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63601.

6. The 4th term of an A P is zero. Prove that the 25th term ot the Al I37. ABC is a triangle. A circle touches sides AB and AC produced and side BC at X, Y and Z respectively.Show, that AX-_ perimeter of AABC.

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From the figure u will get thatAX=AYBX=BZCY=CZnowper of triangle ABC =ab+bc+ac=ab+(bz+zc)+ac=ab+bx+ac+cy=ax+ay=2axper of triangle ABC =2axtherefore1/2perimeter of triangle =ax .

7)Tangents from an exterior point to a circle are equal in length.∴ BP = BQ ----- -(i) CP = CR ------(ii) and, AQ = AR ----- (iii)AQ = AR⇒ AB + BQ = AC + CR⇒ AB + BP = AC + CP...(iv) [From eqn (i) and eqn (ii)]Perimeter of ΔABC= AB + BC + AC= AB + (BP + PC) + AC= (AB + BP) + (AC + PC)= 2(AB + BP) [From eqn (iv)]= 2 AQ [From eqn (i)]⇒ AQ = 1/2(Perimeter of ΔABC)

63602.

If coordinates of two adjacent vertices of a parallelogram are (3, 2), ,diagonals bisect each other at (2, -5), find coordinates of the other two verticesORunits, find x.

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63603.

1.Use Euclid's division algorithm to find the HCF of0) 135 and 225(ii)196 and 38220

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63604.

1 हा e & हि ५ त तन पल 1) ८. ५:नए e 00 Mo X B L pasdbk ) ex above a ntubrania[Z35p7 Lo loto P एल एपकल, 9६ नि wesls कि | iR damer bo a bot J__u_flflm_u ok !

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63605.

6/A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle。elevation from his eyes to the top of the building increases from 30° to 60° as he walktowards the building. Find the distance he walked towards the building.

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63606.

V6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle ofelevation from his eyes to the top of the building increases from 30° to 60° as he walktowards the building. Find the distance he walked towards the building.

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63607.

The value of (11S

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(11111)(11111) = 123454321

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63608.

s ug that there is no slack in the string.6. 1.5 m tall boy is standing at some distance from a 30 m tall building. The angelevation from his eyes tohis eyes to the top of the building increases from 30° to 60° as he walkstowards the building. Find the distance he walked towards the building.7. From a noint on

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A| | | 30 mt | B----------C 60°----D 30° |1.5 m EAE=height of building (30m)BE= 1.5m boy height

in ∆ABCTAN60=AB/BC√3= 28.5/BC√3BC=30BC= 30/√3= 30√3/√3√3=10√3 IN ∆ABD TAN30=AB/BD1/√3= 30/BDBD= 30√3 CD= BD - BC = 30√3 - 10√3 = 20√3 MT = 20×1.732 = 34.64MTHe moves 34.64mt towards building

63609.

6. A boy, 1.6 m tall, is 20 m away from a towerand observes the angle of elevation of the topof the tower to be (i) 45째 (ii) 60째. Find theheight of the tower in each case.

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63610.

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle ofelevation from his eyes to the top of the building increases from 30° to 60° as he walkstowards the building. Find the distance he walked towards the building.

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a

a

63611.

sm tall girl stands at a distante osen from a land post 8 rasta shadase length 4.5m on the ground, thenfind the height of tempost.

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2.5 is the correct answer of the given question

63612.

PRACTICE EXERCISE1. Find the (i) curved surface area and(ii) total surface area of a closed righteylinder, whose height is 7 em and theradius of the base being 3 em.

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63613.

Exercise1. Find the sum ofSolve any two5 (-32

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5/2+(-3/4)

=5/2-3/4

=5*2-3/4

=10-3/4

=7/4

63614.

. In fig., a circle inscribed in triangle ABC touches its sides AB, BC and AC at points D,E and F respectively. If AB12 cm, BC-8 cm and AC 10 cm, then find the lengths ofAD, BE and CF

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63615.

जि FentAT 3 kg of wheotiswidh by the cont9 of crhent

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Cost of 8kg wheat= ₹72By using unitary method, we will calculate the cost of 1kg wheat= 72÷8= 9Cost of 1kg wheat =₹9Cost of 15kg wheat= 9×15= ₹135 Answer

63616.

Exercise1. A sum of 1000 is lent for 1 year at the rate of 7%p.a. Find the interest and the amount.

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Interest= PTR/100hereP= 1000R= 17 %time 1 year

= 1000*7/100== 70amount = 1070

63617.

EXERCISE 1.Use Euclid's division algorithm to find the HCF(i) 196 and 38220() 135 and 225

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I don't think you are of 1st Grade

63618.

008 o1 ofF mid. p; oy PQrye e IR

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63619.

4. Which term of the AP: 3, s, 13, 1S,.. is 78?cont

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a (first term of A.P.) = 3d (common difference) = 8-3 = 5Let the nthterm of A.P. be an.an(nthterm of an A.P.) = 78an= a + ( n - 1 ) d78 = 3 + ( n - 1 ) 578 - 3 = ( n - 1 ) 575 = ( n - 1 ) 575 / 5 = ( n - 1 )n - 1 = 15n = 15 + 1n = 16Therefore 16thterm of the given A.P. is 78.

ty

63620.

(iv)D 5 cm C2. In the adjoining figure, show that ABCD isa parallelogram.Calculate the area of lgm ABCD卜,A 5 cm B

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thanks but it is too long but nice

63621.

