Two equal circles touch each other externally at C and AB is
Question: Two equal circles touch each other externally at C and AB is a common tangent to the circles . Then, ACB=(a) 60(b) 45(c) 30(d) 90 Solution: Let us first put the given data in the form of a diagram. Let us draw radiusOA, such that it touches the circle with centerOat point A. Let us draw radiusOBsuch that it touches the circle with centerOat pointB. Let us also draw radii from each circle to the pointC. We know that the radius is always perpendicular to the tangent at the point of conta...
Read More →Find the sum of all integers between 50 and 500 which are divisible by 7.
Question: Find the sum of all integers between 50 and 500 which are divisible by 7. Solution: The integers between 50 and 500 that are divisible by 7 are: 56, 63...497 Here, we have: $a=56$ $d=7$ $a_{n}=497$ $\Rightarrow 56+(n-1) 7=497$ $\Rightarrow 7 n-7=441$ $\Rightarrow 7 n=448$ $\Rightarrow n=64$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{64}=\frac{64}{2}[2 \times 56+(64-1) 7]$ $\Rightarrow S_{64}=32[2 \times 56+63 \times 7]=17696$...
Read More →Find the sum of all integers between 84 and 719,
Question: Find the sum of all integers between 84 and 719, which are multiples of 5. Solution: The integers between 84 and 719, which are multiples of 5 are: 85, 90...715 Here, we have: $a=85$ $d=5$ $a_{n}=715$ $\Rightarrow 85+(n-1) 5=715$ $\Rightarrow 5 n-5=630$ $\Rightarrow n=127$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{127}=\frac{127}{2}[2 \times 85+(127-1) 5]$ $\Rightarrow S_{127}=\frac{127}{2}[800]=50800$...
Read More →In the adjoining figure, M is the midpoint of side BC of a parallelogram
Question: In the adjoining figure,Mis the midpoint of sideBCof a parallelogramABCDsuch that BAM= DAM. Prove thatAD= 2CD. Solution: Given:parallelogramABCD,MisthemidpointofsideBCand BAM=DAM.Toprove:AD=2CDProof: Since, $A D \| B C$ and $A M$ is the transversal. So, $\angle D A M=\angle A M B \quad$ (Alternate interior angles) But, $\angle D A M=\angle B A M$ (Given) Therefore, $\angle A M B=\angle B A M$ $\Rightarrow A B=B M$ Now, $A B=C D \quad($ Opposite sides of a parallelogram are equal.) $\Ri...
Read More →Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.
Question: Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667. Solution: The odd integers between 1 and 1000 that are divisible by 3 are: 3, 9, 15, 21...999 Here, we have: $a=3, d=6$ $a_{n}=999$ $\Rightarrow 3+(n-1) 6=999$ $\Rightarrow 3+6 n-6=999$ $\Rightarrow 6 n=1002$ $\Rightarrow n=167$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{167}=\frac{167}{2}[2 \times 3+(167-1) 6]$ $\Rightarrow S_{167}=\frac{167}{2}[1002]=83667$ Hence proved....
Read More →Write the value
Question: Write the value of $\tan ^{-1} \frac{a}{b}-\tan ^{-1}\left(\frac{a-b}{a+b}\right)$. Solution: We know that $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$. Now, $\tan ^{-1} \frac{a}{b}-\tan ^{-1}\left(\frac{a-b}{a+b}\right)=\tan ^{-1}\left(\frac{\frac{a}{b}-\frac{a-b}{a+b}}{1+\frac{a}{b} \frac{a-b}{a+b}}\right)$ $=\tan ^{-1}\left(\frac{\frac{a^{2}+a b-a b+b^{2}}{b(a+b)}}{\frac{a b+b^{2}-a b+a^{2}}{b(a+b)}}\right)$ $=\tan ^{-1}(1)$ $=\tan ^{-1}\left(\tan \frac{\pi}{...
Read More →Find the sum of all odd numbers between 100 and 200.
Question: Find the sum of all odd numbers between 100 and 200. Solution: All the odd numbers between 100 and 200 are: 101, 103...199 Here, we have: $a=101$ $d=2$ $a_{n}=199$ $\Rightarrow 101+(n-1) \times 2=199$ $\Rightarrow 2 n-2=98$ $\Rightarrow 2 n=100$ $\Rightarrow n=50$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{50}=\frac{50}{2}[2 \times 101+(50-1) 2]$ $\Rightarrow S_{50}=25[202+98]$ $\Rightarrow S_{50}=7500$...
Read More →PQ is a tangent to a circle with centre O at the point P.
Question: PQis a tangent to a circle with centreOat the pointP. If ΔOPQis an isosceles triangle, then OQPis equal to(a) 30(b) 45(c) 60(d) 90 Solution: Let us first put the given data in the form of a diagram. We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore,OPis perpendicular toQP. Therefore, $\angle O P Q=90^{\circ}$ The side opposite toisOQ.OQwill be the longest side of the triangle. So, in the isosceles right triangleΔOPQ, OP =...
