332354 + Interview Questions in Secondary School in Math

1.	If the edge of a cube is increased by 100%, the surface area of the cube is increased by how much percent?
Answer» ANSWER: The SURFACE area of the CUBE is increased by 300%. Step-by-step explanation: GIVEN : If the edge of a cube is increased by 100%. To find : The surface area of the cube is increased by how much percent? Solution : Let edge of the cube be 's'. The surface area of the original cube is $S_o=6s^2$ If the edge of a cube is increased by 100%. i.e. The increased edge is $e=s+s=2s$ The surface area of the increased edge cube is $S_n=6(2s)^2$ $S_n=6(4s^2)$ $S_n=24s^2$ The increased percentage of surface area of the cube is $P\%=\frac{S_n-S_o}{S_o}\times 100$ $P\%=\frac{24s^2-6s^2}{6s^2}\times 100$ $P\%=\frac{18s^2}{6s^2}\times 100$ $P\%=3\times 100$ $P\%=300\%$ Therefore, The surface area of the cube is increased by 300%.

1.

If the edge of a cube is increased by 100%, the surface area of the cube is increased by how much percent?

Answer»

ANSWER:

The SURFACE area of the CUBE is increased by 300%.

Step-by-step explanation:

GIVEN : If the edge of a cube is increased by 100%.

To find : The surface area of the cube is increased by how much percent?

Solution :

Let edge of the cube be 's'.

The surface area of the original cube is $S_o=6s^2$

If the edge of a cube is increased by 100%.

i.e. The increased edge is $e=s+s=2s$

The surface area of the increased edge cube is $S_n=6(2s)^2$

$S_n=6(4s^2)$

$S_n=24s^2$

The increased percentage of surface area of the cube is

$P\%=\frac{S_n-S_o}{S_o}\times 100$

$P\%=\frac{24s^2-6s^2}{6s^2}\times 100$

$P\%=\frac{18s^2}{6s^2}\times 100$

$P\%=3\times 100$

$P\%=300\%$

Therefore, The surface area of the cube is increased by 300%.

Discussion

2.	A circular flower bed has a diameter of 100 cm .a metal boundary costing rs 2per meter placed around it. findlength of metal boundary that will be required to fence the flower bed twice.
Answer» HIII!!! here's ur ANSWER... given the diameter of the circular flower bed = 100CM = 1m therefore radius of the flower bed = 1/2 = 0.5m circumference of the circular flower bed = 2πr = 2 × 22/7 × 0.5 = 22/7m ATQ, the cost of fencing is RS w at the rate per m. hence, total cost of fencing = 22/7 × 2 = 44/7 = rs 6.28 (approx.) hope this helps..!!

Discussion

3.	Eka mansa kade 16 cows astat .tya cows tyanchya kramanka nusar 5 letter milk 8 number vali 8 litter milk dete..tya mansala 4 mul astat .tyala tya mulana same litter denari cows dyaychya astat tar to pratekala kiti cows deil
Answer» असा प्रश्न अगदी सोपा असतो, ह्याचे उत्तर शोधताना थोडा विचार करावा,ह्याचे उत्तर असे कि, १ ते १६ च्या आकड्यांची बेरीज १३६ अशी येते, तर ते चार जण मध्ये वाटायचे आहे. मग 136 ला 4 ने भागले तर 34 असे उत्तर येते. तर आता प्रत्येकाला 34 लिटर दुध भेटले पाहिजे. 1 ते 16 च्या अनुक्रमांमध्ये सुरुवातीच्या दोन संख्या आणि शेवटच्या दोन संख्या अशी बेरीज करीत गेल्यास आपल्याला समान गट भेटतील. उत्तर: १) १+२+१६+१५=३४ २) ३+४+१४+१३=३४ ३) ५+६+१२+११=३४ ४) ७+८+१०+९=३४

Discussion

4.	Express 36 as the sum of twin prime
Answer» HEY friend here is your ANSWER 17 and 19 are TWIN prime . 17 + 19 = 36. hope it HELPS you.

Discussion

5.	The cost of a machine is ₹ 2,50,000 . It depreciates at the rate of 4% per annum. Find the cost of the machine after three years.
Answer» $\huge \underline \mathfrak {Solution:-}$ Cost of MACHINE (P) = 250000 rupees It will be depreciated for two YEARS at 10% per year. Time = 3 years Rate of depreciation = 4 % Amount $= P {(1 - \frac{Rate}{100}) }^{Time} \\ \\ = 250000 {( 1-\frac{4}{100}) }^{3} \\ \\ = 250000 {( 1- 0.04) }^{3} \\ \\ = 250000 \times {0.96}^{3} \\ \\ =221184$ After 3 years it's cost will be 221184 rupees. Depreciation = (250000-221184) rupees Depreciation = 28816 rupees

5.

