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In the given figure, if quadrilateral PQRS circumscribes a circle, then PD + QB =

Question: In the given figure, if quadrilateralPQRScircumscribes a circle, thenPD+QB= (a)PQ(b)QR(c)PR(d)PS Solution: We know that tangents drawn to a circle from the same external point will be equal in length. Therefore, PD = PA (1) QB = QA (2) Adding equations (1) and (2), we get, PD + QB = PA + QA By looking at the figure we can say, PD + QB = PQ Therefore option (a) is the correct answer....

The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2.

Question: The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms. Solution: Given: $a_{3}=7, a_{7}-3 a_{3}=2$ We have: $a_{3}=7$ $\Rightarrow a+(3-1) d=7$ $\Rightarrow a+2 d=7$ ...(i) Also, $a_{7}-3 a_{3}=2$ $\Rightarrow a_{7}-21=2$ (Given) $\Rightarrow a+(7-1) d=23$ $\Rightarrow a+6 d=23$ ...(ii) From (i) and (ii), we get: $4 d=16$ $\Rightarrow d=4$ Putting the value in (i), we get: $a...

In the given figure, if PR is tangent to the circle at P and Q

Question: In the given figure, ifPRis tangent to the circle atPandQis the centre of the circle, then POQ= (a) 110(b) 100(c) 120(d) 90 Solution: We know that the radius is always perpendicular to the tangent at the point of contact. Therefore, we have, $\angle O P R=90^{\circ}$ It is given that, $\angle Q P R=60^{\circ}$ That is, $\angle O P R-\angle O P Q=60^{\circ}$ $90^{\circ}-\angle O P Q=60^{\circ}$ $\angle O P Q=30^{\circ}$ Now, consider $\triangle O P Q$. We have, OP=OQ(Radii of the same c...

In each of the figures given below, ABD is a rectangle.

Question: In each of the figures given below,ABDis a rectangle. Find the values ofxandyin each case. Solution: (i)ABCDis a rectangle.The diagonals of a rectangle are congruent and bisect each other. Therefore, in ∆ AOB, we have: OA = OBOAB =OBA =35ox= 90o 35o= 55oAnd AOB= 180o (35o+ 35o) = 110oy=AOB=110o [Vertically opposite angles]Hence,x=55oandy=110o(ii) In∆AOB, we have: OA=OB Now, $\angle O A B=\angle O B A=\frac{1}{2} \times\left(180^{\circ}-110^{\circ}\right)=35^{\circ}$ y=BAC=35...

In the given figure, if AP = 10 cm, then BP =

Question: In the given figure, if AP = 10 cm, then BP = (a) $\sqrt{91} \mathrm{~cm}$ (b) $\sqrt{127} \mathrm{~cm}$ (c) $\sqrt{119} \mathrm{~cm}$ (d) $\sqrt{109} \mathrm{~cm}$ Solution: Since the radius is always perpendicular to the tangent at the point of contact, $\angle O A P=90^{\circ}$ Therefore, $O P^{2}=O A^{2}+A P^{2}$ $O P^{2}=6^{2}+10^{2}$ $O P^{2}=36+100$ $O P^{2}=136$ $O P=\sqrt{136}$ Now, consider. Here also,OBis perpendicular toPBsince the radius will be perpendicular to the tangen...

Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm.

Question: Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm. Find the length of the other diagonal and hence find the area of the rhombus. Solution: Let $A B C D$ be a rhombus. $\therefore A B=B C=C D=D A=10 \mathrm{~cm}$ Let $A C$ and $B D$ be the diagonals of $A B C D$. Let $A C=x$ and $B D=16 \mathrm{~cm}$ and $\mathrm{O}$ be the intersection point of the diagonals. We know that the diagonals of a rhombus are perpendicular bisectors of each other. $\therefore \trian...

The sum of first 7 terms of an A.P. is 10 and that of next 7 terms is 17.