Read More →Write the value
Question: Write the value of $2 \sin ^{-1} \frac{1}{2}+\cos ^{-1}\left(-\frac{1}{2}\right)$. Solution: $2 \sin ^{-1} \frac{1}{2}+\cos ^{-1}\left(-\frac{1}{2}\right)=\sin ^{-1} 2 \times \frac{1}{2} \sqrt{1-\left(\frac{1}{2}\right)^{2}}+\cos ^{-1}\left(-\frac{1}{2}\right)$ $=\sin ^{-1} \frac{\sqrt{3}}{2}+\cos ^{-1}\left(-\frac{1}{2}\right)$ $=\sin ^{-1}\left(\sin \frac{\pi}{3}\right)+\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)$ $=\frac{\pi}{3}+\frac{2 \pi}{3}$ $=\pi$...
Read More →In the adjoining figure, ABCD is a parallelogram in which ∠DAB = 80° and ∠DBC = 60°.
Question: In the adjoining figure,ABCDis a parallelogram in whichDAB= 80 and DBC= 60. Calculate CDBand ADB. Solution: Given: ABCDis parallelogram andDAB= 80 andDBC= 60To find: Measure ofCDBand ADBIn parallelogramABCD,AD|| BCDBC= ADB= 60o(Alternate interior angles)...(i)As DABandADCare adajcent angles, DAB+ADC = 180oADC=180oDAB ⇒ADC= 180o 80o= 100oAlso, ADC =ADB+ CDBADC =100o⇒ADB+ CDB =100o ...(ii)From (i) and (ii), we get:60o+CDB =100o⇒ CDB =100o 60o= 40oHence,CDB=40oandADB=...
Read More →Find the sum of first n odd natural numbers.
Question: Find the sum of firstnodd natural numbers. Solution: The firstnodd natural numbers are: 1, 3, 5, 7, 9... a= 1,d= 2, Total terms =n $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{n}=\frac{n}{2}[2 \times 1+(n-1) 2]$ $\Rightarrow S_{n}=\frac{n}{2}[2+(n-1) 2]$ $\Rightarrow S_{n}=\frac{n}{2}[2 n]$ $\Rightarrow S_{n}=n^{2}$...
Read More →Write the value
Question: Write the value of $\tan ^{-1}\left\{\tan \left(\frac{15 \pi}{4}\right)\right\}$. Solution: We have $\tan ^{-1}\left\{\tan \left(\frac{15 \pi}{4}\right)\right\}=\tan ^{-1}\left\{\tan \left(4 \pi-\frac{\pi}{4}\right)\right\}$ $\tan ^{-1}\left\{-\tan \left(\frac{\pi}{4}\right)\right\} \quad[\because \tan (4 \pi-x)=-\tan x]$ $=\tan ^{-1}\left\{\tan \left(-\frac{\pi}{4}\right)\right\}$ $=-\frac{\pi}{4} \quad\left[\because \tan ^{-1}(\tan x)=x\right]$ $\therefore \tan ^{-1}\left\{\tan \left...
Read More →If TP and TQ are two tangents to a circle with centre O so that
Question: If TP and TQ are two tangents to a circle with centre O so that POQ = 110, then, PTQ is equal to(a) 60(b) 70(c) 80(d) 90 Solution: Let us first put the given data in the form of a diagram. It is given that, $\angle P O Q=110^{\circ}$ We have to find $\angle P T Q=110^{\circ}$ We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore, $\angle O Q T=90^{\circ}$ $\angle O P T=90^{\circ}$ Also, we know that sum of all angles of a q...
Read More →In the adjoining figure, ABCD is a parallelogram in which ∠A = 72°.
Question: In the adjoining figure,ABCDis a parallelogram in whichA= 72. Calculate B, Cand D. Solution: ABCD is parallelogram andA= 72.We know that opposite angles of a parallelogram are equal.A=CandB=DC = 72oAandB are adajcent angles.i.e., A+B = 180o⇒ B = 180oA⇒ B = 180o72o= 108oB =D=108oHence,B =D=108o andC =72o...
Read More →Find the sum of all natural numbers between 1 and 100,
Question: Find the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5. Solution: We have to find the sum of all the natural numbers that are divisible by 2 or 5 Required Sum = Sum of the natural numbers between 1 and 100 that are divisible by 2 + Sum of the natural numbers between 1 and 100 that are divisible by 5 Sum of the natural numbers between 1 and 100 that are divisible by 2 and 5, i.e by 10 $=(2+4+6+8+\ldots+98)+(5+10+15+\ldots+95)-(10+20+30+\ldots+90)$ $=\frac{5...
Read More →Write the value of
Question: Write the value of $\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right\}$. Solution: We have $\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right\}=\sin \left\{\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right\}$ $=\sin \left\{\frac{\pi}{3}+\frac{\pi}{6}\right\}$ $=\sin \frac{\pi}{2}$ $=1$ $\therefore \sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right\}=1$...
Read More →Prove that the sum of all the angles of a quadrilateral is 360°.