The cost of a machine is ₹ 2,50,000 . It depreciates at the rate of 4% per annum. Find the cost of the machine after three years.

Answer»

$\huge \underline \mathfrak {Solution:-}$

Cost of MACHINE (P) = 250000 rupees

It will be depreciated for two YEARS at 10% per year.

Time = 3 years

Rate of depreciation = 4 %

Amount

$= P {(1 - \frac{Rate}{100}) }^{Time} \\ \\ = 250000 {( 1-\frac{4}{100}) }^{3} \\ \\ = 250000 {( 1- 0.04) }^{3} \\ \\ = 250000 \times {0.96}^{3} \\ \\ =221184$

After 3 years it's cost will be 221184 rupees.

Depreciation = (250000-221184) rupees

Depreciation = 28816 rupees

Discussion

6.	Hello,solve this: 355+4888 , first will be the brainlist
Answer» HEY... here is your ANSWER 355+4888=5243

Discussion

7.	Is it ok please tell any thing left
Answer» YES.....its OK✌✌✌ well... LOOKING GOOD ENOUGH...

Discussion

8.	The length of a chord of a circle of 16.8 cm, radius is 9.1 cm. Find its distance from the centre.
Answer» Your QUESTION ANSWER is in the PHOTO

Discussion

9.	s t and u can complete a work in 40 48 and 60 days respectively. they received rs.10800 to complete the work.they begin the work together but t left 2 days before the completion of work and u left 5 days before the completion of work.s has completed the ramaining work alone. what is the share of s from the total money?
Answer» Total work=lcm of 40,48,60=240 S 1 DAYS work=240/40=6 T 1 DAY work=240/48=5 U 1 days work=240/60=4 (S+T+U) 1 day work=6+5+4=15 let X days be the time to complete the work working days of S =x working days of T=x-2 days working days of U=x-5 days now 6x+5(x-2)+4(x-5)=240 6x+5x-10+4x-20=240 15x-30=240 15x=270 x=270/15=18 days

Discussion

10.	एक हेक्टेयर कितने एकड के बराबर होती है..?
Answer» HEY here is your ANSWER : ) 10,000 वर्ग मीटर के बराबर क्षेत्र की एक इकाई. 2.471 एकड़ जमीन के बराबर

Discussion

11.	IF COSEC THETA = 2X AND COT THETA =2/X ,FIND VALUE OF 2(X^2-1/X^2)
Answer» Given: cosec theta =2x and COT there=2/x cosec theta = 2x. cot theta =2/x squaring both sides. squaring both side cosec² theta =4x². cot² theta =4/x² cosec² theta - cot² theta =1 4x²-4/xx=1 4(x²-1/x²) =1 2(x²-1/x²) =1/2 hence PROVED PLS MARK it as BRAINLIST

Discussion

12.	Prove that 3 minus 2 root 5 is irrational number
Answer» Heya! Here is yr answer...... => 3-2√5 LET US assume 3-2√5 is rational => 3-2√5 = a/b (a, b are any INTEGERS) => 2√5 = 3-a/b => 2√5 = 3b-a/b => √5 = 3b-a/2b For any two integers RHS (3b-a/2b) is rational but, LHS (√5) is irrational! A rational and irrational are never EQUAL. So, our ASSUMPTION is false Hence, 3-2√5 is irrational Hope it helps.... :)

Discussion

13.	Exact method
Answer» I don't no(hc+by+f)=AX

Discussion

14.	Find the mean of first n natural number
Answer» Hey dear Your answer is..... We know that 〰〰〰〰〰〰〰 MEAN = sum of the observations _______________________ total NUMBER of observations.. The first TEN natural numbers are: 1⃣, 2⃣, 3⃣, 4⃣, 5⃣, 6⃣, 7⃣, 8⃣, 9⃣, 1⃣0⃣ Then we have to add the total ten numbers to find its sum. Let's add: 1+2+3+4+5+6+7+8+9+10 =55 = 55 ------ 10 = 5.5 The mean of the first 10 natural numbers is 5.5.. Hope this helps. Thank you.

Discussion

15.	Three digit numbers such that its digit at tens place is one more than the digit at hundreds place the sum of the digits is 13 the number obtained by reversing the digits can be obtained by multiplying the original number with two and subtracting 49 from this product find the original number
Answer» so the ORIGINAL NUMBER is 346. .

