Divide the first polynomial by the second in each of the following.
Question: Divide the first polynomial by the second in each of the following. Also, write the quotient and remainder: (i) 3x2+ 4x+ 5,x 2 (ii) 10x2 7x+ 8, 5x 3 (iii) 5y3 6y2+ 6y 1, 5y 1 (iv)x4x3+ 5x,x 1 (v)y4+y2,y2 2 Solution: $(i) \frac{3 x^{2}+4 x+5}{x-2}$ $=\frac{3 x(x-2)+10(x-2)+25}{(x-2)}$ $=\frac{(x-2)(3 x+10)+25}{(x-2)}$ $=(3 \mathrm{x}+10)+\frac{25}{(\mathrm{x}-2)}$ Therefore, quotient $=3 x+10$ and remainder $=25$. $(i i) \frac{10 x^{2}-7 x+8}{5 x-3}$ $=\frac{2 \mathrm{x}(5 \mathrm{x}-3)...
Read More →A B C D is a parallelogram and X is the mid-point
Question: $A B C D$ is a parallelogram and $X$ is the mid-point of $A B$. If $\operatorname{ar}(A X C D)=24 \mathrm{~cm}^{2}$, then $\operatorname{ar}(A B C)=24 \mathrm{~cm}^{2}$. Solution: False Given, $A B C D$ is a parallelogram and ar $(A X C D)=24 \mathrm{~cm}^{2}$ Let area of parallelogram $A B C D$ is $2 y \mathrm{~cm}^{2}$ and join $A C$. We know that, diagonal divides the area of parallelogram in two equal areas. $\therefore \quad \operatorname{ar}(\Delta A B C)=\operatorname{ar}(A C D)...
Read More →ABCD is a trapezium with parallel sides AB = a cm and DC = b cm.
Question: ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD)is (a)a: b (b)(3a + b): (a + 3b) (c)(a + 3b): (3a + b) (d)(2a +b): (3a + b) Solution: (b) Given, $A B=a \mathrm{~cm}, D C=b \mathrm{~cm}$ and $A B \| D C$. Also, $E$ and $F$ are mid-points of $A D$ and $B C$, respectively. So, distance between $C D, E F$ and $A B, E F$ will be same say $h$. Join $B D$ which intersect $E F$ at $M$. No...
Read More →The inside perimeter of a running track shown in the figure is 400 m.
Question: The inside perimeter of a running track shown in the figure is 400 m. The length of each of the straight portions is 90 m, and the ends are semicircles. If the track is 14 m wide everywhere, find the area of the track. Also, find the length of the outer boundary of the track. Solution: Lenath of the inner curved portion $(400-2 \times 90)=220 \mathrm{~m}$ $\therefore$ Length of each inner curved path $=\frac{220}{2}=110 \mathrm{~m}$ Thus, we have: $\Rightarrow \pi \mathrm{r}=110$ $\Ri...
Read More →The inside perimeter of a running track shown in the figure is 400 m.
Question: The inside perimeter of a running track shown in the figure is 400 m. The length of each of the straight portions is 90 m, and the ends are semicircles. If the track is 14 m wide everywhere, find the area of the track. Also, find the length of the outer boundary of the track. Solution: Lenath of the inner curved portion $(400-2 \times 90)=220 \mathrm{~m}$ $\therefore$ Length of each inner curved path $=\frac{220}{2}=110 \mathrm{~m}$ Thus, we have: $\Rightarrow \pi \mathrm{r}=110$ $\Ri...
Read More →What must be added to
Question: What must be added tox4+ 2x3 2x2+x 1 , so that the resulting polynomial is exactly divisible byx2+ 2x 3? Solution: Thus, $(x-2)$ should be added to $\left(x^{4}+2 x^{3}-2 x^{2}+x-1\right)$ to make the resulting polynomial exactly divisible by $\left(x^{2}+2 x-3\right)$...
