For what value of x, the following matrix is singular?
Question: For what value ofx, the following matrix is singular? $\left[\begin{array}{cc}5-x x+1 \\ 2 4\end{array}\right]$ Solution: If a matrix $A$ is singular, then $|A|=0$ $\therefore\left|\begin{array}{cc}5-x x+1 \\ 2 4\end{array}\right|=0$ $\Rightarrow 4(5-x)-2(x+1) 0$ $\Rightarrow 20-4 x-2 x-2$ $\Rightarrow 18-6 x=0$ $\Rightarrow 18=6 x$ $\Rightarrow x=3$...
Read More →Factorize the following:
Question: Factorize the following:4a2+ 4ab 4ca Solution: The greatest common factor of the terms $-4 a^{2}, 4 a b$ and $-4 c a$ of the expression $-4 a^{2}+4 a b-4 c a$ is $-4 a$. Also, we can write $-4 a^{2}=-4 a \times a, 4 a b=-4 a \times(-b)$ and $4 c a=-4 a \times c$. $\therefore-4 a^{2}+4 a b-4 c a=-4 a \times a+(-4 a) \times(-b)-4 a \times c$ $=-4 a(a-b+c)$...
Read More →The wheels of the locomotive of a train are 2.1 m in radius.
Question: The wheels of the locomotive of a train are 2.1 m in radius. They make 75 revolutions in one minute. Solution: Radius of the wheel = 2.1 m Circumference of the wheel $=2 \pi \mathrm{r}$ $=2 \times \frac{22}{7} \times 2.1$ $=13.2 \mathrm{~m}$ Distance covered by the wheel in 1 revolution = 13.2 m Distance covered by the wheel in 75 revolutions $=(13.2 \times 75)=\left(990 \times \frac{1}{100}\right) \mathrm{m}$ $=\left(990 \times \frac{1}{1000}\right) \mathrm{km}$ Distance covered by th...
Read More →Factorize the following:
Question: Factorize the following:16m 4m2 Solution: The greatest common factor of the terms $16 m$ and $4 m^{2}$ of the expression $16 m-4 m^{2}$ is $4 m$. Also, we can write $16 m=4 m \times 4$ and $4 m^{2}=4 m \times m$. $\therefore 16 m-4 m^{2}=4 m \times 4-4 m \times m$ $=4 m(4-m)$...
Read More →If A is a singular matrix, then write the value of |A|.
Question: IfAis a singular matrix, then write the value of |A|. Solution: Given:Ais a singular matrix.Thus,|A|=0...
Read More →Factorize the following:
Question: Factorize the following:9x2y+ 3axy Solution: The greatest common factor of the terms $9 x^{2} y$ and 3 axy of the expression $9 x^{2} y+3 a x y$ is $3 x y$. Also, we can write $9 x^{2} y=3 x y \times 3 x$ and $3 a x y=3 x y \times a$. $\therefore 9 x^{2} y+3 a x y=3 x y \times 3 x+3 x y \times a$ $=3 x y(3 x+a)$...
Read More →Solve this
Question: If $\left|\begin{array}{lll}2-x 2+x 2+x \\ 2+x 2-x 2+x \\ 2+x 2+x 2-x\end{array}\right|=0$, then $x=$ Solution: Given: $\left|\begin{array}{lll}2-x 2+x 2+x \\ 2+x 2-x 2+x \\ 2+x 2+x 2-x\end{array}\right|=0$ $\left|\begin{array}{lll}2-x 2+x 2+x \\ 2+x 2-x 2+x \\ 2+x 2+x 2-x\end{array}\right|=0$ Applying $R_{2} \rightarrow R_{2}-R_{1}$ $\Rightarrow\left|\begin{array}{ccc}2-x 2+x 2+x \\ 2+x-2+x 2-x-2-x 2+x-2-x \\ 2+x 2+x 2-x\end{array}\right|=0$ $\Rightarrow\left|\begin{array}{ccc}2-x 2+x...