Discussion

16.	root 2 X square + 7 x + 5 root 2 is equal to zero solve the following quadratic equation and find the root by factorization
Answer» $\textbf{Given:}$ $\sqrt{2}x^2+7x+5\sqrt{2}=0$ $\textbf{To find:}$ $\text{Roots of the given quadratic equation}$ $\textbf{Solution:}$ $\text{Consider,}$ $\sqrt{2}x^2+7x+5\sqrt{2}=0$ $\sqrt{2}x^2+2x+5x+5\sqrt{2}=0$ $\sqrt{2}x(x+\sqrt{2})+5(x+\sqrt{2})=0$ $(\sqrt{2}x+5)(x+\sqrt{2})=0$ $\sqrt{2}x+5=0\;\text{(or)}\;x+\sqrt{2}=0$ $\implies\,x=\dfrac{-5}{\sqrt{2}}\;\text{(or)}\;x=-\sqrt{2}$ $\therefore\textbf{Roots the required roots are $\bf\dfrac{-5}{\sqrt{2}},\,-\sqrt{2}$}$

16.

root 2 X square + 7 x + 5 root 2 is equal to zero solve the following quadratic equation and find the root by factorization

Answer»

$\textbf{Given:}$

$\sqrt{2}x^2+7x+5\sqrt{2}=0$

$\textbf{To find:}$

$\text{Roots of the given quadratic equation}$

$\textbf{Solution:}$

$\text{Consider,}$

$\sqrt{2}x^2+7x+5\sqrt{2}=0$

$\sqrt{2}x^2+2x+5x+5\sqrt{2}=0$

$\sqrt{2}x(x+\sqrt{2})+5(x+\sqrt{2})=0$

$(\sqrt{2}x+5)(x+\sqrt{2})=0$

$\sqrt{2}x+5=0\;\text{(or)}\;x+\sqrt{2}=0$

$\implies\,x=\dfrac{-5}{\sqrt{2}}\;\text{(or)}\;x=-\sqrt{2}$

$\therefore\textbf{Roots the required roots are $\bf\dfrac{-5}{\sqrt{2}},\,-\sqrt{2}$}$

Discussion

17.	Find the largest number which when divided by 825 and 975 leaves remainder 5 in each case
Answer» The numbers that are exactly divisible are 825 - 5 = 800 and 975 - 5 = 970 Prime factorization of 800 = 2×2×2×2×2×5×5 Prime factorization of 970 =2×5×97 HCF(800, 970) = 2×5=10 Therefore the largest number which divides 825 and 975 LEAVING REMAINDER 5 in each CASE is 10

Discussion

18.	ABC is an isosceles triangle in which altitudes BD and CE are drawn to equal sides AC and AB respectively (see figure) Show that these altitudes are equal.
Answer» Here we USE ASA congruenceASA(angle side angle):Two TRIANGLES are congruent if two angles and the included side of One triangle are equal to two angles & the included side of the other triangle. GIVEN: ΔABC is an isosceles∆ with AB = AC, BD and CE are altitudes. To prove: BD = CE Proof: In ΔAEC and ΔADB ∠A = ∠A. (Common) ∠AEC = ∠ADB (each 90°) AB = AC (Given) Therefore, ΔAEC ≅ ΔADB[by AAS congruence rule] Thus, BD = CE (by CPCT) Hence , altitudes BD & CE are equal. HOPE THIS ANSWER WILL HELP YOU...

18.

ABC is an isosceles triangle in which altitudes BD and CE are drawn to equal sides AC and AB respectively (see figure) Show that these altitudes are equal.

Answer»

Here we USE ASA congruenceASA(angle side angle):Two TRIANGLES are congruent if two angles and the included side of One triangle are equal to two angles & the included side of the other triangle.

GIVEN:

ΔABC is an isosceles∆ with AB = AC, BD and CE are altitudes.

To prove:

BD = CE

Proof:

In ΔAEC and ΔADB

∠A = ∠A. (Common)

∠AEC = ∠ADB (each 90°)

AB = AC (Given)

Therefore, ΔAEC ≅ ΔADB[by AAS congruence rule]

Thus, BD = CE (by CPCT)

Hence , altitudes BD & CE are equal.

HOPE THIS ANSWER WILL HELP YOU...

Discussion

19.	3) If sec(theta+36)=cosec theta, find theta is equals to
Answer» $\sec( \theta + 36 \degree) = \cosec( \theta) \\ \\ => \sec( \theta + 36 \degree) = \sec(90 - \theta ) \\ \\ = > \theta + 36 = 90 - \theta \\ \\ = > 2 \theta = 54 \\ \\ = > \theta = 27 \degree$

Discussion

20.	Factorise 2x square + Y square + 8 z square - 2√2 xy + 4√2 Y Z - 8 x z
Answer» This can be written in the FORM of (-√2x)^2 + (y)^2 + (2√2z)^2 + 2(-√2)(1)xy + 2(1)(2√2)yz + 2(2√2)(-√2)xz This is in the form of a^2+b^2+c^2 +2ab + 2bc + 2ca which is EQUAL to (a+b+c)^2 Here a= -√2x b=y c=2√2 So it is simplified as Ans. (-√2x+y+2√2z)^2

Discussion

21.	If length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its perimeter.
Answer» Length of Diagonal 2= 16CM SIDE of rhombus= 17cm Perimeter of rhombus=68cm Hope it HELPS...!!!