In the given figure, AB BC, M is mid-point of AB and N is mid-print of BC. Using Euclid'saxioms, prove that AM = NC.

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Given :

AB = BC .....(1)

MB = NB .....(2) ( due to midpoint)

So,

Eq(1) - Eq(2)

AB - MB = BC - NB

AM = NC ( proved)

Third Axiom: If equals be subtracted from equals, the remainders are equal

63622.

2. Construct a parallelogram ABCD in which AB = 4 cm, BC = 5 cm and ZB = 60°.

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63623.

a boy 2m tall is standing at some distance from a 30 m tall building. the angle of elevation from his eyes of the top of the building increases from 30° to 60°. find the distance he walked towards the building.

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tan 60 =30-2/ab

root 3 =28/ab

ab = 28/root 3

tan 30 = 28/cb

1/root 3 = 28/ ac + 28/rt 3

ac =28 rt 3 - 28/rt 3

ac =84 - 28/rt 3

ac =56/rt 3

32.34m.

Am i right

no u r wrong sonu😑

Q show me answer

63624.

2. In the adjoining figure, show that ABCD isD 5 cm Ca parallelogram.Calculate the area of llgm ABCD.A 5cm B

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63625.

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of theparallelogram is 600. Find the angles of the parallelogranm.

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63626.

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of thepatallelogram is 60. Find the angles of the parallelogram

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63627.

Are the following sets equal ?(x : χ is a letter in the word reap);

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63628.

51. In the given figure, AD is the bisector of6 cm. andZBAC. If AB 10 cm., ACBC 12 cm., find BD12 cm.

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page 2

63629.

Exercise 1Find the rule which gives the number of matchsticks required to make the followingpatterns-(i) A pattern of letter 'H'(ii)Apattern of letter"V

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a) 3 matchsticks

b) 2 matchsticks

wrong answerplease post correct answer

63630.

аQ.7.If each of the vowels in the word AMERICANchanged to next letter in the English alphabet thenwhich of the following will be the sixth letter fromthe Right end?(1) F(2) X (3) R(5) Q.15.Ifas(4) Plo

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F is the correct answer for this question

63631.

1) Describe the following sets in Roster formi){x/ x is a letter of the word MARRIAGE)

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{M,A,R,I,G,E} is the required set

63632.

In tiq ABCD L a T and P n mid contA AB,BD and ap ane

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63633.

A 150cm tall boy stands at a distance of 8m from a lamp post and cast a shadow of 2m. Find the height of the lamp post

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63634.

52522 दि कट न e .४ 2y pamsier (24 ajeeie 2gu a)’ “ane e

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63635.

हमला 2 रपट. o e Mflw\\«j aneProposdvn

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28 × 36 = 1008 24 × 48 = 1152 1008 not equal to 1152vThey are not in proportion.

63636.

A 5 cm Bparallelogram ABCD, it is beingthat AB 10 cm and the altitudesponding to the sides AB and AD are Mcm and BM 8 cm, respectively.D.

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63637.

0. The angle between two altitudes of a parallelogram through the vertexof an obtuse angle of the parallelogram is 60°. Find the angles of theparallelogram.

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63638.

The angle between two altitudes of a parallelogram drawn from vertex of an obtuse angle of parallelogram is 60 degree. Find all the angles of this parallelogram.

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63639.

X2-7x+12=0

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x*x - 7x + 12 = 0 x*x - 4x - 3x + 12 = 0 x(x-4)-3(x-4) = 0 (x-4)(x-3) = 0 x = 3,4

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63640.

(i) x2-7x+2

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63641.

iii) x2 + 7x-6-0(

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63642.

3%x2+7x+3-0

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63643.

flag post stands on the top of a building. From a point on the ground, the angles ofof the top and bottom of a flag post are 609 and 45 respectively. If the heightof the flag post is 10 m, find the height of the building.

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63644.

7, ABE ane ints on the same side of AB sach tha

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63645.

igh Order Thinking Skills (HOTS)41 and 42 are complementary angles. Z223 are supplementary angles. If 21 45o, fnthe measure of 23and

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63646.

2. In a parallelogram ABCD, it is beinggiven that AB 10 cm and the altitudescorresponding to the sides AB and AD Mare DL = 6 cm and BM = 8 cm,respectively. Find AD

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63647.

11. In the adjoining figure PQRS is a kite.Q 120°Find the values of x and y50°

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63648.

l and BM are the heights on sides AB and AD respectively oflelogram ABCD (Fig 11.24). If the area of the parallelogram1470cm, AB-35 cm and AD = 49 cm, find the length ofBM

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63649.

2 cubes sach af sotume 64 aom are joined end to end. Find the surfacubond

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63650.

Find the zeros of the following qbetween the zeros and the coef1. x2+7x +123. x2 + 3x-105. 5x2 -4-8x [CBSE 2008]7. 2x2-11x + 15

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1)x^2 + 7x + 12 = 0 x^2 + 4x + 3x + 12 = 0 x(x + 4) + 3(x + 4) = 0 (x + 3)(x + 4) = 0 x = - 3, - 4

2) x^2 + 3x - 10 = 0 x^2 + 5x - 2x - 10 = 0 x(x + 5) - 2(x + 5) = 0 (x - 2)(x +5) = 0 x = 2, - 5