If ABCD is a parallelogram, then prove that

Question: If ABCD is a parallelogram, then prove that Ar (ΔABD)= Ar(ΔBCD)= Ar(ΔABC)= Ar(ΔACD)=(1/2) Ar (//gmABCD). Solution: Given:- ABCD is a parallelogram, To prove: - Ar (ΔABD)= Ar(ΔBCD)= Ar(ΔABC)= Ar(ΔACD)=(1/2) Ar (//gmABCD). Proof:- We know that diagonal of a parallelogram divides it into two equilaterals . Since, AC is the diagonal. Then,Ar (ΔABC)= Ar(ΔACD)=(1/2) Ar(//gmABCD) (1) Since, BD is the diagonal. Then, Ar(ΔABD)= Ar(ΔBCD)=(1/2) Ar(//gmABCD) (2) Compare equation (1) and (2) Ar(ΔAB...

A function f :

Question: A functionf: R R is defined byf(x) =x2. Determine (a) range off, (b) {x:f(x) = 4}, (c) [y:f(y) = 1]. Solution: (a) Given: f(x) =x2 Range off= R+ (Set of all real numbers greater than or equal to zero) (b) Given: f(x) =x2 ⇒x2= 4 ⇒x= 2 {x:f(x) = 4 } = {-2, 2}. (c) {y:f(y) =-1} ⇒f(y) =1 It is clear that $x^{2}=-1$ but $x^{2} \geq 0$. $\Rightarrow f(y) \neq-1$ $\therefore\{y: f(y)=-1\}=\Phi$...

Let ABCD be a parallelogram of area 124 cm2.

Question: Let $A B C D$ be a parallelogram of area $124 \mathrm{~cm}^{2}$. If $E$ and $F$ are the mid-points of sides $A B$ and $C D$ respectively, then find the area of parallelogram AEFD. Solution: Given, Area of a parallelogram $A B C D=124 \mathrm{~cm}^{2}$ Construction: Draw APDC Proof:- Area of a parallelogram AFED = DF AP (1) Andarea of parallelogram EBCF = FCAP (2) And DF = FC (3) [F is the midpoint of DC] Compare equation (1), (2)and (3) Area of parallelogram AEFD = Area of parallelogra...

A function f :

Question: A functionf: R R is defined byf(x) =x2. Determine (a) range off, (b) {x:f(x) = 4}, (c) [y:f(y) = 1]. Solution: (a) Given: f(x) =x2 Range off= R+ (Set of all real numbers greater than or equal to zero) (b) Given: f(x) =x2 ⇒x2= 4 ⇒x= 2 {x:f(x) = 4 } = {-2, 2}. (c) {y:f(y) =-1} ⇒f(y) =1 It is clear that $x^{2}=-1$ but $x^{2} \geq 0$. $\Rightarrow f(y) \neq-1$ $\therefore\{y: f(y)=-1\}=\Phi$...

In Q1, if AD = 6 cm, CF = 10 cm, and AE = 8 cm, Find AB.

Question: In Q1, if AD = 6 cm, CF = 10 cm, and AE = 8 cm, Find AB. Solution: We know that, Area of a parallelogram ABCD = ADCF (1) Again area of parallelogram ABCD = CDAE (2) Compare equation (1) and equation (2) ADCF= CDAE ⇒ 6 10 = D 8 ⇒ D = 60/8= 7.5 cm AB = DC = 7.5 cm [∵ Opposite side of a parallelogram are equal]...

If figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm, and CF = 10 cm, Find AD.

Question: If figure, ABCD is a parallelogram, AE DC and CF AD. If AB = 16 cm, AE = 8 cm, and CF = 10 cm, Find AD. Solution: Given that, In parallelogram ABCD, CD = AB = 16 cm [∵ Opposite side of a parallelogram are equal] We know that, Area of parallelogram = BaseCorresponding altitude Area of parallelogram ABCD = CDAE = ADCF 16 cm cm = AD 10 cm $\mathrm{AD}=\frac{16 \times 8}{10} \mathrm{~cm}=12.8 \mathrm{~cm}$ Thus, The length of AD is 12.8 cm....

Which of the following figures lie on the same base and between the same parallels. In such a case,

Question: Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels: Solution: (i) ΔPCD and trapezium ABCD are on the same base CD and between the same parallels AB and DC. (ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ. (iii) Parallelogram ABCD andΔPQR are between the same parallels AD and BC but they are not on the same base. (iv)ΔQQRT and parallelogram PQRS are...

Ramesh travels 760 km to his home partly by train and partly by car.

Question: Ramesh travels 760 km to his home partly by train and partly by car. He takes 8 hours if he travels 160 km. by train and the rest by car. He takes 12 minutes more if the travels 240 km by train and the rest by car. Find the speed of the train and car respectively. Solution: Let the speed of the train be x km/hour that of the car be y km/hr, we have the following cases Case I: When Ramesh travels 760 Km by train and the rest by car Time taken by Ramesh to travel $160 \mathrm{Km}$ by tra...