Discussion

22.	Sin square 15 plus sin square 75/cos square 36 plus cos square 54
Answer» $\frac{ \sin^{2} (15) + \sin^{2} (75) }{ \cos ^{2} (36) + \cos ^{2} (54) } \\ \\ \\ => \frac{ \sin ^{2} (90 - 75) + \sin^{2} (75) }{ \cos ^{2} (90 - 54) + \cos^{2} (54) } \\ \\ \\ \\ => \frac{ \cos^{2} (75) + \sin^{2} (75) }{ \sin^{2} (54) + \cos^{2} (54) } \\ \\ \\ \\ \mathbf{ \textcolor{blue}{we \: \: know \: \: \: ( \: \: \: \: \: \sin ^{2} ( \theta) \: + \: \cos ^{2} ( \theta ) = 1 \: \: \: \: \: )}} \\ \\ \\ \\ \\ hence \\ \\ \rightarrow \frac{1}{1} \\ \\ \rightarrow1$

Discussion

23.	a particle of mass 100g is thrown vertically upwards with a speed of 5m/s.the work done by the force of gravity during the time particle goes up is?
Answer» Work DONE =f•s m=100g v=5m/s force=mass×ACCELERATION 0.1×(-G)=-1N since acceleration is CONSTANT so ${v}^{2} - {u}^{2} = 2as$ -25=2(-g)s s=1.25 now work done=-1N×1.25m -1.25 joules is answer.

Discussion

24.	find the area of a trapezium whose parallel sides are 20 cm and 18 cm long and the distance between them is 15 cm.
Answer» 19 × 15 = 285 cm SQUARE will be the ANSWER

Discussion

25.	two places aurangabad and ellora are 120 km apart.two cyclist leave Ellora for aurangabad such that second leaves 4 hours after the first.both cyclist arrive at aurangabad at the same time.find the time taken by the slower cyclist on the trip if the speed of one cyclist is 5km/h more than the other.
Answer» You can SOLVE this my RELATIVE VELOCITY

Discussion

26.	the angle are (5x),(3x+10)degree,(6x-20)degree and (x+25)degree .Find the value of x and the measures of each angle of the quadrilateral.
Answer» 360= 5x+3x+10+6x-20+X+25 X=23

Discussion

27.	Cube root of 216000 by prime factorization method
Answer» 2 \| 216000 2 \| 108000 2 \| 54000 2 \| 27000 2 \| 13500 2 \| 6750 3 \| 3375 3 \| 1125 3 \| 375 5 \| 125 5 \| 25 5 \| 5 216000 = 2^6 × 3^3 × 5^3 Cube ROOT ( 216000) = 2^2 × 3 × 5 = 4 × 3 × 5 = 60

Discussion

28.	in a poultry farm, every 100 birds provide an average of 72 egg. how many birds will produce 324 eggs?
Answer» 72 EGGS = 100 BIRDS Find 1 egg: 1 egg = 100/72 = 25/18 Find 324 eggs: 324 eggs = 25/18 X 324 = 450 birds Answer: 450 birds will produce 324 eggs

28.

in a poultry farm, every 100 birds provide an average of 72 egg. how many birds will produce 324 eggs?

Answer»

72 EGGS = 100 BIRDS

Find 1 egg:

1 egg = 100/72 = 25/18

Find 324 eggs:

324 eggs = 25/18 X 324 = 450 birds

Answer: 450 birds will produce 324 eggs

Discussion

29.	A carrom board has a perimeter of 320 cm how much is its area
Answer» A CARROM board is square Perimeter of square= SIDE4 320= s 4 s=80 Area=ss Area=8080 6400 square CM

Discussion

30.	One of the angles of a triangle is 80 degrees and the other two angles are equal. Find the measure of each of the equal angles.
Answer» LET the two ANGLES be EQUAL to x. Since they are equal we assume both of them to be x. So according to angle sum property of triangle, the sum of all the INTERIOR angles of a triangle sum up to 180°. => Angle 1 + Angle 2 + Angle 3 = 180° => x + x + 80° = 180° => 2x + 80° = 180° => 2x = 180° - 80° => 2x = 100° => x = 100° / 2 => x = 50° Hence the angle measure of the two equal angles are 50°.

30.

One of the angles of a triangle is 80 degrees and the other two angles are equal. Find the measure of each of the equal angles.

Answer»

LET the two ANGLES be EQUAL to x. Since they are equal we assume both of them to be x.

So according to angle sum property of triangle, the sum of all the INTERIOR angles of a triangle sum up to 180°.

=> Angle 1 + Angle 2 + Angle 3 = 180°

=> x + x + 80° = 180°

=> 2x + 80° = 180°

=> 2x = 180° - 80°

=> 2x = 100°

=> x = 100° / 2

=> x = 50°

Hence the angle measure of the two equal angles are 50°.