Question: The sum of first 7 terms of an A.P. is 10 and that of next 7 terms is 17. Find the progression. Solution: We have: $S_{7}=10$ $\Rightarrow \frac{7}{2}[2 a+(7-1) d]=10$ $\Rightarrow \frac{7}{2}[2 a+6 d]=10$ $\Rightarrow a+3 d=\frac{10}{7} \ldots(\mathrm{i})$ Also, the sum of the next seven terms $=S_{14}-S_{7}=17$ $\Rightarrow \frac{14}{2}[2 a+(14-1) d]-\frac{7}{2}[2 a+(7-1) d]=17$ $\Rightarrow 7[2 a+13 d]$ $-\frac{7}{2}[2 a+6 d]=17$ $\Rightarrow 14 a+91 d-7 a-21 d=17$ $\Rightarrow 7 a+...

At one end of a diameter PQ of a circle of radius 5 cm, tangent XPY is drawn to the circle.

Question: At one end of a diameterPQof a circle of radius 5 cm, tangentXPYis drawn to the circle. The length of chordABparallel toXYand at a distance of 8 cm fromPis(a) 5 cm(b) 6 cm(c) 7 cm(d) 8 cm Solution: Consider the figure. In right angled $\triangle \mathrm{OMB}$, By pythagoras theorem. $\mathrm{OB}^{2}=\mathrm{MB}^{2}+\mathrm{OM}^{2}$ $\Rightarrow 5^{2}=\mathrm{MB}^{2}+3^{2}$ $\Rightarrow 25-9=\mathrm{MB}^{2}$ $\Rightarrow 16=\mathrm{MB}^{2}$ $\Rightarrow 4=\mathrm{MB}$ Therefore, AM = 2M...

In the given figure, if AP = PB, then

Question: In the given figure, if AP =PB, then (a)AC=AB(b)AC=BC(c)AQ=QC(d) AB = BC Solution: The given figure is below It is given that, AP = PB We know that PB = BR(tangents from an external point are equal) From the above two equations, we can say that, AP = BR (1) Also, AP = AQ(tangents from an external point are equal) (2) From equations (1) and (2), we have, AQ = BR (3) Also, QC = CR(tangents from an external point are equal) (4) Adding equations (3) and (4), we get, AQ + QC = BR + CR By lo...

What is the principal value

Question: What is the principal value of $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right) ?$ Solution: Let $y=\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ Then, $\sin y=-\frac{\sqrt{3}}{2}=\sin \left(-\frac{\pi}{3}\right)$ $y=-\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ Here, $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is the range of the principal value branch of inverse sine function. $\therefore \sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}$...

The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively.

Question: The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively. Find the length of each side of the rhombus. Solution: LetABCDbe a rhombus.AB = BC = CD = DAHere, ACandBDare the diagonals ofABCD,whereAC= 24 cm andBD= 18 cm.Let the diagonals intersect each other at O.We know that the diagonals of a rhombus are perpendicular bisectors of each other.∆AOBis a right angle triangle in whichOA = AC/2 = 24/2 = 12 cm andOB = BD/2 = 18/2 = 9 cm.Now,AB2= OA2+ OB2 [Pythagoras theorem]⇒...

In the adjacent figure, If AB = 12 cm, BC = 8 cm and AC = 10 cm, then AD =

Question: In the adjacent figure, If AB = 12 cm, BC = 8 cm and AC = 10 cm, then AD = (a) 5 cm(b) 4 cm(c) 6 cm(d) 7 cm Solution: By looking at the given figure We can write the following equations: AD + DB= 12 We know that tangents form an external point will be of equal length. Therefore, DB = BE (1) Hence we have, AD + BE= 12 From the figure, we have BE + EC= 8 (2) Let us subtract equation (2) from equation (3). We get, AD EC= 4 (3) Since tangents from an external point will be equal, we have, ...

In each of the figures given below, ABCD is a rhombus. Find the value of x and y in each case.

Question: In each of the figures given below,ABCDis a rhombus. Find the value ofxandyin each case. Solution: ABCD is a rhombus and a rhombus is also a parallelogram. A rhombus has four equal sides. (i) $\ln \triangle A B C, \angle B A C=\angle B C A=\frac{1}{2}(180-110)=35^{\circ}$ i.e.,x= 35oNow, B+ C= 180o (Adjacent angles are supplementary) But C =x+y= 70o⇒y =70o x ⇒y= 70o 35o= 35oHence, x =35o;y =35o (ii) The diagonals of a rhombus are perpendicular bisectors of each other.So,in∆AOB, ...

