🔁 Repeated Questions — Class 10 Maths

Solution: (a) Let $\sqrt{5}$ be a rational number such that $\sqrt{5} = \frac{p}{q}$ ($p$ and $q$ are co-prime numbers, $q \neq 0$) [1/2 mark]
$\sqrt{5}q = p \Rightarrow 5q^2 = p^2$
$5$ divides $p^2 \Rightarrow 5$ divides $p$ as well [1 mark]
$p = 5m$ (for some integer $m$)
$5q^2 = 25m^2 \Rightarrow q^2 = 5m^2$
$5$ divides $q^2 \Rightarrow 5$ divides $q$ as well [1 mark]
$p$ and $q$ have a common factor $5$ which is a contradiction as $p$ and $q$ are co-prime.
$\therefore \text{our assumption is wrong}$
Hence, $\sqrt{5}$ is an irrational number [1/2 mark]

ABCD is a parallelogram touching the circle at P, Q, R, S by sides AB, BC, CD, DA respectively.
We know that tangents drawn from the external point to a circle are equal.
$\therefore AP = AS$
quad --------(i)
$PB = BQ$
quad --------(ii)
$CR = CQ$
quad --------(iii)
$DR = DS$
quad ---------(iv)
Adding (i), (ii), (iii), (iv)
$(AP + PB) + (CR + DR) = (AS + DS) + (BQ + CQ)$
$AB + CD = AD + BC$
ABCD is a parallelogram
$\Rightarrow AB = CD, AD = BC$
$\Rightarrow 2AB = 2AD \Rightarrow AB = AD$
$\Rightarrow \text{ABCD is a rhombus.}$

Let $\sqrt{3}$ be a rational number.
$\therefore \sqrt{3} = \frac{p}{q}$, let $p \& q$ be co-primes and $q \neq 0$
$3q^2 = p^2 \Rightarrow p^2$ is divisible by $3 \Rightarrow p$ is divisible by $3$
$\Rightarrow p = 3a$, where 'a' is some integer
quad ----- (i)
$9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q^2$ is divisible by $3 \Rightarrow q$ is divisible by $3$
$\Rightarrow q = 3b$, where 'b' is some integer
quad ----- (ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{3}$ is an irrational number.

LHS = $\frac{\sin \theta - \cos \theta + 1}{\cos \theta + \sin \theta - 1}$
Dividing Numerator and Denominator by $\cos \theta$,
$\frac{\tan \theta - 1 + \sec \theta}{1 + \tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta)-(\sec^2\theta-\tan^2\theta)}{1+\tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta)-(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{1+\tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta) (1-\sec\theta+\tan\theta)}{1+\tan \theta - \sec \theta}$
$= (\tan \theta + \sec \theta)$
Multiplying & dividing by $(\sec \theta – \tan \theta)$
$= (\tan \theta + \sec \theta) \times \frac{(\sec \theta - \tan \theta)}{(\sec \theta - \tan \theta)}$
$= \frac{(\sec^2\theta-\tan^2\theta)}{\sec \theta - \tan \theta} = \frac{1}{\sec \theta - \tan \theta} = \text{RHS}$

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. (Correct Statement: $1$ mark)
Given: In $\Delta ABC, DE \parallel BC$
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$
Construction: Draw $DM \perp AC, EN \perp AB$, join $BE$ and $CD$ (Given + To prove + Construction + Figure: $1$ mark)
Proof: $\frac{ar(\Delta ADE)}{ar(\Delta DBE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \dots (i)$ [$1$ mark]
$\frac{ar(\Delta ADE)}{ar(\Delta ECD)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \dots (ii)$ [$1$ mark]
as $\Delta DBE$ and $\Delta DCE$ lie on the same base $DE$ and between same parallels $BC$ and $DE$
$\therefore ar(\Delta DBE) = ar(\Delta ECD)$ or $\frac{ar(\Delta ADE)}{ar(\Delta DBE)} = \frac{ar(\Delta ADE)}{ar(\Delta ECD)} \dots (iii)$ [$\frac{1}{2}$ mark]
From $(i), (ii)$ and $(iii)$, we get $\frac{AD}{DB} = \frac{AE}{EC}$ [$\frac{1}{2}$ mark]

