From a point Q, the length of the tangent to a circle is 24 cm and
Question: From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is(a) 7 cm(b) 12 cm(c) 15 cm(d) 24.5 cm Solution: Let us first put the given data in the form of a diagram. The given data is as follows: QP= 24 cm QO= 25 cm We have to find the length ofOP, which is the radius of the circle. We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore,OPis perpendi...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: A tangentPQat a pointPof a circle of radius 5 cm meets a line through the centreOat a pointQsuch thatOQ= 12 cm. LengthPQis(a) 12 cm(b) 13 cm(c) 8.5 cm (d) $\sqrt{119} \mathrm{~cm}$ Solution: Let us first put the given data in the form of a diagram. Given data is as follows: OQ= 12 cm OP= 5 cm We have to find the length ofQP. We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. The...
Read More →Write the value
Question: Write the value of $\cos ^{2}\left(\frac{1}{2} \cos ^{-1} \frac{3}{5}\right)$. Solution: Let $y=\cos ^{-1}\left(\frac{3}{5}\right)$ $\Rightarrow \cos y=\frac{3}{5}$ Now, $\cos ^{2}\left(\frac{1}{2} \cos ^{-1} \frac{3}{5}\right)=\cos ^{2}\left(\frac{1}{2} y\right)$ $=\frac{\cos y+1}{2} \quad\left[\because \cos 2 x=2 \cos ^{2} x-1\right]$ $=\frac{\frac{3}{5}+1}{2}$ $=\frac{\frac{8}{5}}{2}$ $=\frac{4}{5}$ $\therefore \cos ^{2}\left(\frac{1}{2} \cos ^{-1} \frac{3}{5}\right)=\frac{4}{5}$...
Read More →Two concentric circles are of radii 5 cm and 3 cm.
Question: Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Solution: Let us first draw whatever is given in the problem so that we can understand the problem better. We have to find the length of AB, which is the chord of the larger circle which touches the smaller circle. Clearly, OC is the radius of the smaller circle and is touching the tangent AB. We know that the radius of the circle will always form a rig...
Read More →Find the sum of the following arithmetic progressions:
Question: Find the sum of the following arithmetic progressions: (i) 50, 46, 42, ... to 10 terms (ii) 1, 3, 5, 7, ... to 12 terms (iii) 3, 9/2, 6, 15/2, ... to 25 terms (iv) 41, 36, 31, ... to 12 terms (v)a+b,ab,a 3b, ... to 22 terms (vi) $(x-y)^{2},\left(x^{2}+y^{2}\right),(x+y)^{2}, \ldots$ to $n$ terms (vii) $\frac{x-y}{x+y}, \frac{3 x-2 y}{x+y}, \frac{5 x-3 y}{x+y}, \ldots$ to $n$ terms Solution: (i) 50, 46, 42 ... to 10 terms We have: $a=50, d=(46-50)=-4$ $n=10$ $S_{n}=\frac{n}{2}[2 a+(n-1)...
Read More →Write the value
Question: Write the value of $\cos ^{-1}\left(\cos 350^{\circ}\right)-\sin ^{-1}\left(\sin 350^{\circ}\right)$ Solution: We have $\cos ^{-1}\left(\cos 350^{\circ}\right)-\sin ^{-1}\left(\sin 350^{\circ}\right)$ $=\cos ^{-1}\left\{\cos \left(360^{\circ}-350^{\circ}\right)\right\}-\sin ^{-1}\left\{\sin \left(360^{\circ}-350^{\circ}\right)\right\}$ $\left[\because \sin \left(360^{\circ}-x\right)=-\sin x, \quad \cos \left(360^{\circ}-x\right)=\cos x\right]$ $=\cos ^{-1}\left\{\cos \left(10^{\circ}\r...