Discussion

31.	two angles of a quadrilateral are 116 degree and 84 degree the remaining two are equal. what is the measurement of the other angle.
Answer» HEY MATE.... HERE'S THE ANSWER _____________ total MEASURE of a QUADRILATERAL is 360° so, from question 84+116+x+x=360° 200+2x= 360 2x= 360-200 2x=160 x=80° is ANSWER _______ hope it will help you

Discussion

32.	In the adjacent figure, AB is a chord of circle with centre O. CD is the diameter perpendicualr to AB. Show that AD=BD.
Answer» Hi , GIVEN : AB is a chord of a circle with center O . CD is the diameter Perpendicular to AB. RTP : AD = BD proof : Let P is the intersecting point of AB and CD. From triangles APD and BPD DP = DP ( Common side ) AP = PB ( diameter bisects chord ) Therefore , TRIANGLE APD congruent to triangle BPD ( SAS creiteria ) AD = BD ( corrospoding parts of congruent triangles ) Hence proved. I hope this helps you. : )

Discussion

33.	IF THE AREAS OF THE ADJACENT FACES OF A RECTANGULAR BLOCK ARE IN THE RATIO 2:3:4 AND ITS VOLUME IS 9000 cm3,THEN find its dimension
Answer» Let the dimensions be length: L, BREADTH: B and height: H Given, LBH = 9000 Given, the ratio of FACE area LB : BH : LH = 2:3:4 Therefore, let the face areas be 2x, 3x, 4x LB × BH × LH = 2x × 3x × 4x (LBH)² = 24x³ 9000² = 24x³ x = 150 Thus, LB = 2×150 = 300 BH = 3×150 = 450 LH = 4×150 = 600 Therefore, length, L = LBH/BH = 9000/450 = 20 cm breadth, B = LBH/LH = 9000/600 = 15 cm height, H = LBH/LB = 9000/300 = 30 cm

Discussion

34.	64 1/2 (64 1/2 +1) find the value
Answer» Hey Friend Here is Your ANSWER = (8 square) 1 upon 2 + 1)] = 8 (8 + 1) = 8 (9) = 72.

Discussion

35.	How many bricks of length 25 cm, breadth 15 cm and height 10 cm are required to build a wall of length 6 m, height 2.5 m and breadth 0.5 m?
Answer» Length of BRICK = 25CM breadth of brick = 15 cm height of brick = 10cm volume of brick = length × breadth × height = 25 cm × 15 cm × 10cm = 3750 cm³ length of wall = 6m = 600 cm breadth of wall = 0.5 m = 50cm height of wall = 2.5 m = 250 cm volume of wall = length × breadth × height = 600 × 250 × 50 = 7500000 cm³ $\textbf{number of bricks} = \frac{\text{volume of wall}}{\text{volume of brick}}$ = 7500000/3750 = 750000/375 = 2000 HENCE, number of bricks = 2000

Discussion

36.	The complement of (90° - a) is _______.
Answer» The complement of (90° - a) is $a^{\circ}$ SOLUTION: Given that we have to find complement of (90° - a) Let US first UNDERSTAND about complement angles Two Angles are Complementary when they ADD up to 90 degrees (a RIGHT Angle) If a and b ( in degrees) are two complement angles, then a + b = 90 degrees From given information, One of complement angle = (90° - a) Let the other complement angle be x Then by above definition of complementary angles, $\begin{aligned}&90^{\circ}-a+x=90^{\circ}\\\\&-a+x=0\\\\&x=a\end{aligned}$ Thus the another complementary angle is $a^{\circ}$ Learn more about complementary angle The ratio of supplementary angle to complementary angle of an angle is 8:3 . Find the angle brainly.in/question/5655818 Find complementary angle of 35 brainly.in/question/744721

36.

The complement of (90° - a) is _______.

Answer»

The complement of (90° - a) is $a^{\circ}$

SOLUTION:

Given that we have to find complement of (90° - a)

Let US first UNDERSTAND about complement angles

Two Angles are Complementary when they ADD up to 90 degrees (a RIGHT Angle)

If a and b ( in degrees) are two complement angles, then a + b = 90 degrees

From given information,

One of complement angle = (90° - a)

Let the other complement angle be x

Then by above definition of complementary angles,

$\begin{aligned}&90^{\circ}-a+x=90^{\circ}\\\\&-a+x=0\\\\&x=a\end{aligned}$

Thus the another complementary angle is $a^{\circ}$

Learn more about complementary angle

The ratio of supplementary angle to complementary angle of an angle is 8:3 . Find the angle

brainly.in/question/5655818

Find complementary angle of 35

brainly.in/question/744721

Discussion

37.	A cuboid shape soap bar has volume 150 cc. Find its thickness if its length is 10 cm and breadth is 5 cm.
Answer» $\textbf{volume of cuboid shape soap bar}$ = length × breadth × thickness. GIVEN, length of soap bar = 10cm breadth of soap bar = 5cm Let the thickness of soap bar = x cm now, 150 = 10 × 5 × x => 150 = 50X => x = 3 cm hence thickness of soap bar = 3 cm

Discussion

38.	Find the volume of a box if its length, breadth and height are 20 cm, 10.5 cm and 8 cm respectively.
Answer» l=20cm B=10.5cm and h=8cm so VOLUME of box = lbh =20 10.58 cm CUBE =1680cm cube is UR answer

38.