Let ABCD be a quadrilateral circumscribing a circle with centre O. Let the sides AB, BC, CD, DA touch the circle at P, Q, R, S respectively.
Join OA, OB, OC, OD, OP, OQ, OR, OS.
In $\triangle AOP$ and $\triangle AOS$:
OP = OS (Radii of the same circle)
OA = OA (Common side)
AP = AS (Tangents from an external point A)
So, $\triangle AOP \cong \triangle AOS$ (SSS congruence criterion)
$\Rightarrow \angle 1 = \angle 2$ (CPCTC)
Similarly, we can prove:
$\triangle BOQ \cong \triangle BOP \Rightarrow \angle 3 = \angle 4$
$\triangle COR \cong \triangle COQ \Rightarrow \angle 5 = \angle 6$
$\triangle DOR \cong \triangle DOS \Rightarrow \angle 7 = \angle 8$
The sum of all angles around the centre O is $360^\circ$.
$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$
$2(\angle 1 + \angle 3 + \angle 5 + \angle 7) = 360^\circ$ (since $\angle 1=\angle 2, \angle 3=\angle 4, \angle 5=\angle 6, \angle 7=\angle 8$)
$\angle 1 + \angle 3 + \angle 5 + \angle 7 = 180^\circ$
Now, consider angles subtended by opposite sides at the centre:
$\angle AOB + \angle COD = (\angle 1 + \angle 4) + (\angle 5 + \angle 8)$
Since $\angle 1 = \angle 2$, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$
$\angle AOB + \angle COD = (\angle 1 + \angle 3) + (\angle 5 + \angle 7)$ (This step is incorrect in the provided solution, it should be $\angle 1+\angle 4$ and $\angle 5+\angle 8$)
Let's re-evaluate: $\angle AOB = \angle 1 + \angle 4$, $\angle COD = \angle 5 + \angle 8$
$\angle BOC = \angle 3 + \angle 6$, $\angle DOA = \angle 2 + \angle 7$
We know $\angle 1 = \angle 2$, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$.
So, $2(\angle 1 + \angle 3 + \angle 5 + \angle 7) = 360^\circ \Rightarrow \angle 1 + \angle 3 + \angle 5 + \angle 7 = 180^\circ$.
$\angle AOB + \angle COD = (\angle 1 + \angle 4) + (\angle 5 + \angle 8) = (\angle 1 + \angle 3) + (\angle 5 + \angle 7) = 180^\circ$.
Similarly, $\angle BOC + \angle DOA = (\angle 3 + \angle 6) + (\angle 2 + \angle 7) = (\angle 3 + \angle 5) + (\angle 1 + \angle 7) = 180^\circ$.
Thus, opposite sides subtend supplementary angles at the centre.

Let $\sqrt{2}$ be a rational number such that $\sqrt{2} = \frac{p}{q}$ [$\frac{1}{2}$ mark]
($p$ and $q$ are co-prime numbers, $q \neq 0$)
$\sqrt{2}q = p \Rightarrow 2q^2 = p^2$
$2$ divides $p^2 \Rightarrow 2$ divides $p$ as well [$1$ mark]
$p = 2m$ (for some integer $m$)
$2q^2 = 4m^2 \Rightarrow q^2 = 2m^2$
$2$ divides $q^2 \Rightarrow 2$ divides $q$ as well
$p$ and $q$ have a common factor $2$ which is a contradiction as $p$ and $q$ are co-prime. [$1$ mark]
$\therefore$ our assumption is wrong
Hence, $\sqrt{2}$ is an irrational number [$\frac{1}{2}$ mark]

LHS = $\frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)}$ (1 Mark)
$= \frac{\sin^2 \theta + 1 + \cos^2 \theta + 2\cos \theta}{\sin \theta (1 + \cos \theta)}$ (1 Mark)
$= \frac{2 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}$ (1/2 Mark)
$= \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta} = 2 \operatorname{cosec} \theta = \text{RHS}$ (1/2 Mark)

Let AB be the tower of height 'h'.
$\therefore$ AC = $\sqrt{3}$ h
In $\triangle ABC$, $\tan \theta = \frac{AB}{AC} = \frac{h}{\sqrt{3} h}$
$\Rightarrow \tan \theta = \frac{1}{\sqrt{3}}$
$\Rightarrow \theta = 30^\circ$

Correct figure
Let height of multi storeyed building AB be H m and CD is a tall building.
Let the distance between the two buildings be X m.
In $\Delta ABD$
$$\begin{aligned}& \tan 45^\circ = 1 = \frac{H}{X} \\ & \Rightarrow H = X \text{---------(i)} \\ & \text{In } \Delta AEC\end{aligned}$$
tan 30^
circ =
frac{1}{
sqrt{3}} =
frac{H-8}{X}$ \\ $X =
sqrt{3} (H-8) ---------(ii)
Solving equations (i) & (ii)
$X = 4 (3 + \sqrt{3})$
and $H = 4 (3 + \sqrt{3})$
The height of the multi storeyed building is $4 (3 + \sqrt{3})$ m and the distance between the two buildings is $4 (3 + \sqrt{3})$ m.