Read More →Find the value of a, if x + 2 is a factor of
Question: Find the value ofa, ifx+ 2 is a factor of 4x4+ 2x3 3x2+ 8x+ 5a. Solution: We have to find the value of a if $(x+2)$ is a factor of $\left(4 x^{4}+2 x^{3}-3 x^{2}+8 x+5 a\right)$. $S$ ubstituting $\mathrm{x}=-2$ in $4 \mathrm{x}^{4}+2 \mathrm{x}^{3}-3 \mathrm{x}^{2}+8 \mathrm{x}+5 \mathrm{a}$, we get : $4(-2)^{4}+2(-2)^{3}-3(-2)^{2}+8(-2)+5 \mathrm{a}=0$ or, $64-16-12-16+5 \mathrm{a}=0$ or, $5 \mathrm{a}=-20$ or, $\mathrm{a}=-4$ $\therefore$ If $(x+2)$ is a factor of $\left(4 x^{4}+2 x^...
Read More →PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal.
Question: PQRSis a diameter of a circle of radius 6 cm. The lengthsPQ,QRandRSare equal. Semicircles are drawn withPQandQSas diameters, as shown in the given figure. IfPS= 12 cm, find the perimeter and area of the shaded region. Solution: Perimeter (circumference of the circle) $=2 \pi \mathrm{r}$ We know: Perimeter of a semicircular $\operatorname{arc}=\pi \mathrm{r}$ Now,For the arc PTS, radius is 6 cm. $\therefore$ Circumference of the semicircle PTS $=\pi \mathrm{r}=6 \pi \mathrm{cm}$ For the...
Read More →PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal.
Question: PQRSis a diameter of a circle of radius 6 cm. The lengthsPQ,QRandRSare equal. Semicircles are drawn withPQandQSas diameters, as shown in the given figure. IfPS= 12 cm, find the perimeter and area of the shaded region. Solution: Perimeter (circumference of the circle) $=2 \pi \mathrm{r}$ We know: Perimeter of a semicircular $\operatorname{arc}=\pi \mathrm{r}$ Now,For the arc PTS, radius is 6 cm. $\therefore$ Circumference of the semicircle PTS $=\pi \mathrm{r}=6 \pi \mathrm{cm}$ For the...
Read More →If a triangle and a parallelogram are on the same
Question: If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is (a)1 : 3 (b)1:2 (c)3 : 1 (d)1 : 4 Solution: (b) We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram. i.e., Area of triangle $=\frac{1}{2}$ Area of parallelogram $\Rightarrow \quad \frac{\text { Area of triangle }}{\text...
Read More →Using division of polynomials, state whether
Question: Using division of polynomials, state whether (i)x+ 6 is a factor ofx2x 42 (ii) 4x 1 is a factor of 4x2 13x 12 (iii) 2y 5 is a factor of 4y4 10y3 10y2+ 30y 15 (iv) 3y2+ 5 is a factor of 6y5+ 15y4+ 16y3+ 4y2+ 10y 35 (v)z2+ 3 is a factor ofz5 9z (vi) 2x2x+ 3 is a factor of 6x5x4+ 4x3 5x2x 15 Solution: (i) Remainder is zero. Hence (x+6) is a factor ofx2-x-42 (ii) As the remainder is non zero . Hence ( 4x-1) is not a factor of 4x2-13x-12(iii) $\because$ The remainder is non zero, $2 y-5$ is...
Read More →ABCD is a quadrilateral whose diagonal
Question: ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD (a)is a rectangle (b)is always a rhombus (c)is a parallelogram (d)need not be any of (a), (b) or (c) Solution: (d)Here, ABCD need not be any of rectangle, rhombus and parallelogram because if ABCD is a square, then its diagonal AC also divides it into two parts which are equal in area....
Read More →Two parallelograms are on equal bases
Question: Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is (a)1 : 2 (b)1 : 1 (c)2 : 1 (d)3 : 1 Solution: (b)We know that, parallelogram on the equal bases and between the same parallels are equal in area. So, ratio of their areas is 1 :1....