Read More →Factorize the following:
Question: Factorize the following:x4y2x2y4x4y4 Solution: The greatest common factor of the terms $x^{4} y^{2}, x^{2} y^{4}$ and $x^{4} y^{4}$ of the expression $x^{4} y^{2}-x^{2} y^{4}-x^{4} y^{4}$ is $x^{2} y^{2}$. Also, we can write $x^{4} y^{2}=x^{2} y^{2} \times x^{2}, x^{2} y^{4}=x^{2} y^{2} \times y^{2}$ and $x^{4} y^{4}=x^{2} y^{2} \times x^{2} y^{2}$. $\therefore x^{4} y^{2}-x^{2} y^{4}-x^{4} y^{4}=x^{2} y^{2} \times x^{2}-x^{2} y^{2} \times y^{2}-x^{2} y^{2} \times x^{2} y^{2}$ $=x^{2} ...
Read More →The radius of the wheel of a vehicle is 42 cm.
Question: The radius of the wheel of a vehicle is 42 cm. How many revolutions will it complete in a 19.8-km-long journey? Solution: Radius of the wheel = 42 cm Circumference of the wheel $=2 \pi \mathrm{r}$ $=2 \times \frac{22}{7} \times 42$ $=264 \mathrm{~cm}$ $=\frac{264}{100} \mathrm{~m}$ $=2.64 \mathrm{~m}$ Distance covered by the wheel in 1 revolution = 2.64 mTotal distance = 19.8 km or 19800 m $\therefore$ Number of revolutions taken by the wheel $=\frac{19800}{2.64}=7500$...
Read More →P, Q, R and S are respectively the mid-points of the sides AB,
Question: P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus, Thinking Process Firstly, use the mid-point theorem in various triangles of a quadrilateral. Further show that the line segments formed by joining the mid-points are equal, which prove the required quadrilateral. Solution: Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = B...
Read More →Factorize the following:
Question: Factorize the following:2l2mn- 3lm2n+ 4lmn2 Solution: The greatest common factor of the terms $2 l^{2} m n, 3 l m^{2} n$ and $4 l m n^{2}$ of the expression $2 l^{2} m n-3 l m^{2} n+4 l m n^{2}$ is lmn. Also, we can write $2 l^{2} m n=l m n \times 2 l, 3 l m^{2} n=l m n \times 3 m$ and $4 l m n^{2}=l m n \times 4 n .$ $\therefore 2 l^{2} n m-3 l m^{2} n+4 l m n^{2}=l m n \times 2 l-l m n \times 3 m+l m n \times 4 n$ $=\operatorname{lm} n(2 l-3 m+4 n)$...
Read More →Factorize the following:
Question: Factorize the following:a4b 3a2b2 6ab3 Solution: The greatest common factor of the terms $a^{4} b, 3 a^{2} b^{2}$ and $6 a b^{3}$ of the expression $a^{4} b-3 a^{2} b^{2}-6 a b^{3}$ is $a b$. Also, we can write $a^{4} b=a b \times a^{3}, 3 a^{2} b^{2}=a b \times 3 a b$ and $6 a b^{3}=a b \times 6 b^{2}$. $\therefore a^{4} b-3 a^{2} b^{2}-6 a b^{3}=a b \times a^{3}-a b \times 3 a b-a b \times 6 b^{2}$ $=a b\left(a^{3}-3 a b-6 b^{2}\right)$...
Read More →The area of a circle inscribed in an equilateral triangle is 154 cm2.
Question: The area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimeter of the triangle. Solution: Let the radius of the inscribed circle bercm.Given: Area of the circle $=154 \mathrm{~cm}^{2}$ We know Area of the circle $=\pi r^{2}$ $\Rightarrow 154=\frac{22}{7} r^{2}$ $\Rightarrow \frac{154 \times 7}{22}=r^{2}$ $\Rightarrow r^{2}=49$ $\Rightarrow r=7$ In a triangle, the centre of the inscribed circle is the point of intersection of the medians and altitudes of the tr...