Read More →If O is a point within a quadrilateral ABCD,
Question: IfOis a point within a quadrilateralABCD, show thatOA+OB+OC+ODAC+BD. Solution: LetABCDbe a quadrilateral whose diagonals areACandBDandOis any point within the quadrilateral.JoinOwithA, B,C, andD.We know that the sum of any two sides of a triangle is greater than the third side.So, in∆AOC,OA + OC AC Also, in ∆BOD,OB+OD BD Adding these inequalities, we get: (OA+OC) + (OB + OD) (AC + BD)⇒OA + OB + OC + ODAC + BD...
Read More →In the given figure, CP and CQ are tangents from an external point C to a circle with centre O.
Question: In the given figure,CPandCQare tangents from an external pointCto a circle with centreO.ABis another tangent which touches the circle atR. IfCP= 11 cm andBR= 4 cm, find the length ofBC. [Hint:We have,CP= 11 cmCP=CQ=CQ= 11 cmNow,BR=BQ⇒BQ= 4 cmBC=CQBQ= (114)cm = 7 cm Solution: We are given the following figure From the figure, we have BC = CQ BQ (1) Let us now find out the values of CQ and BQ separately. From the property of tangents we know that length of two tangents drawn from the sam...
Read More →Write the value
Question: Write the value of $\cos \left(2 \sin ^{-1} \frac{1}{2}\right)$. Solution: We have, $\cos \left(2 \sin ^{-1} \frac{1}{2}\right)$ $=\cos \left(2 \times \frac{\pi}{6}\right)$ $=\cos \left(\frac{\pi}{3}\right)=\frac{1}{2}$...
Read More →In the given figure, ABCD is a square and ∠PQR = 90°. If PB = QC = DR, prove that
Question: In the given figure,ABCDis a square andPQR= 90. IfPB=QC=DR, prove that(i)QB=RC,(ii)PQ=QR,(iii)QPR= 45. Solution: Given: ABCDis a square andPQR= 90.Also,PB=QC=DR(i) We have: BC = CD (Sides of square)CQ = DR (Given) BC = BQ + CQ⇒ CQ = BC BQ DR= BC BQ ...(i) Also, CD = RC+ DR DR = CD RC = BC RC ...(ii)From (i) and (ii), we have:BC BQ =BC RC BQ = RC(ii) In ∆RCQ and∆QBP, we have:PB=QC (Given)BQ = RC (Proven above)RCQ =QBP (90oeach)i.e., ∆RCQ ∆QBP (SAS congruence rule) QR = PQ (By CPCT)(iii...
Read More →In the given figure, Δ ABC is circumscribing a circle. Find the length of BC.
Question: In the given figure, ΔABCis circumscribing a circle. Find the length ofBC. Solution: We are given the following figure From the figure we get, BC=BP+PC (1) Now, let us find BP and PC separately. From the property of tangents we know that when two tangents are drawn to a circle from a common external point, the length of the two tangents from the external point to the respective points of contact will be equal. Therefore we have, BR = BP It is given in the problem that BR = 3 cm. Theref...
Read More →In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC.
Question: In the given figure,ABCDis a quadrilateral in whichAB=ADandBC=DC. Prove that(i) AC bisectsAand C,(ii)BE=DE,(iii) ABC= ADC. Solution: Given:ABCDis a quadrilateral in which AB = AD and BC = DC (i)In∆ABCand∆ADC,we have:AB = AD (Given) BC = DC (Given)AC is common.i.e.,∆ABC ∆ADC (SSS congruence rule) BAC=DACandBCA=DCA (By CPCT)Thus, AC bisectsA and C.(ii)Now, in∆ABEand ∆ADE,we have: AB = AD (Given)BAE=DAE (Proven above)AE is common.∆ABE∆ADE (SAS congruence rule)⇒ BE = DE (By CPCT)(iii) ...
Read More →Write the value
Question: Write the value of $\cos ^{-1}\left(\tan \frac{3 \pi}{4}\right)$. Solution: We have $\cos ^{-1}\left(\tan \frac{3 \pi}{4}\right)=\cos ^{-1}\left\{-\tan \left(\pi-\frac{3 \pi}{4}\right)\right\} \quad[\because \tan (\pi-x)=-\tan x]$ $=\cos ^{-1}\left\{\tan \left(-\frac{\pi}{4}\right)\right\}$ $=\cos ^{-1}\left\{-\tan \left(\frac{\pi}{4}\right)\right\}$ $=\cos ^{-1}(-1)$ $=\cos ^{-1}(\cos \pi) \quad[\because \cos \pi=-1]$ $=\pi$ $\therefore \cos ^{-1}\left(\tan \frac{3 \pi}{4}\right)=\pi$...
