If A and B are square matrices of order 2, then det (A + B) = 0 is possible only when
Question: If $A$ and $B$ are square matrices of order 2, then $\operatorname{det}(A+B)=0$ is possible only when (a) $\operatorname{det}(A)=0$ or $\operatorname{det}(B)=0$ (b) $\operatorname{det}(A)+\operatorname{det}(B)=0$ (c) $\operatorname{det}(A)=0$ and $\operatorname{det}(B)=0$ (d) $A+B=0$ Solution: (d) $A+B=O$ Let $A=\left[a_{i j}\right]$ and $B=\left[b_{i j}\right]$ be a square matrix of order 2 . As their orders are same, $\mathrm{A}+\mathrm{B}$ is defined as $\mathrm{A}+\mathrm{B}=\left[...
Read More →Find the following products:
Question: Find the following products: (i) (x+ 4) (x+ 7) (ii) (x 11) (x+ 4) (iii) (x+ 7) (x 5) (iv) (x 3) (x 2) (v) (y2 4) (y2 3) (vi) $\left(x+\frac{4}{3}\right)\left(x+\frac{3}{4}\right)$ (vii) (3x+ 5) (3x+ 11) (viii) (2x2 3) (2x2+ 5) (ix) (z2+ 2) (z2 3) (x) (3x 4y) (2x 4y) (xi) (3x2 4xy) (3x2 3xy) (xii) $\left(x+\frac{1}{5}\right)(x+5)$ (xiii) $\left(z+\frac{3}{4}\right)\left(z+\frac{4}{3}\right)$ (xiv) (x2+ 4) (x2+ 9) (xv) (y2+ 12) (y2+ 6) (xvi) $\left(y^{2}+\frac{5}{7}\right)\left(y^{2}-\fr...
Read More →Find the following products:
Question: Find the following products: (i) (x+ 4) (x+ 7) (ii) (x 11) (x+ 4) (iii) (x+ 7) (x 5) (iv) (x 3) (x 2) (v) (y2 4) (y2 3) (vi) $\left(x+\frac{4}{3}\right)\left(x+\frac{3}{4}\right)$ (vii) (3x+ 5) (3x+ 11) (viii) (2x2 3) (2x2+ 5) (ix) (z2+ 2) (z2 3) (x) (3x 4y) (2x 4y) (xi) (3x2 4xy) (3x2 3xy) (xii) $\left(x+\frac{1}{5}\right)(x+5)$ (xiii) $\left(z+\frac{3}{4}\right)\left(z+\frac{4}{3}\right)$ (xiv) (x2+ 4) (x2+ 9) (xv) (y2+ 12) (y2+ 6) (xvi) $\left(y^{2}+\frac{5}{7}\right)\left(y^{2}-\fr...
Read More →If a, b, c are non-zero real numbers and if the system of equations
Question: If $a, b, c$ are non-zero real numbers and if the system of equations $(a-1) x=y+z$ $(b-1) y=z+x$ $(c-1) z=x+y$ has a non-trivial solution, then prove that $a b+b c+c a=a b c$. Solution: The three equations can be expressed as $(a-1) x-y-z=0$ $-x+(b-1) y-z=0$ $-x-y+(c-1) z=0$ Expressing this as a determinant, we get $\Delta=\left|\begin{array}{ccc}(a-1) -1 -1 \\ -1 (b-1) -1 \\ -1 -1 (c-1)\end{array}\right|$ If the matrix has a non-trivial solution, then $\left|\begin{array}{ccc}(a-1) -...
Read More →In ΔPQR, if ∠R = ∠P and QR = 4 cm
Question: In ΔPQR, if R = P and QR = 4 cm and PR = 5 cm. Then, the length of PQ is (a) 4 cm (b) 5 cm (c) 2 cm (d) 2.5 cm Solution: (a) Given, $\triangle P Q R$ such that $\angle R=\angle P, Q R=4 \mathrm{~cm}$ and $P R=5 \mathrm{~cm}$ In $\triangle P Q R$, $\angle R=\angle P$ $\Rightarrow \quad P Q=Q R \quad$ [sides opposite to equal anqles are equal] $\Rightarrow \quad P Q=4 \mathrm{~cm} \quad[\because Q R=4 \mathrm{~cm}]$ Hence, the length of $P Q$ is $4 \mathrm{~cm}$....