Read More →Factorize the following:
Question: Factorize the following:28a2+ 14a2b2 21a4 Solution: The greatest common factor of the terms $28 a^{2}, 14 a^{2} b^{2}$ and $21 a^{4}$ of the expression $28 a^{2}+14 a^{2} b^{2}-21 a^{4}$ is $7 a^{2}$. Also, we can write $28 a^{2}=7 a^{2} \times 4,14 a^{2} b^{2}=7 a^{2} \times 2 b^{2}$ and $21 a^{4}=7 a^{2} \times 3 a^{2}$. $\therefore 28 a^{2}+14 a^{2} b^{2}-21 a^{4}=7 a^{2} \times 4+7 a^{2} \times 2 b^{2}-7 a^{2} \times 3 a^{2}$ $=7 a^{2}\left(4+2 b^{2}-3 a^{2}\right)$...
Read More →In a parallelogram ABCD, AB = 10 cm
Question: In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of A meets DC in E. AE and BC produced meet at F. Find the length of CF. Solution: Given, a parallelogram ABCD in which AB = 10 cm and AD = 6 cm. Now, draw a bisector of A meets DC in E and produce it to F and produce BC to meet at F. Also, produce $A D$ to $H$ and join $H F$, so that $A B F H$, is a parallelogram. Since, $H F \| A B$ Since, CFHD is a parallelogram. Therefore, opposite sides are equal. $\therefore \quad D ...
Read More →Factorize the following:
Question: Factorize the following:2a4b4 3a3b5+ 4a2b5 Solution: The greatest common factor of the terms $2 a^{4} b^{4},-3 a^{3} b^{5}$ and $4 a^{2} b^{5}$ of the expression $2 a^{4} b^{4}-3 a^{3} b^{5}+4 a^{2} b^{5}$ is $a^{2} b^{4}$. Now, $2 a^{4} b^{4}=a^{2} b^{4} \times 2 a^{2}$ $-3 a^{3} b^{5}=a^{2} b^{4} \times-3 a b$ $4 a^{2} b^{5}=a^{2} b^{4} \times 4 b$ Hence, (2a4b4-3a3b5+4a2b5) can be factorised as [a2b4(2a2 -3ab+4b)]....
Read More →The value of the determinant
Question: The value of the determinant $\Delta=\left|\begin{array}{ccc}\sin ^{2} 33^{\circ} \sin ^{2} 57^{\circ} \cos 180^{\circ} \\ -\sin ^{2} 57^{\circ} -\sin ^{2} 33^{\circ} \cos ^{2} 180^{\circ} \\ \cos 180^{\circ} \sin ^{2} 33^{\circ} \sin ^{2} 57^{\circ}\end{array}\right|$ is equal to___________ Solution: Let $\Delta=\left|\begin{array}{ccc}\sin ^{2} 33^{\circ} \sin ^{2} 57^{\circ} \cos 180^{\circ} \\ -\sin ^{2} 57^{\circ} -\sin ^{2} 33^{\circ} \cos ^{2} 180^{\circ} \\ \cos 180^{\circ} \si...
Read More →Factorize the following:
Question: Factorize the following:10m3n2+ 15m4n 20m2n3 Solution: The greatest common factor of the terms $10 m^{3} n^{2}, 15 m^{4} n$ and $-20 m^{2} n^{3}$ of the expression $10 m^{3} n^{2}+15 m^{4} n-20 m^{2} n^{3}$ is $5 m^{2} n$. Now, $10 m^{3} n^{2}=5 m^{2} n \times 2 m n$ $15 m^{4} n=5 m^{2} n \times 3 m^{2}$ $-20 m^{2} n^{3}=5 m^{2} n \times-4 n^{2}$ Hence,10m3n2+15m2n-20m2n3can be factorised as5m2n(2mn+ 3m2-4n2)....