Read More →In the given figure, ∆ABC is right-angled at A.
Question: In the given figure, ∆ABCis right-angled atA. Find the area of the shaded region ifAB= 6 cm,BC= 10 cm andOis the centre of the incircle of ∆ABC. Solution: Using Pythagoras' theorem for triangleABC, we have: $C A^{2}+A B^{2}=B C^{2}$ $C A=\sqrt{B C^{2}-A B^{2}}$ $=\sqrt{100-36}$ $=\sqrt{64}$ $=8 \mathrm{~cm}$ Now, we must find the radius of the incircle. DrawOE,ODandOFperpendicular toAC,ABandBC,respectively. Consider quadrilateralAEOD.Here, $E O=O D$ (Both are radii.) Because the circle...
Read More →Divide the following Write down the coefficients of the terms in the quotient.
Question: Divide $15 y^{4}+16 y^{3}+\frac{10}{3} y-9 y^{2}-6$ by $3 y-2$. Write down the coefficients of the terms in the quotient. Solution: $\therefore$ Quotient $=5 y^{3}+(26 / 3) y^{2}+(25 / 9) y+(80 / 27)$ Remainder $=(-2 / 27)$ Coefficientofy3= 5 Coefficientofy2= (26/3) Coefficientofy= (25/9) Constant = (80/27)...
Read More →The mid-point of the sides of a triangle
Question: The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to (a)ar (ABC) (b)1/3 ar (ABC) (c)ar (ABC) (d)ar (ABC) Solution: (a) We know that, if $D, E$ and $F$ are respectively the mid-points of the sides $B C, C A$ and $A B$ of a $\triangle A B C$, then all four triangles has equal area $i . e .$, $\operatorname{ar}(\triangle A F E)=\operatorname{ar}(\triangle B F D)=\operatorname{ar}(\triangle E D C)=\operatorname{ar...
Read More →In the given figure, ∆ABC is right-angled at A.
Question: In the given figure, ∆ABCis right-angled atA. Find the area of the shaded region ifAB= 6 cm,BC= 10 cm andOis the centre of the incircle of ∆ABC. Solution: Using Pythagoras' theorem for triangleABC, we have: $C A^{2}+A B^{2}=B C^{2}$ $C A=\sqrt{B C^{2}-A B^{2}}$ $=\sqrt{100-36}$ $=\sqrt{64}$ $=8 \mathrm{~cm}$ Now, we must find the radius of the incircle. DrawOE,ODandOFperpendicular toAC,ABandBC,respectively. Consider quadrilateralAEOD.Here, $E O=O D$ (Both are radii.) Because the circle...
Read More →Verify the division algorithm i.e. Dividend = Divisor × Quotient + Remainder, in each of the following.
Question: Verify the division algorithm i.e. Dividend = Divisor Quotient + Remainder, in each of the following. Also, write the quotient and remainder. Solution: (i) Quotient =2x+ 3 Remainder $=-3$ Divisor $=7 x-4$ Divisor $\times$ Quotient $+$ Remainder $=(7 x-4)(2 x+3)-3$ $=14 x^{2}+21 x-8 x-12-3$ $=14 x^{2}+13 x-15$ $=$ Dividend Thus, Divisor $\times$ Quotient $+$ Remainder $=$ Dividend Hence verified. (ii) Quotient $=5 z^{2}+\frac{10}{3} z+11$ Remainder $=54$ Divisor $=3 z-6$ Divisor $\times...
Read More →In figure, if parallelogram
Question: In figure, if parallelogram $\mathrm{ABCD}$ and rectangle $\mathrm{ABEM}$ are of equal area, then (a)perimeter of ABCD = perimeter of ABEM (b)perimeter of ABCD perimeter of ABEM (c)perimeter of ABCD perimeter of ABEM (d)perimeter of ABCD = (perimeter of ABEM) Solution: (c)In rectangle ABEM, AB = EM [sides of rectangle] and in parallelogram ABCD, CD = AB On adding, both equations, we get AB + CD = EM + AB (i) We know that, the perpendicular distance between two parallel sides of a paral...