How many terms are there in the A.P.

Question: How many terms are there in the A.P. whose first and fifth terms are 14 and 2 respectively and the sum of the terms is 40? Solution: We have: $a=-14$ and $S_{n}=40 \ldots$ (i) $a_{5}=2$ $\Rightarrow a+(5-1) d=2$ $\Rightarrow-14+4 d=2$ $\Rightarrow 4 d=16$ $\Rightarrow d=4$ ...(ii) Also, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow 40=\frac{n}{2}[2(-14)+(n-1) \times 4] \quad($ From $(\mathrm{i})$ and $(\mathrm{ii}))$ $\Rightarrow 80=n[-28+4 n-4]$ $\Rightarrow 80=4 n^{2}-32 n$ $\Rightar...

if

Question: If $x0, y0$ such that $x y=1$, then write the value of $\tan ^{-1} x+\tan ^{-1} y$. Solution: We know $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $x0, y0$ such that $x y=1$ Let $x=-a$ and $y=-b$ where both $a$ and $b$ are positive. $\therefore \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $=\tan ^{-1}\left(\frac{-a-a}{1-1}\right)$ $=\tan ^{-1}(-\infty)$ $=\tan ^{-1}\left\{\tan \left(-\frac{\pi}{2}\right)\right\}$ $=-\frac{\pi}{2}$...

If PT is tangent drawn from a point P to a circle touching it at T and

Question: If PT is tangent drawn from a point P to a circle touching it at T and O is the centre of the circle, then OPT + POT =(a) 30(b) 60(c) 90(d) 180 Solution: Let us first put the given data in the form of a diagram. We know that the radius will always be perpendicular to the tangent at the point of contact. Therefore, $O T \perp P T$ $\angle O T P=90^{\circ}$ Consider $\triangle O T P$. We know that sum of all angles of a triangle will be $180^{\circ}$. Therefore, $\angle O T P+\angle P O ...

ABCD is a parallelogram in which AB = 9.5 cm and its perimeter is 30 cm.

Question: ABCDis a parallelogram in whichAB= 9.5 cm and its perimeter is 30 cm. Find the length of each side of the parallelogram. Solution: ABCDis a parallelogram.The opposite sides of a parallelogram are parallel and equal.AB=DC= 9.5 cmLetBC = AD=x Perimeter ofABCD=AB + BC + CD + DA =30 cm⇒ 9.5 +x+ 9.5 +x= 30⇒ 19 + 2x= 30⇒ 2x= 11⇒x= 5.5 cmHence,AB = DC =9.5 cm andBC = DA= 5.5 cm...

Find the measure of each angle of a parallelogram, if one of its angles is 30°

Question: Find the measure of each angle of a parallelogram, if one of its angles is 30 less than twice the smallest angle. Solution: LetABCDbe a parallelogram.A= CandB= D (Opposite angles)LetAbe the smallest angle whose measure isxo. B= (2x 30)oNow, A+ B=180o (Adjacent angles are supplementary) ⇒x+ 2x 30o= 180o ⇒ 3x= 210o ⇒x= 70o B= 2 ⨯ 70o 30o= 110oHence,A=C=70o; B=D=110o...