Area of sector $= \frac{\theta}{360^\circ} \pi r^2 = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times \frac{22}{7} \times 196 = 154$ cm$^2$.
Area of triangle $= \frac{1}{2} r^2 \sin \theta = \frac{1}{2} \times 14^2 \times \sin 90^\circ = \frac{1}{2} \times 196 \times 1 = 98$ cm$^2$.
Area of segment = Area of sector - Area of triangle
Area of segment $= \frac{22}{7} \times 14 \times 14 \times \frac{90}{360} - \frac{1}{2} \times 14 \times 14$
$= 154 - 98 = 56$
Hence area of segment = $56$ cm$^2$

(B) $10$

$PM = PN$
$QS = QM$
$RS = RN$ (I) (1/2 Mark)
$PM + PN = PQ + QM + PR + RN$ (II) (1/2 Mark)
$2 PM = PQ + QS + PR + RS$
$= PQ + QS + RS + PR$
$= PQ + QR + PR$ (III) (1/2 Mark)
$\therefore PM = \frac{1}{2} (PQ + QR + PR)$ (IV) (1/2 Mark)

(D) Assertion (A) is not true but Reason (R) is true.

OA = OP
$\therefore$ In $\triangle OAP$, $\angle OPA = \angle OAP$ ... (i)
$\angle OPA + \angle APR = 90^\circ$
$\Rightarrow \angle OAP + \angle APR = 90^\circ$ Using (i)
$\Rightarrow \angle QAP + \angle APR = 90^\circ$

$TP = TQ$
$\Rightarrow \angle TPQ = \angle TQP$
Let $\angle PTQ$ be $\theta$
$\Rightarrow \angle TPQ = \angle TQP = \frac{180^{\circ} - \theta}{2} = 90^{\circ} - \frac{\theta}{2}$
Now $\angle OPT = 90^{\circ}$
$\Rightarrow \angle OPQ = 90^{\circ} - (90^{\circ} - \frac{\theta}{2}) = \frac{\theta}{2}$
$\angle PTQ = 2 \angle OPQ$

(B) $4$

(C) Assertion (A) is true but Reason (R) is false

Let the fraction be $\frac{x}{y}$ [$\frac{1}{2}$ mark]
$\frac{x - 1}{y} = \frac{1}{3} \Rightarrow 3x - y = 3 \dots (i)$ [$\frac{1}{2}$ mark]
$\frac{x}{y + 8} = \frac{1}{4} \Rightarrow 4x - y = 8 \dots (ii)$ [$\frac{1}{2}$ mark]
On solving the equations $(i)$ and $(ii)$, we get $x = 5, y = 12$ [$1$ mark]
Required fraction is $\frac{5}{12}$ [$\frac{1}{2}$ mark]

(c) $\frac{1}{3}$, $-4$

Sol. (D)$\frac{2}{3}$

Here $\alpha+\beta = -\frac{11}{6}$ and $$\begin{aligned}& \alpha\beta = -\frac{10}{6} \\ & \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} \\ & = \frac{(-\frac{11}{6})^2-2\times(-\frac{10}{6})}{-\frac{10}{6}} \\ & = \frac{\frac{121}{36}+\frac{20}{6}}{-\frac{10}{6}} = \frac{\frac{121+120}{36}}{-\frac{10}{6}} = \frac{241}{36} \times -\frac{6}{10} = -\frac{241}{60}\end{aligned}$$

Sol. (c) Assertion (A) is true but Reason (R) is false

Ans. (d) $\frac{9}{25}$

(C) $\frac{3}{4}$

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

(B) $- 2$

Sol. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Ans. (c) Assertion (A) is true but Reason (R) is false

(d) $10$

In $\triangle ABC$, $AB = AC$ (Given)
$\therefore \angle ACB = \angle ABC$ ----- (1)
In $\triangle ABD$ and $\triangle ECF$
$\angle ADB = \angle EFC$ (each $90^\circ$)
$\angle ABD = \angle ACD$ (from (1))
$\therefore \triangle ABD \sim \triangle ECF$ (AA rule)

Given $\triangle ABE \cong \triangle ACD$
$\therefore AE = AD$ or $AD = AE$ ---- (1) ($1/2$)
and $AB = AC$ ---- (2) ($1/2$)
Dividing (1) by (2), we have
$\frac{AD}{AB} = \frac{AE}{AC}$ ($1/2$)
and $\angle DAE = \angle BAC$
$\therefore \triangle ADE \sim \triangle ABC$ ($1/2$)