Read More →If A is a square matrix of order n × n
Question: If $A$ is a square matrix of order $n \times n$ such that $|A|=\lambda$, then write the value of $|-A|$. Solution: $|A|=\lambda \quad[$ Order of $A$ is $n]$ $\Rightarrow|-A|=(-1)^{n}|A|=(-1)^{n} \lambda$...
Read More →In the given figure, ∆ABC is right-angled at A.
Question: In the given figure, ∆ABCis right-angled atA. Find the area of the shaded region ifAB= 6 cm,BC= 10 cm andOis the centre of the incircle of ∆ABC. Solution: Using Pythagoras' theorem for triangleABC, we have:...
Read More →In the given figure, PQ = 24, PR = 7 cm and O is the centre of the circle.
Question: In the given figure,PQ= 24,PR= 7 cm andOis the centre of the circle. Find the area of the shaded region. Solution: In the rightΔRPQ, we have: $R Q=\sqrt{R P^{2}+P Q^{2}}$ $=\sqrt{7^{2}+24^{2}}$ $=\sqrt{49+576}$ $=25 \mathrm{~cm}$ OR=OQ= 12.5 cmNow, Area of the circle $=\pi r^{2}$ $=3.14 \times 12.5 \times 12.5$ $=490.625$ sq. $\mathrm{cm}$ Area of the semicircle $=\frac{490.625}{2}=245.31$ sq. $\mathrm{cm}$ Area of the triangle $=\frac{1}{2} \times b \times h=\frac{1}{2} \times 7 \time...
Read More →If the solve matrix
Question: If the matrix $\left[\begin{array}{cc}5 x 2 \\ -10 1\end{array}\right]$ is singular, find the value of $x$. Solution: A matrix is said to be singular if its determinant is zero. Since the given matrix is singular, we get $A=\left[\begin{array}{ll}5 x 2\end{array}\right.$ $\left.\begin{array}{ll}-10 1\end{array}\right]$ $\Rightarrow|A|=\mid 5 x \quad 2$ $-10 \quad 1 \mid=0$ $\Rightarrow 5 x+20=0 \quad$ [Expanding] $\Rightarrow x=-\frac{20}{5}=-4$...
Read More →Verify the division algorithm i.e. Dividend = Divisor × Quotient + Remainder, in each of the following.
Question: Verify the division algorithm i.e. Dividend = Divisor Quotient + Remainder, in each of the following. Also, write the quotient and remainder. Solution: Quotient =2x+ 3 Remainder $=-3$ Divisor $=7 x-4$ Divisor $\times$ Quotient $+$ Remainder $=(7 x-4)(2 x+3)-3$ $=14 x^{2}+21 x-8 x-12-3$ $=14 x^{2}+13 x-15$ $=$ Dividend Thus, Divisor $\times$ Quotient $+$ Remainder $=$ Dividend Hence verified. (ii) Quotient $=5 z^{2}+\frac{10}{3} z+11$ Remainder $=54$ Divisor $=3 z-6$ Divisor $\times$ Qu...
Read More →In the figure, the area of parallelogram $A B C D$ is
Question: In the figure, the area of parallelogram $A B C D$ is (a) $A B \times B M$ (b) $\mathrm{BC} \times \mathrm{BN}$ (c) $D C \times D L$ (d) $A D \times D L$ Thinking Process Use the formula, area of parallelogram =B asex Altitude to get the required result Solution: (c)We know that, area of parallelogram is the product of its any side and the corresponding altitude (or height). Here, when AB is base, then height is DL. Area of parallelogram = AB x DL and when AD is base, then height is BM...
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