: We know that a quadrilateral is said to a rhombus if all sides of the quadrilateral are equal. Diagonals of a rhombus bisect each other at right angles. Quadrilateral ABCD is a rhombus and diagonals AC and BD intersect at point O. As we defined above, we get $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}, \mathrm{AO}=\mathrm{OC}, \mathrm{BO}=\mathrm{OD}$ and angle $\angle \...
Read More →Suppose you are given a circle. Give a construction to find its centre.
Question: Suppose you are given a circle. Give a construction to find its centre. Solution: Steps of Construction: (1) Take three points A, B and C on the given circle. (2) Join AB and BC. (3) Draw the perpendicular bisectors of chord AB and BC which intersect each other at O. (4) Point O will be the required centre of the circle because we know that the perpendicular bisector of the chord always passes through the centre....
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $\operatorname{cosec} x(\sec x-1)-\cot x(1-\cos x)=\tan x-\sin x$ Solution: $\mathrm{LHS}=\operatorname{cosec} x(\sec x-1)-\cot x(1-\cos x)$ $=\frac{1}{\sin x}\left(\frac{1}{\cos x}-1\right)-\frac{\cos x}{\sin x}(1-\cos x)$ $=\frac{1}{\sin x}\left(\frac{1-\cos x}{\cos x}\right)-\frac{\cos x}{\sin x}(1-\cos x)$ $=\left(\frac{1-\cos x}{\sin x}\right)\left(\frac{1}{\cos x}-\cos x\right)$ $=\left(\frac{1-\cos x}{\sin x}\right)\left(\frac{1-\cos ^{2} x}...
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Question: $\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$ is equal to A. $\frac{-1}{\sin x+\cos x}+\mathrm{C}$ B. $\log |\sin x+\cos x|+\mathrm{C}$ C. $\log |\sin x-\cos x|+\mathrm{C}$ D. $\frac{1}{(\sin x+\cos x)^{2}}$ Solution: Let $I=\frac{\cos 2 x}{(\cos x+\sin x)^{2}}$ $I=\int \frac{\cos ^{2} x-\sin ^{2} x}{(\cos x+\sin x)^{2}} d x$ $=\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)^{2}} d x$ $=\int \frac{\cos x-\sin x}{\cos +\sin x} d x$ Let $\cos x+\sin x=t \Rightarrow(\cos x-\s...
Read More →Draw different pairs of circles. How many points does each pair have in common?
Question: Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? Solution: Each pair of circles have 0, 1 or 2 points in common. The maximum number of points in common is 2....
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $(\operatorname{cosec} x-\sin x)(\sec x-\cos x)(\tan x+\cot x)=1$ Solution: $\mathrm{LHS}=(\operatorname{cosec} x-\sin x)(\sec x-\cos x)(\tan x+\cot x)$ $=\left(\frac{1}{\sin x}-\sin x\right)\left(\frac{1}{\cos x}-\cos x\right)\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)$ $=\left(\frac{1-\sin ^{2} x}{\sin x}\right)\left(\frac{1-\cos ^{2} x}{\cos x}\right)\left(\frac{\sin ^{2} x+\cos ^{2} x}{\cos x \sin x}\right)$ $=\left(\frac{\cos ^{2}...
Read More →Given an arc of a circle, complete the circle.
Question: Given an arc of a circle, complete the circle. Solution: Steps of Construction: (1) Take three points A, B and C on the given arc. (2) Join AB and BC. (3) Draw the perpendicular bisectors of chords AB and BC which intersect each other at point O. Then O will be the required centre of the required circle. (4) Join OA. (5) With centre O and radius OA, complete the circle....
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Question: $\int \frac{d x}{e^{x}+e^{-x}}$ is equal to A. $\tan ^{-1}\left(e^{x}\right)+\mathrm{C}$ B. $\tan ^{-1}\left(e^{-x}\right)+\mathrm{C}$ C. $\log \left(e^{x}-e^{-x}\right)+\mathrm{C}$ D. $\log \left(e^{x}+e^{-x}\right)+\mathbf{C}$ Solution: Let $I=\int \frac{d x}{e^{x}+e^{-x}} d x=\int \frac{e^{x}}{e^{2 x}+1} d x$ Also, let $e^{x}=t \Rightarrow e^{x} d x=d t$ $\therefore I=\int \frac{d t}{1+t^{2}}$ $=\tan ^{-1} t+\mathrm{C}$ $=\tan ^{-1}\left(e^{x}\right)+\mathrm{C}$ Hence, the correct a...
Read More →The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.