Read More →Find the area of the quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90°
Question: Find the area of the quadrilateralABCDin whichAD= 24 cm, BAD= 90 andBCDforms an equilateral triangle whose each side is equal to 26 cm. Also find the perimeter of the quadrilateral. Take$\sqrt{3}=1.73$ Solution: ∆BDCis an equilateral triangle with sidea= 26 cm. Area of $\Delta B D C=\frac{\sqrt{3}}{4} a^{2}$ $=\frac{\sqrt{3}}{4} \times 26^{2}$ $=\frac{1.73}{4} \times 676$ $=292.37 \mathrm{~cm}^{2}$ By using Pythagoras' theorem in the right-angled triangle $\triangle D A B$, we get: $A ...
Read More →In ΔABC,if BC = AB and ∠B = 80°,
Question: In ΔABC,if BC = AB and B = 80, then A is equal to (a) 80 (b) 40 (c) 50 (d) 100 Solution: (c) Given, $\triangle A B C$ such that $B C=A B$ and $\angle B=80^{\circ}$ $\ln \triangle A B C$, $A B=B C$ $\Rightarrow \quad \angle C=\angle A$ $\ldots(1)$ [angles opposite to equal sides are equal] We know that, the sum of all the angles of a triangle is $180^{\circ}$. $\therefore \quad \angle A+\angle B+\angle C=180^{\circ}$ $\Rightarrow \quad \angle A+80^{\circ}+\angle A=180^{\circ}$ $\Rightar...
Read More →Find the real values
Question: Find the real values of $\lambda$ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions $2 \lambda x-2 y+3 z=0$ $x+\lambda y+2 z=0$ $2 x+\lambda z=0$ Solution: The given system of equations can be written as $2 \lambda x-2 y+3 z=0$ $x+\lambda y+2 z=0$ $2 x+0 y+\lambda z=0$ The given system of equations will have non-trivial solution $s$ if $D=0$. $\Rightarrow\left|\begin{array}{ccc}2 \lambda -2 3 \\ 1 \lambda 2 \\ 2 0 \lambda...
Read More →In ΔABC,if AB = AC and ∠B = 50°,
Question: In ΔABC,if AB = AC and B = 50, then C is equal to (a) 40 (b) 50 (c) 80 (d)130 Solution: (b) Given, $\triangle A B C$ such that $A B=A C$ and $\angle B=50^{\circ}$. In $\triangle A B C$, $A B=A C$ [given] $\Rightarrow \quad \angle C=\angle B \quad$ [anqles opposite to equal sides are equal] $\Rightarrow \quad \angle C=50^{\circ} \quad\left[\because \angle B=50^{\circ}\right.$ (given)]...
Read More →Find the area of the quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90°
Question: Find the area of the quadrilateralABCDin whichAD= 24 cm, BAD= 90 andBCDforms an equilateral triangle whose each side is equal to 26 cm. Also find the perimeter of the quadrilateral. Take$\sqrt{3}=1.73$ Solution: ∆BDCis an equilateral triangle with sidea= 26 cm. Area of $\Delta B D C=\frac{\sqrt{3}}{4} a^{2}$ $=\frac{\sqrt{3}}{4} \times 26^{2}$ $=\frac{1.73}{4} \times 676$ $=292.37 \mathrm{~cm}^{2}$ By using Pythagoras' theorem in the right-angled triangle $\triangle D A B$, we get: $A ...
Read More →If AB = QR, BC = PR
Question: If AB = QR, BC = PR and CA = PQ, then (a) ΔABC ΔQRP (b) ΔCBA ΔPRQ (c) ΔBAC ΔRQP (d) ΔPQR ΔBCA Solution: (b) We know that, if ΔRST is congruent to ΔUVW i.e., ΔRST = ΔUVW, then sides of ΔRST fall on corresponding equal sides of ΔUVW and angles of ΔRST fall on corresponding equal angles of ΔUVW. Here, given AB = QR, BC = PR and CA = PQ, which shows that AB covers QR, BC covers PR and CA covers PQ i.e., A correspond to Q, B correspond to R and C correspond to P. or AQ,BR,CP Under this corr...
Read More →Solve each of the following system of homogeneous linear equations.
Question: Solve each of the following system of homogeneous linear equations. $3 x+y+z=0$ $x-4 y+3 z=0$ $2 x+5 y-2 z=0$ Solution: Given: $3 x+y+z=0$ $x-4 y+3 z=0$ $2 x+5 y-2 z=0$ $D=\left|\begin{array}{rrr}3 1 1 \\ 1 -4 3 \\ 2 5 -2\end{array}\right|=0$ The system has infinitely many solutions. Putting $z=k$ in the first two equations, we get $3 x+y=-k$ $x-4 y=-3 k$ Solving these equations by Cramer's rule, we get $x=\frac{D_{1}}{D}=\frac{\left|\begin{array}{rr}-k 1 \\ -3 k -4\end{array}\right|}{...