Find the volume of a box if its length, breadth and height are 20 cm, 10.5 cm and 8 cm respectively.

Answer»

l=20cm

B=10.5cm

and h=8cm

so VOLUME of box = l*b*h

=20 *10.5*8 cm CUBE

=1680cm cube is UR answer

Discussion

39.	The area of parallelogram ABCD is 36 cm². Calculate the height of parallelogram ABEF if AB = 4.2 cm.
Answer» Hi , It is given that.Area of parallelogram ABCD is 36 sq cm AB = 4.2 cm Let height of parallogram ABEF = h cm area of parallelogram ABEF = area of parallelogram ABCD [ SINCE area of parallelograms have same base AB LIE between same parallel lines DE and AB are equal ] = base × height = AB × h 36 = 4.2 × h h = 36/4.2 h = 8.57 cm I hope this HELPS you. : )

Discussion

40.	find tha linear relation between x and y such that p(x, y). is equidistant from the points A(1, 4) and b (-1, 2)
Answer» I HOPE this will HELP.

Discussion

41.	Write the coordinates the reflections of point (3 ,5) in x axis and y axis
Answer» MX(3,5)=(3,-5) My(3,5)=(-3,5)

41.

Write the coordinates the reflections of point (3 ,5) in x axis and y axis

Answer»

MX(3,5)=(3,-5)

My(3,5)=(-3,5)

Discussion

42.	In the figure,ΔABC, D, E, F are the midpoints of sides BC, CA and AB respectively. Show that(i) BDEF is a parallelogram(ii) ar(ΔDEF)=1/4ar(ΔABC)(iii) ar(BDEF)=1/2ar(ΔABC)
Answer» Parallelogram :A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram. In a parallelogram diagonal divides it into two triangles of equal areas. Mid point THEOREM:The line segment joining the midpoints of any two sides of a triangle is parallel to the third SIDE and equal to half of it. SOLUTION : Given:ABC is a Triangle in which the midpoints of sides BC ,CA and AB are D, E and F. To show:(i) BDEF is a parallelogram. (ii) ar (DEF) = 1/4ar (ABC) (iii) ar (BDEF) =1/2 ar (ABC) Proof: i)Since E and F are the midpoints of AC and AB. BC\|\|FE & FE= ½ BC= BD (By mid point theorem) BD \|\| FE & BD= FE Similarly, BF\|\|DE & BF= DE Hence, BDEF is a parallelogram .[A pair of opposite sides are equal and parallel] (ii) Similarly, we can prove that FDCE & AFDE are also parallelograms. Now, BDEF is a parallelogram so its diagonal FD divides its into two Triangles of equal areas. ∴ ar(ΔBDF) = ar(ΔDEF) — (i) In Parallelogram AFDE ar(ΔAFE) = ar(ΔDEF) (EF is a diagonal) — (ii) In Parallelogram FDCE ar(ΔCDE) = ar(ΔDEF) (DE is a diagonal) — (iii) From (i), (ii) and (iii) ar(ΔBDF) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF).....(iv) ar(ΔBDF) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = ar(ΔABC) 4 ar(ΔDEF) = ar(ΔABC)(From eq iv) ar(∆DEF) = 1/4 ar(∆ABC)........(v) (iii) AREA (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDF)ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF) ar(parallelogram BDEF) = 2× ar(ΔDEF)(From eq iv) ar(parallelogram BDEF) = 2× 1/4 ar(ΔABC)(From eq v) ar(parallelogram BDEF) = 1/2 ar(ΔABC) Hope this will help you...

42.

In the figure,ΔABC, D, E, F are the midpoints of sides BC, CA and AB respectively. Show that(i) BDEF is a parallelogram(ii) ar(ΔDEF)=1/4ar(ΔABC)(iii) ar(BDEF)=1/2ar(ΔABC)

Answer»

Parallelogram :A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.

In a parallelogram diagonal divides it into two triangles of equal areas.

Mid point THEOREM:The line segment joining the midpoints of any two sides of a triangle is parallel to the third SIDE and equal to half of it.

SOLUTION :

Given:ABC is a Triangle in which the midpoints of sides BC ,CA and AB are D, E and F.

To show:(i) BDEF is a parallelogram. (ii) ar (DEF) = 1/4ar (ABC) (iii) ar (BDEF) =1/2 ar (ABC)

Proof: i)Since E and F are the midpoints of AC and AB.