Question: The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus. Solution: We know that a quadrilateral is said to a rhombus if all sides of the quadrilateral are equal. Diagonals of a rhombus bisect each other at right angles. Quadrilateral ABCD is a rhombus and diagonals AC and BD intersect at point O. As we defined above, we get $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}, \mathrm{AO}=\mathrm{OC}, \mathrm{BO}=\mathrm{OD}$ and angle $\angle \mathrm{A...
Read More →An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.
Question: An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle. Solution: Let ABC be an equilateral triangle of side 9 cm and let AD is one of its median. Let G be the centroid ofΔABC. Then AG: GD = 2 : 1 We know that in an equilateral triangle, centroid coincides with the circum centre. Therefore, G is the centre of the circumference with circum radius GA. Also G is the centre and GD is perpendicular to BC. Therefore, In right triangle ADB, we have $A B^{...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $\sin ^{6} x+\cos ^{6} x=1-3 \sin ^{2} x \cos ^{2} x$ Solution: $\mathrm{LHS}=\sin ^{6} x+\cos ^{6} x$ $=\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}$ $=\left(\sin ^{2} x+\cos ^{2} x\right)\left[\left(\sin ^{2} x\right)^{2}+\left(\cos ^{2} x\right)^{2}-\sin ^{2} x \cos ^{2} x\right]$ $\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]$ $=1 \times\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x-\...
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Question: Evaluate $\int_{0}^{1} e^{2-3 x} d x$ as a limit of a sum. Solution: Let $I=\int_{0}^{1} e^{2-3 x} d x$ It is known that, $\int_{0}^{b} f(x) d x=(b-a) \lim _{n \rightarrow \infty} \frac{1}{n}[f(a)+f(a+h)+\ldots+f(a+(n-1) h)]$ Where, $h=\frac{b-a}{n}$ Here, $a=0, b=1$, and $f(x)=e^{2-3 x}$ $\Rightarrow h=\frac{1-0}{n}=\frac{1}{n}$ $\therefore \int_{0}^{1} e^{2-3 x} d x=(1-0) \lim _{n \rightarrow \infty} \frac{1}{n}[f(0)+f(0+h)+\ldots+f(0+(n-1) h)]$ $=\lim _{n \rightarrow \infty} \frac{1...
Read More →A line segment AB is of length 5 cm. Draw a circle of radius 4 cm passing through A and B.
Question: A line segment AB is of length 5 cm. Draw a circle of radius 4 cm passing through A and B. Can you draw a circle of radius 2 cm passing through A and B? Give reason in support of your answer. Solution: (1) Draw a line segment AB of 5cm. (2) Draw the perpendicular bisectors of AB. (3) With centre A and radius of 4cm, draw an arc which intersects the perpendicular bisector at point O. The point O will be the required centre. (4) Join OA. (5) With centre O and radius OA, draw a circle. No...
Read More →In a ∆ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that
Question: In a $\triangle A B C, A B=B C=C A=2 a$ and $A D \perp B C$. Prove that (i) $\mathrm{AD}=a \sqrt{3}$ (ii) Area $(\Delta A B C)=\sqrt{3} a^{2}$ Solution: In $\triangle A D B$ and $\triangle A D C$ $\angle B=\angle C \quad\left(60^{\circ}\right.$ each $)$ $A D=A D$ $\angle A D B=\angle A D C \quad\left(90^{\circ}\right.$ each $)$ $\triangle A D B \cong \triangle A D C \quad$ (AAS congruence theorem) $\therefore B D=D C$ $\therefore B C=2 B D$ But $B C=2 a$ therefore, we get, $2 a=2 B D$....
Read More →Prove that two different circles cannot intersect each other at more than two points.
Question: Prove that two different circles cannot intersect each other at more than two points. Solution: Suppose two circles intersect in three points A, B, C. Then A, B, C are non-collinear so a unique circle passes through these three points. This is contradiction to the face that two given circles are passing through A, B, C. Hence, two circles cannot intersect each other at more than two points....
Read More →stion Prove the following identities (1-16)
Question: stionProve the following identities (1-16) $\sec ^{4} x-\sec ^{2} x=\tan ^{4} x+\tan ^{2} x$ Solution: $\mathrm{LHS}=\sec ^{4} x-\sec ^{2} x$ $=\sec ^{2} x\left(\sec ^{2} x-1\right)$ $=\left(\tan ^{2} x+1\right)\left(\tan ^{2} x\right) \quad\left(\because \sec ^{2} x-\tan ^{2} x=1\right)$ $=\tan ^{4} x+\tan ^{2} x$ $=$ RHS Hence proved....