Read More →Solve each of the following system of homogeneous linear equations.
Question: Solve each of the following system of homogeneous linear equations. $3 x+y+z=0$ $x-4 y+3 z=0$ $2 x+5 y-2 z=0$ Solution: Given: $3 x+y+z=0$ $x-4 y+3 z=0$ $2 x+5 y-2 z=0$ $D=\left|\begin{array}{rrr}3 1 1 \\ 1 -4 3 \\ 2 5 -2\end{array}\right|=0$ The system has infinitely many solutions. Putting $z=k$ in the first two equations, we get $3 x+y=-k$ $x-4 y=-3 k$ Solving these equations by Cramer's rule, we get $x=\frac{D_{1}}{D}=\frac{\left|\begin{array}{rr}-k 1 \\ -3 k -4\end{array}\right|}{...
Read More →Which of the following is not a criterion
Question: Which of the following is not a criterion for congruence of triangles? (a) SAS (b) ASA (c) SSA (d) SSS Thinking Process For triangle to be congruent it is very important that the equal angles are included between the pairs of equal sides. So, SAS congruence rule holds but not ASS or SSA rule. Solution: (c) We know that, two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle. Also, criterion for congruen...
Read More →In the given figure, ∠Q > ∠R,
Question: In the given figure, Q R, PA is the bisector of QPR and PM QR.Prove that APM = (Q R). Solution: Given In $\triangle P Q R, \angle Q\angle R, P A$ is the bisector of $\angle Q P R$ and $P M \perp Q R$. To prove $\angle A P M=\frac{1}{2}(\angle Q-\angle R)$ Proof Since, $P A$ is the bisector of $\angle Q P R$. $\therefore$ $\angle Q P A=\angle A P R$ $\ldots$ (i) In $\triangle P Q M$, $\angle P Q M+\angle P M Q+\angle Q P M=180^{\circ}$ [by angle sum property of a triangle] $\Rightarrow$...
Read More →Solve each of the following system of homogeneous linear equations.
Question: Solve each of the following system of homogeneous linear equations. $2 x+3 y+4 z=0$ $x+y+z=0$ $2 x-y+3 z=0$ Solution: $D=\left|\begin{array}{rrr}2 3 4 \\ 1 1 1 \\ 2 -1 3\end{array}\right|$ $=2(3+1)-3(3-2)+4(-1-2)$ $=8-3-12$ $=-7$ So, the given system of equations has only the trivial solution i. e. $x=0, y=0, z=0$...
Read More →Show that:
Question: Show that: (i) (3x+ 7)2 84x= (3x 7)2 (ii) (9a 5b)2+ 180ab= (9a+ 5b)2 (iii) $\left(\frac{4 m}{3}-\frac{3 n}{4}\right)^{2}+2 m n=\frac{16 m^{2}}{9}+\frac{9 n^{2}}{16}$ (iv) (4pq+ 3q)2 (4pq 3q)2= 48pq2 (v) (a b)(a + b) + (b c)(b + c) + (c a)(c + a) = 0 Solution: (i) $\mathrm{LHS}=(3 x+7)^{2}-84 x$ $=(3 x+7)^{2}-4 \times 3 x \times 7$ $=(3 x-7)^{2} \quad\left[\because(a+b)^{2}-4 a b=(a-b)^{2}\right]$ $=$ RHS Because LHS is equal to RHS, the given equation is verified. (ii) LHS $=(9 a-5 b)^...
Read More →In the given figure, ABCD is a quadrilateral in which diagonal BD = 24 cm, AL ⊥ BD and CM ⊥ BD such that AL = 9 cm and CM = 12 cm.
Question: In the given figure, ABCD is a quadrilateral in which diagonal BD = 24 cm, ALBD andCM BD such that AL = 9 cm andCM = 12 cm. Calculate the area of thequadrilateral. Solution: We have, $\mathrm{BD}=24 \mathrm{~cm}, \mathrm{AL}=9 \mathrm{~cm}, \mathrm{CM}=12 \mathrm{~cm}, \mathrm{AL} \perp \mathrm{BD}$ and $\mathrm{CM} \perp \mathrm{BD}$ Area of the quadrilateral $=\operatorname{ar}(\Delta \mathrm{ABD})+\operatorname{ar}(\Delta \mathrm{BCD})$ $=\frac{1}{2} \times \mathrm{BD} \times \mathr...
Read More →Solve each of the following system of homogeneous linear equations.