Given $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$
Since AD and PM are medians, $BC = 2BD$ and $QR = 2QM$
So, $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM} \Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$
Therefore, $\triangle ABD \sim \triangle PQM$ (SSS similarity criterion)
$\Rightarrow \angle B = \angle Q$ (Corresponding angles of similar triangles)
Now, in $\triangle ABC$ and $\triangle PQR$
$\frac{AB}{PQ} = \frac{BC}{QR}$ (Given)
$\angle B = \angle Q$ (Proved above)
Therefore, $\triangle ABC \sim \triangle PQR$ (SAS similarity criterion)

(i) $\triangle ABC \sim \triangle PQR$
$\angle A=\angle P$
and $\frac{AB}{PQ} = \frac{AC}{PR}$
$\Rightarrow \frac{2AD}{2PS} = \frac{AC}{PR}$
$\Rightarrow \frac{AD}{PS} = \frac{AC}{PR}$ and $\angle A = \angle P$
Therefore $\triangle ADC \sim \triangle PSR$
(ii)Hence $\frac{AD}{PS} = \frac{AC}{PR}$
$\Rightarrow AD \times PR = AC \times PS$

Sol. $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} = \frac{5(1/2)^2 + 4(2/\sqrt{3})^2 - (1)^2}{1}$
$= \frac{5/4 + 16/3-1}{1} = \frac{67}{12}$

(C) $1$

LHS $= \frac{\sqrt{\sec A-1}}{\sqrt{\sec A+1}} + \frac{\sqrt{\sec A+1}}{\sqrt{\sec A-1}}$
$= \frac{(\sec A-1) + (\sec A+1)}{\sqrt{(\sec A+1)(\sec A-1)}}$
$= \frac{2 \sec A}{\sqrt{\sec^2 A-1}}$
$= \frac{2 \sec A}{\sqrt{\tan^2 A}}$
$= \frac{2 \sec A}{\tan A}$
$= \frac{2/\cos A}{\sin A/\cos A}$
$= \frac{2}{\sin A}$
$= 2 \operatorname{cosec} A = \text{RHS}$

LHS = $$\begin{aligned}& 2(\sin^6\theta + \cos^6\theta) - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = 2[(\sin^2\theta)^3 + (\cos^2\theta)^3] - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = 2[(\sin^2\theta + \cos^2\theta)(\sin^4\theta - \sin^2\theta \cos^2\theta + \cos^4\theta)] - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = 2[\sin^4\theta + \cos^4\theta - \sin^2\theta \cos^2\theta] - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = -[\sin^4\theta + \cos^4\theta + 2 \sin^2\theta \cos^2\theta] + 1 \\ & = -(\sin^2\theta + \cos^2\theta)^2 + 1 \\ & = -1 + 1 = 0\end{aligned}$$

Sol. LHS = $\sec A (1 – \sin A) (\sec A + \tan A)$
$= \frac{1}{\cos A} (1 – \sin A) (\frac{1}{\cos A} + \frac{\sin A}{\cos A})$
$= \frac{1}{\cos A} (1 – \sin A) (\frac{1 + \sin A}{\cos A})$
$= \frac{1 - \sin^2 A}{\cos^2 A} = \frac{\cos^2 A}{\cos^2 A} = 1=RHS$

LHS $= \frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$
$= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$
$= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$
$= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$
$= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
Using $a^3 - b^3 = (a-b)(a^2+ab+b^2)$:
$= \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
$= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \operatorname{cosec} \theta \sec \theta + 1$
$= 1 + \sec \theta \operatorname{cosec} \theta = \text{RHS}$

LHS $= \frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{(\sin A - \cos A)(\sin A + \cos A)} = \frac{\sin^2 A + \cos^2 A + 2 \sin A \cos A + \sin^2 A + \cos^2 A - 2 \sin A \cos A}{\sin^2 A - \cos^2 A} = \frac{1 + 1}{\sin^2 A - (1 - \sin^2 A)} = \frac{2}{2 \sin^2 A - 1} = \text{RHS}$.

LHS $$\begin{aligned}& = \frac{\tan A(1 - \sec A) - \tan A(1 + \sec A)}{1 - \sec^2 A} \\ & = \frac{-2 \tan A \sec A}{-\tan^2 A} \\ & = 2 \times \frac{\cos A}{\sin A} \times \frac{1}{\cos A} \\ & = 2 \operatorname{cosec} A = \text{RHS}\end{aligned}$$