Question: Prove that the sum of all the angles of a quadrilateral is 360. Solution: LetABCDbe a quadrilateraland1, 2, 3 and 4 are its four angles as shown in the figure.JoinBDwhich dividesABCDin two triangles, ∆ABDand ∆BCD.In∆ABD, we have:1 + 2 +A= 180o ...(i)In∆BCD, we have:3+ 4+C= 180o ...(ii)On adding (i) and (ii), we get: (1+ 3) +A + C + (4+ 2)= 360o⇒ A+C+B+D= 360o [ ∵1+ 3 = B; 4+ 2 =D]A+C+B+D=360o...
Read More →If tangents PA and PB from a point P to a circle with centre
Question: If tangents PA and PB from a point P to a circle with centre O are inclined to each other at(a) 50(b) 60(c) 70(d) 80 Solution: Let us first put the given data in the form of a diagram. Given data is as follows: $\angle A P B=80^{\circ}$ We have to find $\angle A O B$. We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore, $\angle O A P=90^{\circ}$ $\angle O B P=90^{\circ}$ Also, we know that sum of all angles of a quadrilat...
Read More →Find the sum of first n natural numbers.
Question: Find the sum of firstnnatural numbers. Solution: The firstnnatural numbers are: 1, 2, 3, 4... a= 1,d= 1, Total terms =n $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{n}=\frac{n}{2}[2 \times 1+(n-1) 1]$ $\Rightarrow S_{n}=\frac{n}{2}[2+(n-1) 1]$ $\Rightarrow S_{n}=\frac{n}{2}[n+1]$...
Read More →Write the value
Question: Write the value of $\sin ^{-1}\left(\cos \frac{\pi}{9}\right)$. Solution: Consider, $\sin ^{-1}\left(\cos \frac{\pi}{9}\right)=\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-\frac{\pi}{9}\right)\right\} \quad\left[\because \cos x=\sin \left(\frac{\pi}{2}-x\right)\right]$ $=\sin ^{-1}\left\{\sin \left(\frac{7 \pi}{18}\right)\right\}$ $=\frac{7 \pi}{18} \quad\left[\because \sin ^{-1}(\sin x)=x\right]$ $\therefore \sin ^{-1}\left(\cos \frac{\pi}{9}\right)=\frac{7 \pi}{18}$...
Read More →In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals.
Question: In the adjoining figure,ABCDis a quadrilateral andACis one of its diagonals. Prove that:(i)AB+BC+CD+DA 2AC(ii)AB+BC+CDDA(iii)AB+BC+CD+DAAC+BD Solution: Given:ABCDis a quadrilateral andACis one of its diagonal.(i) We know that the sum of any two sides of a triangle is greater than the third side.In∆ABC,AB+BCAC...(1) In∆ACD,CD+DAAC ...(2) Adding inequalities (1) and (2), we get: AB+BC+CD+DA 2AC(ii)In∆ABC,we have :AB+BCAC ...(1)We also know that the length of each side of a triangle is g...
Read More →The length of the tangent from a point A at a circle, of radius 3 cm,
Question: The length of the tangent from a point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is(a) $\sqrt{7} \mathrm{~cm}$(b) 7 cm(c) 5 cm(d) 25 cm Solution: Let us first put the given data in the form of a diagram. Given data is as follows: OP= 3 cm AP= 4 cm We have to find the length ofOA. We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore,OPis perpendicular toAP. We can now use Pythagor...
Read More →Find the sum of the following series:
Question: Find the sum of the following series: (i) 2 + 5 + 8 + ... + 182 (ii) 101 + 99 + 97 + ... + 47 (iii) (ab)2+ (a2+b2) + (a+b)2+ ... + [(a+b)2+ 6ab] Solution: (i) 2 + 5 + 8 + ... + 182 Here, the series is an A.P. where we have the following: $a=2$ $d=(5-2)=3$ $a_{n}=182$ $\Rightarrow 2+(n-1)(3)=182$ $\Rightarrow 2+3 n-3=182$ $\Rightarrow 3 n-1=182$ $\Rightarrow 3 n=183$ $\Rightarrow n=61$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{61}=\frac{61}{2}[2 \times 2+(61-1) \times 3]$ $=\frac...
Read More →Write the value
Question: Write the value of $\cos ^{-1}(\cos 6)$. Solution: We know that $\cos ^{-1}(\cos x)=x$. Now, $\cos ^{-1}(\cos 6)=\cos ^{-1}\{\cos (2 \pi-6)\}$ $=2 \pi-6$...
Read More →if
Question: If $\tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4}$, then write the value of $x+y+x y .$ Solution: We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ Now, $\tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4}$ $\Rightarrow \tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\frac{\pi}{4}$ $\Rightarrow \frac{x+y}{1-x y}=\tan \frac{\pi}{4}$ $\Rightarrow \frac{x+y}{1-x y}=1$ $\Rightarrow x+y=1-x y$ $\Rightarrow x+y+x y=1$ $\therefore x+y+x y=1$...
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