Question: Solve each of the following system of homogeneous linear equations. $x+y-2 z=0$ $2 x+y-3 z=0$ $5 x+4 y-9 z=0$ Solution: Given: $x+y-2 z=0$ $2 x+y-3 z=0$ $5 x+4 y-9 z=0$ $D=\left|\begin{array}{rrr}1 1 -2 \\ 2 1 -3 \\ 5 4 -9\end{array}\right|$ $=1(-9+12)-1(-18+15)-2(8-5)$ $=0$ So, the system has infinitely many solutions. Putting $z=k$ in the first two equations, we get $x+y=2 k$ $2 x+y=3 k$ Using Cramer's rule, we get $x=\frac{D_{1}}{D}=\frac{\left|\begin{array}{ll}2 k 1 \\ 3 k 1\end{ar...
Read More →The cost of fencing a square lawn at Rs 14 per metre is Rs 28000.
Question: The cost of fencing a square lawn at Rs 14 per metre is Rs 28000. Find the cost of mowing the lawn at Rs 54 per 100 m2. Solution: Cost of fencing the lawn = Rs 28000Letlbe the length of each side of the lawn. Then, the perimeter is 4l.We know: Cost $=$ Rate $\times$ Perimeter $\Rightarrow 28000=14 \times 4 \mathrm{l}$ $\Rightarrow 28000=56 l$ $O r$ $l=\frac{28000}{56}=500 \mathrm{~m}$ Area of the square lawn $=500 \times 500=250000 \mathrm{~m}^{2}$ Cost of mowing 100 m2of the lawn = Rs...
Read More →Prove that a triangle must have atleast two acute angles.
Question: Prove that a triangle must have atleast two acute angles. Solution: Given ΔABC is a triangle. To prove ΔABC must have two acute angles Proof Let us consider the following cases Case IWhen two angles are 90. Suppose two angles are B = 90 and C = 90 We know that, the sum of all three angles is $180^{\circ}$. $\therefore \quad \angle A+\angle B+\angle C=180^{\circ}$ .....(i) $\therefore$ $\angle A+90^{\circ}+90^{\circ}=180^{\circ}$ $\Rightarrow$ $\angle A=180^{\circ}-180^{\circ}=0$ So, no...
Read More →An automobile company uses three types of steel S1, S2 and S3
Question: An automobile company uses three types of steel $S_{1}, S_{2}$ and $S_{3}$ for producing three types of cars $C_{1}, C_{2}$ and $C_{3}$. Steel requirements (in tons) for each type of cars are given below Using Cramer's rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively. Solution: Let $x, y$ and $z$ denote the number of cars that can be produced of each type. Then, $2 x+3 y+4 z=29$ $x+y+2 z=13$ $3 x+2 y+z=16$ Us...
Read More →The cost of harvesting a square field at ₹900 per hectare is ₹8100.
Question: The cost of harvesting a square field at₹900 per hectare is₹8100. Findthe cost of putting a fence around it at₹18 per metre. Solution: As, the rate of the harvesting $=$ ₹ 900 per hectare And, the cost of harvesting $=₹ 8100$ So, the area of the square field $=\frac{8100}{900}=9$ hectare $\Rightarrow$ the area $=90000 \mathrm{~m}^{2} \quad\left(\right.$ As, 1 hectare $\left.=10000 \mathrm{~m}^{2}\right)$ $\Rightarrow(\text { side })^{2}=90000$ $\Rightarrow$ side $=\sqrt{90000}$ So, sid...
Read More →Simplify:
Question: Simplify: (i) (x y)(x + y) (x2+y2)(x4+y2) (ii) (2x 1)(2x+ 1)(4x2+ 1)(16x4+ 1) (iii) (4m 8n)2+ (7m+ 8n)2 (iv) (2.5p 1.5q)2 (1.5p 2.5q)2 (v) (m2n2m)2+ 2m3n2 Solution: To simplify, we will proceed as follows: (i) $(x-y)(x+y)\left(x^{2}+y^{2}\right)\left(x^{4}+y^{4}\right)$ $=\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)\left(x^{4}+y^{4}\right) \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$ $=\left(x^{4}-y^{4}\right)\left(x^{4}+y^{4}\right) \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\r...
Read More →The area of a square field is 8 hectares.
Question: The area of a square field is 8 hectares. How long would a man take to cross it diagonally by walking at the rate of 4 km per hour? Solution: We know that 1 hectare = 10000 m2.So, Area of square field = 8 hectares = 80000 m2. $\Rightarrow \frac{1}{2}(\text { diagonal })^{2}=80000$ $\Rightarrow(\text { diagonal })^{2}=160000$ $\Rightarrow$ diagonal $=\sqrt{160000}=400 \mathrm{~m} .$ Now, distance to be travelled = 400 m speed is given to be $4 \mathrm{~km} / \mathrm{h}=4 \times \frac{10...
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