Show that

Question: $\frac{1}{x-x^{3}}$ Solution: $\frac{1}{x-x^{3}}=\frac{1}{x\left(1-x^{2}\right)}=\frac{1}{x(1-x)(1+x)}$ Let $\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{1+x}$ ...(1) $\Rightarrow 1=A\left(1-x^{2}\right)+B x(1+x)+C x(1-x)$ $\Rightarrow 1=A-A x^{2}+B x+B x^{2}+C x-C x^{2}$ Equating the coefficients of $x^{2}, x$, and constant term, we obtain $-A+B-C=0$ $B+C=0$ $A=1$ On solving these equations, we obtain $A=1, B=\frac{1}{2}$, and $C=-\frac{1}{2}$ From equation (1), we obtai...

Fill in the blanks to make the following statements correct:

Question: Fill in the blanks to make the following statements correct: (i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is ____________. (ii) The triangle formed by joining the mid-points of the sides of a right triangle is ____________. (iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is ____________. Solution: (i) Isosceles (ii) Right triangle (iii) Parallelogram...

If a function f :

Question: If a functionf: R R be defined by $f(x)=\left\{\begin{array}{cc}3 x-2, x0 \\ 1, x=0 \\ 4 x+1, x0\end{array}\right.$ find:f(1),f(1),f(0) andf(2). Solution: f(1) = 4 1 + 1 = 5 [By usingf(x) = 4x+ 1,x 0] $f(-1)=3 \times(-1)-2 \quad$ [By using $f(x)=3 x-2, x0]$ $=-3-2=-5$ f(0) = 1 [By usingf(x) = 1,x= 0] f(2) = 4 2 + 1 [By usingf(x) = 4x+ 1,x 0] = 9 Hence, f(1) = 5,f(-1) =-5,f(0) = 1 andf(2) = 9....

Show that, the line segments joining the mid-points of opposite sides of a quadrilateral bisects each other.

Question: Show that, the line segments joining the mid-points of opposite sides of a quadrilateral bisects each other. Solution: Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively. So, by using mid-point theorem we can say that SP ∥ BD and SP = (1/2) BD .... (i) Similarly inΔBCD QR ∥ BD and QR = (1/2) BD .... (ii) From equations (i) and (ii), we have SP ∥ QR and SP = QR As in quadrilateral SPQR, one pair of opposite sides is equal and parall...

BM and CN are perpendiculars to a line passing through the vertex A of triangle ABC.

Question: BM and CN are perpendiculars to a line passing through the vertex A of triangle ABC. If L is the mid-point of BC, prove that LM = LN. Solution: To prove LM = LN Draw LS as perpendicular to line MN. Therefore, the lines BM, LS and CN being the same perpendiculars on line MN are parallel to each other. According to intercept theorem, If there are three or more parallel lines and the intercepts made by them on a transversal are equal, then the corresponding intercepts on any other transve...

Let A = {−2, −1, 0, 1, 2} and f :

Question: Let $A=\{-2,-1,0,1,2\}$ and $t: A \rightarrow Z$ be a function defined by $f(x)=x^{2}-2 x-3$. Find: (a) range off, i.e.f(A). (b) pre-images of 6, 3 and 5. Solution: (a) Given: f(x) =x2 2x 3 f(2) = ( 2)2 2( 2) 3 = 4 + 4 3 = 8 3 = 5 f(1) = (1)2 2(1) 3 = 1+ 2 3 = 3 3 = 0 f(0) = (0)2 2(0) 3 = 0 0 3 = 3 f(1) = (1)2 2(1) 3 = 1 2 3 =1 5 = 4 f(2) = (2)2 2(2) 3 = 4 4 3 (b) Letxbe the pre-image of 6. Then, f(6) =x2 2x 3 = 6 ⇒x2 2x 9 = 0 = 4 7 = 3 Thus, range off(A) = ( 4, 3, 0, 5). $\Rightarrow ...

Let A = {−2, −1, 0, 1, 2} and f :

Question: Let $A=\{-2,-1,0,1,2\}$ and $t: A \rightarrow Z$ be a function defined by $f(x)=x^{2}-2 x-3$. Find: (a) range off, i.e.f(A). (b) pre-images of 6, 3 and 5. Solution: (a) Given: f(x) =x2 2x 3 f(2) = ( 2)2 2( 2) 3 = 4 + 4 3 = 8 3 = 5 f(1) = (1)2 2(1) 3 = 1+ 2 3 = 3 3 = 0 f(0) = (0)2 2(0) 3 = 0 0 3 = 3 f(1) = (1)2 2(1) 3 = 1 2 3 =1 5 = 4 f(2) = (2)2 2(2) 3 = 4 4 3 (b) Letxbe the pre-image of 6. Then, f(6) =x2 2x 3 = 6 ⇒x2 2x 9 = 0 = 4 7 = 3 Thus, range off(A) = ( 4, 3, 0, 5). $\Rightarrow ...

A man walks a certain distance with certain speed.