BC||FE & FE= ½ BC= BD

(By mid point theorem)

BD || FE & BD= FE

Similarly, BF||DE & BF= DE

Hence, BDEF is a parallelogram

.[A pair of opposite sides are equal and parallel]

(ii) Similarly, we can prove that FDCE & AFDE are also parallelograms.

Now, BDEF is a parallelogram so its diagonal FD divides its into two Triangles of equal areas.

∴ ar(ΔBDF) = ar(ΔDEF) — (i)

In Parallelogram AFDE

ar(ΔAFE) = ar(ΔDEF) (EF is a diagonal) — (ii)

In Parallelogram FDCE

ar(ΔCDE) = ar(ΔDEF) (DE is a diagonal) — (iii)

From (i), (ii) and (iii)

ar(ΔBDF) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF).....(iv)

ar(ΔBDF) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = ar(ΔABC)

4 ar(ΔDEF) = ar(ΔABC)(From eq iv)

ar(∆DEF) = 1/4 ar(∆ABC)........(v)

(iii) AREA (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDF)ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF)

ar(parallelogram BDEF) = 2× ar(ΔDEF)(From eq iv)

ar(parallelogram BDEF) = 2× 1/4

ar(ΔABC)(From eq v)

ar(parallelogram BDEF) = 1/2 ar(ΔABC)

Hope this will help you...

Discussion

43.	If perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral?
Answer» Answer: The area of QUADRILATERAL is 336 cm². Step-by-step EXPLANATION: A rhombus has four EQUAL SIDES. The perimeter of a rhombus is 100 cm. $4a=100$ $a=25$ The measure of each side is 25. $4a^2=d_1^2+d_2^2$ Where, a is side length and d₁ and d₂ are diagonals. $4(25)^2=(48)^2+d_2^2$ $2500=2304+d_2^2$ $d_2=\sqrt{2500-2304}$ $d_2=14$ The area of rhombus is $A=\frac{1}{2}(d_1d_2)$ $A=\frac{1}{2}\times 48\times 14$ $A=336$ Therefor the area of quadrilateral is 336 cm².

43.

If perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral?

Answer»

Answer:

The area of QUADRILATERAL is 336 cm².

Step-by-step EXPLANATION:

A rhombus has four EQUAL SIDES. The perimeter of a rhombus is 100 cm.

$4a=100$

$a=25$

The measure of each side is 25.

$4a^2=d_1^2+d_2^2$

Where, a is side length and d₁ and d₂ are diagonals.

$4(25)^2=(48)^2+d_2^2$

$2500=2304+d_2^2$

$d_2=\sqrt{2500-2304}$

$d_2=14$

The area of rhombus is

$A=\frac{1}{2}(d_1d_2)$

$A=\frac{1}{2}\times 48\times 14$

$A=336$

Therefor the area of quadrilateral is 336 cm².

Discussion

44.	Find the height of the cylinder whose volume is 1.54 m3 and diameter of base is 140 cm.(those who will give me answer i will mark them as brainiest)
Answer» Answer:Given:- VOLUME of the cylinder = 1.54m³ Diametre of the BASE = 140cm Radius of the base = 140/2 = 70CM = 0.7m Volume of cylinder = πr²h *1.54 = 22/70.7²h* *1.547/220.70.7 = h h = 1 Height = 1M Ans. Ans.) The height of the cylinder is 1m. Step-by-step explanation:**

44.

Find the height of the cylinder whose volume is 1.54 m3 and diameter of base is 140 cm.(those who will give me answer i will mark them as brainiest)

Answer»

Answer:Given:-

VOLUME of the cylinder = 1.54m³

Diametre of the BASE = 140cm

Radius of the base = 140/2

= 70CM

= 0.7m

Volume of cylinder = πr²h

1.54 = 22/7*0.7²*h

1.54*7/22*0.7*0.7 = h

h = 1

Height = 1M Ans.

Ans.) The height of the cylinder is 1m.

Step-by-step explanation:

Discussion

45.	Quadraticequation. by factorisation method.4y² +4y+ 1 =0
Answer» Step-by-step explanation: 4y^2+4y+1=0 =>4y^2+2Y+2y+1=0 =>2y(2y+1)+1(2y+1)=0 =>(2y+1)(2y+1)=0 2y+1=0 =>2y= -1 =>y= -1/2

45.

Quadraticequation. by factorisation method.4y² +4y+ 1 =0

Answer»

Step-by-step explanation:

4y^2+4y+1=0

=>4y^2+2Y+2y+1=0

=>2y(2y+1)+1(2y+1)=0

=>(2y+1)(2y+1)=0

2y+1=0

=>2y= -1

=>y= -1/2

Discussion

46.	What is the value of (vectorA)^2 ?
Answer» Step-by-step explanation: vector2(1,5) is a DIRECTION with the RATIO of 1 part x, and 5 parts y. E.G a LINE 1/6th to the RIGHT, and 5/6th's up. a negative would make the line left and down respectively. Magnitude show's the "strength" of the direction.