Read More →In the given figure, CP and CQ are tangents to a circle with centre O.
Question: In the given figure,CPandCQare tangents to a circle with centreO.ARBis another tangent touching the circle atR. IfCP= 11 cm andBC= 7 cm, then find the length ofBR. Solution: CP and CQ are tangents drawn from an external point C to the circle. CP = CQ (Length of tangents drawn from an external point to the circle are equal) ⇒ CQ = 11 cm (CP = 11 cm) BQ = CQ BC = 11 cm 7 cm = 4 cm BR and BQ are tangents drawn from an external point B to the circle. BR = BQ ⇒ BR = 4 cm ( BQ = 4 cm)...
Read More →In the adjoining figure, BM ⊥ AC and DN ⊥ AC. If BM = DN,
Question: In the adjoining figure,BMACandDNAC. IfBM=DN, prove that AC bisectsBD. Solution: Given: A quadrilateral ABCD, in whichBM AC and DN AC and BM = DN.To prove: AC bisects BD; or DO = BOProof:Let AC and BD intersect at O.Now, in∆ONDand∆OMB, we have:OND = OMB (90oeach)DON =BOM (Vertically opposite angles)Also, DN = BM (Given)i.e.,∆OND ∆OMB (AAS congurence rule)OD = OB (CPCT)Hence, AC bisects BD....
Read More →In Q.No. 1, if PB = 10 cm, what is the perimeter of Δ PCD?
Question: In Q.No. 1, ifPB= 10 cm, what is the perimeter of ΔPCD? Solution: Here, we have to find the perimeter of triangle PCD. Perimeter is nothing but sum of all sides of the triangle. Therefore we have, Perimeter of= In the given figure we can see that, = Therefore, Perimeter of= We know that the two tangents drawn to a circle from a common external point will be equal in length. From this property we have, $\mathrm{CQ}=\mathrm{CA}$ $\mathrm{QD}=\mathrm{DB}$ Now let us replace CQ and QD with...
Read More →In the adjoining figure, ABCD is a square and ∆EDC is an equilateral triangle.
Question: In the adjoining figure,ABCDis a square and∆EDCis an equilateral triangle. Prove that (i)AE=BE, (ii) DAE= 15. Solution: Given: $A B C D$ is a square in which $A B=B C=C D=D A . \triangle E D C$ is an equilateral triangle in which $E D=E C=D C$ and $\angle E D C=\angle D E C=\angle D C E=60^{\circ}$. To prove: $A E=B E$ and $\angle D A E=15^{\circ}$ Proof: In $\triangle A D E$ and $\triangle B C E$, we have: $A D=B C \quad$ [Sides of a square] $\mathrm{DE}=\mathrm{EC} \quad$ [Sides of a...
Read More →What the distance between two parallel tangents to a circle of radius 5 cm?
Question: What the distance between two parallel tangents to a circle of radius 5 cm? Solution: Two parallel tangents can exist at the two ends of the diameter of the circle. Therefore, the distance between the two parallel tangents will be equal to the diameter of the circle. In the problem the radius of the circle is given as 5 cm. Therefore, Diameter = 5 2 Diameter = 10 cm Hence, the distance between the two parallel tangents is 10 cm....
Read More →Two tangents TP and TQ are drawn from an external point T to a circle with centre O
Question: Two tangents TP and TQ are drawn from an external point T to a circle with centre O as shown in Fig. 10.73. If they are inclined to each other at an angle of 100, then what is the value of POQ? Solution: Consider the quadrilateral OPTQ. It is given thatPTQ = 100. From the property of the tangent we know that the tangent will always be perpendicular to the radius at the point of contact. Therefore we have, We know that the sum of all angles of a quadrilateral will always be equal to 360...