Question: A man walks a certain distance with certain speed. If he walks 1/2 km an hour faster, he takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours. Find the distance covered by the man and his original rate of walking. Solution: Let the actual speed of the train bex Km/hrand the actual time taken by y hours. Then, Distance covered= speed $\times$ dis tance $=x \times y$ $=x y \ldots(i)$ If the speed is increased by $\frac{1}{2} \mathrm{Km} / \mathrm{hr}$, then time...

Question: The value of $\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is A. 2 B. $\frac{3}{4}$ C. 0 D. $-2$ Solution: Let $I=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ ...(1) $\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right] d x$ $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$ $\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \log \left...

ABCD is a parallelogram; E and f are the mid-points of AB and CD respectively.

Question: ABCD is a parallelogram; E and f are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH. Solution: Since E and F are mid-points of AB and CD respectively AE = BE = (1/2) AB AndCF = DF = (1/2) CD But, AB = CD (1/2)AB = (1/2) CD ⟹ BE = CF Also,BE ∥ CF [ AB ∥ CD] Therefore, BEFC is a parallelogram BC ∥ EFand BE = PH .... (i) Now,BC ∥ EF ⟹ AD ∥ EF [∵ BC ∥ AD as ABCD is a parallelogram] Therefore, AEFD is a para...

In figure, ABCD and PQRC are rectangles and Q is the mid-point of AC. Prove that

Question: In figure, ABCD and PQRC are rectangles and Q is the mid-point of AC. Prove that (i) DP = PC (ii) PR = (1/2) AC Solution: (i) InΔADC, Q is the mid-point of AC such thatPQ∥AD Therefore, P is the mid-point of DC. ⟹ DP = DC [Using mid-point theorem] (ii) Similarly, R is the mid-point of BC PR = (1/2) BD PR = (1/2) AC [Diagonal of rectangle are equal, BD = AC]...

What is the fundamental difference between a relation and a function?

Question: What is the fundamental difference between a relation and a function? Is every relation a function? Solution: Differences between relation and function 1.IfRis a relation fromAtoB, then domain ofRmay be a subset ofA. But iffis a function fromAtoB, then domainfis equal toA 2. In a relation from $A$ to $B$, an element of $A$ may be related to more than one element in $B$. But in a function from $A$ to $B$, each element of $A$ must be associated to one and only one element of $B$. Thus, e...

In Figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = (1/2) AC.

Question: In Figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = (1/2) AC. If PQ produced meets BC at R, prove that R is a mid-point of BC. Solution: Join B and D. Suppose AC and BD intersect at O. Then OC = (1/2) AC Now, CQ = (1/4) AC ⇒ CQ = 1/2((1/2) AC) =(1/2) OC InΔDCO, P and Q are mid points of DC and OC respectively. PQ ∥ DO Also inΔCOB, Q is the mid-point of OC andQR ∥ OB Therefore, R is the mid-point of BC....

While covering a distance of 30 km. Ajeet takes 2 hours more than Amit.

Question: While covering a distance of 30 km. Ajeet takes 2 hours more than Amit. If Ajeet doubles his speed, he would take 1 hour less than Amit. Find their speeds of walking. Solution: Let the speed of Ajeet and Amit bexKm/hrrespectively. Then, Time taken by Ajeet to cover $30 \mathrm{Km}=\frac{30}{x} \mathrm{hrs}$ Time taken by Amit to cover $30 \mathrm{Km}=\frac{30}{y} \mathrm{hrs}$ By the given conditions, we have $\frac{30}{x}-\frac{30}{y}=2 \cdots(i)$ If Ajeet doubles his speed, then spee...

ABC is a triangle. D is a point on AB such that AD = (1/4)

Question: ABC is a triangle. D is a point on AB such that AD = (1/4) AB and E is a point on AC such that AE = (1/4) AC. Prove that DE = (1/4) BC. Solution: Let P and Q be the mid-points of AB and AC respectively. Then PQ ∥ BC PQ = (1/2) BC .... (i) InΔAPQ, D and E are the mid-points of AP and AQ respectively. (1/2) DE ∥ PQ, and DE = (1/2) PQ .... (ii) From (i) and (ii):DE = (1/2) PQ = (1/2) ((1/2) BC) DE = (1/2) BC Hence proved....

Show that

Question: The value of$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x$ is A. 0 B. 2 C. $\pi$ D. 1 Solution: It is known that if $f(x)$ is an even function, then $\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$ and iff(x) is an odd function, then$\int_{-a}^{a} f(x) d x=0$ $I=0+0+0+2 \int_{0}^{\pi} 2^{2} 1 \cdot d x$ $=2[x]_{0}^{\frac{\pi}{2}}$ $=\frac{2 \pi}{2}$ $\pi$ Hence, the correct answer is C....

Define a function as a correspondence between two sets.

Question: Define a function as a correspondence between two sets. Solution: Afunctionis a correspondence between two sets of elements, such that for each element in the first set there is only one corresponding element in the second set. The first set is called the domain and the set of all corresponding elements in the second set is called the range. LetA= {1, 2, 3} andB= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Letf:ABbe the correspondence which assigns to each element inAits square. Hence, f(1) = 12= ...