Read More →A square is inscribed in an isosceles
Question: A square is inscribed in an isosceles right triangle, so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse. Solution: Given In isosceles triangle ABC, a square ΔDEF is inscribed. To prove CE = BE Proof In an isosceles ΔABC, A = 90 and AB=AC (i) Since, ΔDEF is a square. AD = AF [all sides of square are equal] (ii) On subtracting Eq.(ii)from Eq.(i), we get AB AD = AC- AF BD = CF .(iii)...
Read More →Factorize the following:
Question: Factorize the following:2x3y2 4x2y3+ 8xy4 Solution: The greatest common factor of the terms $2 x^{3} y^{2},-4 x^{2} y^{3}$ and $8 x y^{4}$ of the expression $2 x^{3} y^{2}-4 x^{2} y^{3}+8 x y^{4} y^{64}$ is $2 x y^{2}$. Now, $2 x^{3} y^{2}=2 x y^{2} \times x^{2}$ $-4 x^{2} y^{3}=2 x y^{2} \times-2 x y$ $8 x y^{4}=2 x y^{2} \times 4 y^{2}$ Hence, the expression 2x3y2-4x2y3+8xy4can be factorised as2xy2(x2-2xy+4y2)....
Read More →In figure, P is the mid-point of side
Question: In figure, P is the mid-point of side BC of a parallelogram ABCD such that BAP = DAP. Prove that AD = 2 CD. Thinking Process Firstly, use the property that sum of cointerior angles is 180. Secondly, use the property that sum of all angles in a triangle is 180 and then prove the required result. Solution: Given in a parallelogram $A B C D, P$ is a mid-point of $B C$ such that $\angle B A P=\angle D A P$. To prove $A D=2 C D$ Proof Since, $A B C D$ is a parallelogram. So, $A D \| B C$ an...
Read More →If a square is inscribed in a circle,
Question: If a square is inscribed in a circle, find the ratio of the areas of the circle and the square. Solution: If a square is inscribed in a circle, then the diagonals of the square are diameters of the circle.Let the diagonal of the square bedcm.Thus, we have: Radius, $r=\frac{d}{2} \mathrm{~cm}$ Area of the circle $=\pi r^{2}$ $=\pi \frac{d^{2}}{4} \mathrm{~cm}^{2}$ We know : $d=\sqrt{2} \times$ Side $\Rightarrow$ Side $=\frac{d}{\sqrt{2}} \mathrm{~cm}$ Area of the square $=(\text { Side ...
Read More →Factorize the following:
Question: Factorize the following:20x3 40x2+ 80x Solution: The greatest common factor of the terms $20 x^{3},-40 x^{2}$ and $80 x$ of the expression $20 x^{3}-40 x^{2}+80 x$ is $20 x$ Now, $20 x^{3}=20 x \times x^{2}$ $-40 x^{2}=20 x \times-2 x$ and $80 x=20 x \times 4$ Hence, the expression20x3-40x2+80xcan be factorised as20x(x2 -2x+ 4)....
Read More →Factorize the following:
Question: Factorize the following:72x6y7 96x7y6 Solution: The greatest common factor of the terms $72 x^{6} y^{7}$ and $-96 x^{7} y^{6}$ of the expression $72 x^{6} y^{7}-96 x^{7} y^{64}$ is $24 x^{6} y^{6}$. Now, $72 x^{6} y^{7}=24 x^{6} y^{6} \times 3 y$ and $-96 x^{7} y^{6}=24 x^{6} y^{6} \times-4 x$ Hence, the expression 72x6y7-96x7y6can be factorised as24x6y6(3y - 4x)....
Read More →The side of a square is 10 cm.
Question: The side of a square is 10 cm. Find (i) the area of the inscribed circle, and (ii) the area of the circumscribed circle. Solution: (i) If a circle is inscribed in a square, then the side of the square is equal to the diameter of the circle.Side of the square = 10 cmSide = Diameter = 10 Radius = 5 cm Area of the inscribed circle $=\pi r^{2}$ $=3.14 \times 5 \times 5$ $=78.5 \mathrm{~cm}^{2}$ (ii) If a circle is circumscribed in a square, then the diagonal of the square is equal to the...
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