Read More →The angles of a quadrilateral are in A.P.
Question: The angles of a quadrilateral are in A.P. whose common difference is 10. Find the angles. Solution: Let the angles be $(A)^{\circ},(A+d)^{\circ},(A+2 d)^{\circ},(A+3 d)^{\circ}$. Here,d= 10 So, $(A)^{\circ},(A+10)^{\circ},(A+20)^{\circ},(A+30)^{\circ}$ are the angles of a quadrilateral whose sum is $360^{\circ}$. $\therefore(A)^{\circ},(A+10)^{\circ},(A+20)^{\circ},(A+30)^{\circ}=360^{\circ}$ $\Rightarrow 4 A=360-60$ $\Rightarrow A=\frac{300}{4}=75^{\circ}$ The angles are as follows: $...
Read More →The length of tangent from a point A at a distance of 5 cm from the
Question: The length of tangent from a point A at a distance of 5 cm from the centre of the circle is 4 cm. What is the radius of the circle? Solution: Let us first draw whatever is given for a better understanding of the problem. LetObe the center of the circle andBbe the point of contact. We know that the radius of the circle will be perpendicular to the tangent at the point of contact. Therefore, we haveas a right triangle and we have to apply Pythagoras theorem to find the radius of the tria...
Read More →Evaluate
Question: Evaluate $\sin \left(\tan ^{-1} \frac{3}{4}\right)$. Solution: We know that $\tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}}$ $\therefore \sin \left(\tan ^{-1} \frac{3}{4}\right)=\sin \left\{\sin ^{-1}\left(\frac{\frac{3}{4}}{\sqrt{1+\frac{9}{16}}}\right)\right\}$ $=\sin \left\{\sin ^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right)\right\}$ $=\sin \left(\sin ^{-1} \frac{3}{5}\right)$ $=\frac{3}{5} \quad\left[\because \sin \left(\sin ^{-1} x\right)=x\right]$ $\therefore \sin \left(\tan...
Read More →What is the distance between two parallel tangents of a circle of radius 4 cm?
Question: What is the distance between two parallel tangents of a circle of radius 4 cm? Solution: Two parallel tangents can exist at the two ends of the diameter of the circle. Therefore, the distance between the two parallel tangents will be equal to the diameter of the circle. In the problem the radius of the circle is given as 4 cm. Therefore, Diameter = Diameter = 8 cm Hence, the distance between the two parallel tangents is 8 cm....
Read More →In the adjoining figure, ABCD is a trapezium in which AB || DC.
Question: In the adjoining figure,ABCDis a trapezium in whichAB||DC. IfA= 55 and B= 70, find Cand D. Solution: We have $A B \| D C$. $\angle \mathrm{A}$ and $\angle \mathrm{D}$ are the interior angles on the same side of transversal line $\mathrm{AD}$, whereas $\angle \mathrm{B}$ and $\angle \mathrm{C}$ are the interior angles on the same side of transversal line $\mathrm{BC}$. Now, $\angle \mathrm{A}+\angle \mathrm{D}=180^{\circ}$ $\Rightarrow \angle \mathrm{D}=180^{\circ}-\angle \mathrm{A}$ $\...
Read More →If the sum of three numbers in A.P. is 24 and their product is 440,
Question: If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers. Solution: Let the three numbers be $(a-d), a,(a+d)$. Sum $=24$ $\Rightarrow(a-d)+a+(a+d)=24$ $\Rightarrow 3 a=24$ $\Rightarrow a=8 \quad \ldots(i)$ Product $=a(a-d)(a+d)=440$ $\Rightarrow a\left(a^{2}-d^{2}\right)=440$ $\Rightarrow 8\left(64-d^{2}\right)=440$ (From (i)) $\Rightarrow\left(64-d^{2}\right)=55$ $\Rightarrow d^{2}=9$ $\Rightarrow d=\pm 3$ With $\mathrm{a}=8, \mathrm{~d}=3$, we have : $5,8,...
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