Read More →Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at
Question: Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle. Solution: Given: PQ is a diameter of circle which bisects the chord AB at C. To Prove: PQ bisectsAOB Proof: InAOC and BOC OA = OB [Radius of circle] OC = OC [Common] AC = BC [Given] Then ΔAOC ΔBOC [By SSS condition] AOC = BOC [C.P.C.T] Hence PQ bisects AOB....
Read More →Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the
Question: Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc. Solution: Given: C is the mid-point of chord AB. To prove: D is the mid-point of arc AB. Proof: InΔOAC and ΔOBC OA = OB [Radius of circle] OC = OC [Common] AC = BC [C is the mid-point of AB] Then ΔOAC ΔOBC [By SSS condition] AOC = BOC $\Rightarrow \mathrm{m} \overline{\mathrm{A}} \mathrm{D} \cong \mathrm{m} \overline{\mathrm{B}} \mathrm{D}$ $\Rig...
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Question: $\int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1$ Solution: Let $I=\int_{0}^{1} \sin ^{-1} x d x$ $\Rightarrow I=\int_{0}^{1} \sin ^{-1} x \cdot 1 \cdot d x$ Integrating by parts, we obtain $I=\left[\sin ^{-1} x \cdot x\right]_{0}^{1}-\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} \cdot x d x$ $=\left[x \sin ^{-1} x\right]_{0}^{1}+\frac{1}{2} \int_{0}^{1} \frac{(-2 x)}{\sqrt{1-x^{2}}} d x$ Let $1-x^{2}=t \Rightarrow-2 x d x=d t$ When $x=0, t=1$ and when $x=1, t=0$ $I=\left[x \sin ^{-1} x\right]_{...
Read More →In an isosceles triangle ABC, if AB = AC = 13
Question: In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC. Solution: We have given an isosceles triangle and we know that the altitude drawn on the unequal side of the isosceles triangle bisects that side. Therefore, in $\triangle A D B$ and $\triangle A D C$ $\angle B=\angle C \quad$ (Equal sides have equal angles opposite to them) $A D=A D$ $\angle A D B=\angle A D C \quad\left(90^{\circ}\right.$ each $)$ $\triangle A D B \cong \triangle A D C \q...
Read More →Give a method to find the centre of a given circle.
Question: Give a method to find the centre of a given circle. Solution: Steps of Construction: (1) Take three points A, B and C on the given circle. (2) Join AB and BC. (3) Draw the perpendicular bisectors of the chord AB and BC which intersect each other at O. (4) Point O will give the required circle because we know that, the Perpendicular bisectors of chord always pass through the centre....
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Question: $\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2$ Solution: Let $I=\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x$ $I=2 \int_{0}^{\frac{\pi}{4}} \tan ^{2} x \tan x d x=2 \int_{0}^{\frac{\pi}{4}}\left(\sec ^{2} x-1\right) \tan x d x$ $=2 \int_{0}^{\frac{\pi}{4}} \sec ^{2} x \tan x d x-2 \int_{0}^{\frac{\pi}{4}} \tan x d x$ $=2\left[\frac{\tan ^{2} x}{2}\right]_{0}^{\frac{\pi}{4}}+2[\log \cos x]_{0}^{\frac{\pi}{4}}$ $=1+2\left[\log \cos \frac{\pi}{4}-\log \cos 0\right]$ $=1+2\left[\log \...
Read More →ABCD is a square. F is the mid-point of AB.
Question: $\mathrm{ABCD}$ is a square. $\mathrm{F}$ is the mid-point of $\mathrm{AB} . \mathrm{BE}$ is one third of $\mathrm{BC}$. If the area of $\triangle \mathrm{FBE}=108 \mathrm{~cm}^{2}$, find the length of $\mathrm{AC}$. Solution: It is given that F is the midpoint of AB. Therefore, we have AF = FB. It is also given that $B E=\frac{1}{3} B C$...(1) Now look at the figure. Quadrilateral ABCD is a square and hence all angles are of 90. In $\triangle F B E, \angle B=90^{\circ}$ and hence it i...
Read More →Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel.
Question: Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle. Solution: Construction: DrawOP CD Chord AB = 5 cm Chord CD = 11cm Distance PQ = 3 cm Let OP = x cm And OC = OA = r cm We know that the perpendicular from centre to chord bisects it. CP = PD = 11/2 cm AndAQ = BQ = 5/2 cm In ΔOCP, by Pythagoras theorem $O C^{2}=O P^{2}+C P^{2}$ $\Rightarrow r^{2}=x^{2}+(11 / 2)^{2} \ldots(\mathrm{i})$ I...
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