46.

What is the value of (vectorA)^2 ?

Answer»

Step-by-step explanation:

vector2(1,5) is a DIRECTION with the RATIO of 1 part x, and 5 parts y. E.G a LINE 1/6th to the RIGHT, and 5/6th's up. a negative would make the line left and down respectively. Magnitude show's the "strength" of the direction.

Discussion

47.	What is answer3100+610+x
Answer» Answer: 360 + X Step-by-step explanation: 3×100 + 6×10 + x 300+60+x 360 + x

47.

What is answer3100+610+x

Answer»

Answer:

360 + X

Step-by-step explanation:

3×100 + 6×10 + x

300+60+x

360 + x

Discussion

48.	एक दुकानदार कुछ वस्तुएं 619977 में खरीदा हैअगर वह प्रत्येक का एक बटे तीन हिस्सा 10% हानि पर बैठता है बताओ वह बचे हुए वस्तुओं को कितने प्रतिशत लाभ पर भेजें कि उसे कुल 20 प्रकरण का लाभ
Answer» Answer: like my answer Step-by-step explanation: क्रय मूल्य : - जिस मूल्य पर कोई वस्तु खरीदी जाती है क्रय मूल्य कहलाता है विक्रय मूल्य : - जिस मूल्य पर कोई वस्तु बेची जाती है वह मूल्य इस वस्तु का विक्रय मूल्य कहलाता है। लाभ = SP-CP हानि = Cp - Sp यदि विक्रय मूल्य X रु है तथा लाभ 20 प्रतिशत है तो NOTE : बट्टा/छुट सदैव अंकित मूल्य पर ही होता है कमिशन हमेशा विक्रय मूल्य पर दिया जाता है। 2 यदि एक व्यक्ति दो वस्तुए प्रत्येक Y रु मे बेचकर एक पर x% लाभ तथा दूसरे पर x% हानि उठाता है तो पूरे लेनदेन मे हमेशा हानि ही होगी उदाहरण 1 : - एक वस्तु का क्र. मू. 110 रु है तथा विक्रय मू. 123.20 है इसे बेचने पर वस्तु विक्रेता को कितने प्रतिशत लाभ होगा ? Sp-Cp = 123.20-110=13.20

48.

एक दुकानदार कुछ वस्तुएं 619977 में खरीदा हैअगर वह प्रत्येक का एक बटे तीन हिस्सा 10% हानि पर बैठता है बताओ वह बचे हुए वस्तुओं को कितने प्रतिशत लाभ पर भेजें कि उसे कुल 20 प्रकरण का लाभ

Answer»

Answer:

like my answer

Step-by-step explanation:

क्रय मूल्य : - जिस मूल्य पर कोई वस्तु खरीदी जाती है क्रय मूल्य कहलाता है

विक्रय मूल्य : - जिस मूल्य पर कोई वस्तु बेची जाती है वह मूल्य इस वस्तु का विक्रय मूल्य कहलाता है।

लाभ = SP-CP

हानि = Cp - Sp

यदि विक्रय मूल्य X रु है तथा लाभ 20 प्रतिशत है तो

NOTE :

बट्टा/छुट सदैव अंकित मूल्य पर ही होता है

कमिशन हमेशा विक्रय मूल्य पर दिया जाता है।

2 यदि एक व्यक्ति दो वस्तुए प्रत्येक Y रु मे बेचकर एक पर x% लाभ तथा दूसरे पर x% हानि उठाता है तो पूरे लेनदेन मे हमेशा हानि ही होगी

उदाहरण 1 : - एक वस्तु का क्र. मू. 110 रु है तथा विक्रय मू. 123.20 है इसे बेचने पर वस्तु विक्रेता को कितने प्रतिशत लाभ होगा ?

Sp-Cp = 123.20-110=13.20

Discussion

49.	What is other name for average in statistics?
Answer» ANSWER: ALSO CALLED the ARITHMETIC MEAN

49.

What is other name for average in statistics?

Answer»

ANSWER:

ALSO CALLED the ARITHMETIC MEAN

Discussion

50.	What is the equivalent ratio of 1/3
Answer» Answer: Note that the RATIO 2 to 6 is said to be EQUIVALENT to the ratio 1 to 3, that is 2:6 = 1:3. Note also that a fraction is a number that stands for "part of something".

50.

What is the equivalent ratio of 1/3

Answer»

Answer:

Note that the RATIO 2 to 6 is said to be EQUIVALENT to the ratio 1 to 3, that is 2:6 = 1:3. Note also that a fraction is a number that stands for "part of something".

Discussion

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