Find the rth term of an A.P., the sum of whose first n terms is

Question: Find therth term of an A.P., the sum of whose firstnterms is 3n2+ 2n. [NCERT EXEMPLAR] Solution: Let $a$ and $d$ be the first term and the common difference of the given A.P., respectively As, $S_{n}=3 n^{2}+2 n$ So, $a=S_{1}=3 \times 1^{2}+2 \times 1=3+2=5$ and $S_{2}=3 \times 2^{2}+2 \times 2=12+4=16$ $\Rightarrow a+a_{2}=16$ $\Rightarrow a+a+d=16$ $\Rightarrow 2 a+d=16$ $\Rightarrow 2 \times 5+d=16$ $\Rightarrow d=16-10$ $\Rightarrow d=6$ Now, $a_{r}=a+(r-1) d$ $=5+(r-1) \times 6$ $...

if the

Question: If $4 \sin ^{-1} x+\cos ^{-1} x=\pi$, then what is the value of $x$ ? Solution: We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ $\therefore 4 \sin ^{-1} x+\cos ^{-1} x=\pi$ $\Rightarrow 4 \sin ^{-1} x+\frac{\pi}{2}-\sin ^{-1} x=\pi \quad\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$ $\Rightarrow 3 \sin ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \sin ^{-1} x=\frac{\pi}{6}$ $\Rightarrow x=\sin \frac{\pi}{6}$ $\Rightarrow x=\frac{1}{2}$ $\therefore x=\frac{1}{2}$...

AP and PQ are tangents drawn from a point A to a circle with

Question: AP and PQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. If OA = 15 cm, then AP + AQ =(a) 12 cm(b) 18 cm(c) 24 cm(d) 36 cm Solution: Let us first put the given data in the form of a diagram. We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore, $O P \perp A P$ $A P^{2}=O A^{2}-O P^{2}$ $A P^{2}=15^{2}-9^{2}$ $A P^{2}=225-81$ $A P^{2}=144$ $A P=\sqrt{144}$ $A P=12$ Since tangents drawn from an ...

If an angle of a parallelogram is four-fifths of its adjacent angle,

Question: If an angle of a parallelogram is four-fifths of its adjacent angle, find the angles of the parallelogram. Solution: LetABCDbe a parallelogram. $\therefore \angle A=\angle C$ and $\angle B=\angle D \quad$ (Opposite angles) Let $\angle A=x^{\circ}$ and $\angle B=\left(\frac{4 x}{5}\right)^{\circ}$ Now, $\angle A+\angle B=180^{\circ}$ (Adjacent angles are supplementary) $\Rightarrow x+\frac{4 x}{5}=180^{\circ}$ $\Rightarrow \frac{9 x}{5}=180^{\circ}$ $\Rightarrow x=100^{\circ}$ Now, $\an...

Write the value

Question: Write the value of $\sin ^{-1}\left(\frac{1}{3}\right)-\cos ^{-1}\left(-\frac{1}{3}\right)$. Solution: We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ and $\cos ^{-1}(-x)=\pi-\cos ^{-1} x$. $\therefore \sin ^{-1}\left(\frac{1}{3}\right)-\cos ^{-1}\left(-\frac{1}{3}\right)=\sin ^{-1}\left(\frac{1}{3}\right)-\left[\pi-\cos ^{-1}\left(\frac{1}{3}\right)\right]$ $=\sin ^{-1}\left(\frac{1}{3}\right)-\pi+\cos ^{-1}\left(\frac{1}{3}\right)$ $=\left[\sin ^{-1}\left(\frac{1}{3}\right)+\c...

In the given figure, AP is a tangent to the circle with centre O such that OP = 4 cm and ∠OPA = 30°. Then, AP =

Question: In the given figure, AP is a tangent to the circle with centre O such that OP = 4 cm and OPA = 30. Then,AP = (a) $2 \sqrt{2} \mathrm{~cm}$ (b) $2 \mathrm{~cm}$ (c) $2 \sqrt{3} \mathrm{~cm}$ (d) $3 \sqrt{2} \mathrm{~cm}$ Solution: In the given figure, if we joinOandA, the lineOAwill be perpendicular toAP. This is because the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore,is a right triangle. We know that, $\cos \theta=\frac{\text { Adja...

if

Question: If $\sin ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1} x=\frac{\pi}{2}$, then find $x$ Solution: We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$. We have $\sin ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \sin ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}-\cos ^{-1} x$ $\Rightarrow \sin ^{-1}\left(\frac{1}{3}\right)=\sin ^{-1} x$ $\Rightarrow x=\frac{1}{3}$ $\therefore x=\frac{1}{3}$...

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