Explore topic-wise InterviewSolutions in .

Thanks 🙂🙂🙂

first question answer is(c) 2200cm and second question's answer is (b)121

thanks

3)

AB=CD=16cmarea of parallelogram=b×harea of ABCD=CD×AE=AD×CF16×8=AD×10128=AD×1012.8=AD

4km/hr = 10/9m/secspeed =distance /time=》10/9=600/time=》600/10/9=time=》5400/10=540 sec.

We know ₹1= 100 paise

a) ₹15 = 1500 p

b) ₹ 79 = 7900 p

c) ₹3 = 300 p

d) ₹ 45 = 4500 p

e) ₹ 35 = 3500 p

f) ₹ 4 = 400 p

1) slope AB = (-0.5-2)/(1-0) = -2.5/1 slope BC = (-3+0.5)/(3-1) = -2.5/1 since they are equal , so collinear

2) slope PQ = (8/5-2)/(2-1) = -2/5 slope QR = (6/5-8/5)/(3-1) = -2/5 since they are equal , so collinear..

Parallelogram :

A quadrilateral inwhich both pairs of opposite sides are parallel is called a parallelogram.

Rhombus:

A quadrilateral in which all four sides are equal and both pairs ofopposite sides are parallel is called a rhombus.

==========================================================

Given:diagonal AC of parallelogram ABCD bisects ∠A.

∠CAB =∠CAD

To show:

(i) It bisects ∠C also,

(ii) ABCD is a rhombus.

Proof:

i)

In ΔADC and ΔCBA,

AD = CB

(Oppositesides of a ||gm)

DC = BA

(Oppositesides of a ||gm)

AC = CA (Common)

Therefore,

ΔADC ≅ ΔCBA

( by SSS congruence rule.)

∠ACD = ∠CAB ……(i)

(by CPCT)

∠BCA= ∠CAD…..(ii)

(By CPCT)

& ∠CAB = ∠CAD…..(iii)

(Given)

From eq i,ii,iii,

All 4 above angles are equal to each other..

Hence, ∠ACD = ∠BCA

Thus, AC bisects ∠C also.

(ii) ∠ACD = ∠CAD (Proved)

AD = CD (Opposite sides of equal anglesof a triangle are equal)

But, AB= CD & AD= BC

[Opposite sides of a parallelogram]

AB = BC = CD = DA

Hence, ABCD is a rhombus.

=========================================================

ty bro

M×T=W40 men× 8 hrs=60 treesnow 32 men×12 hrs=x treesso x=(60×32×12)/(40×8)=72 trees

when a perfect square leaves a remainder 0 or 1 when it is divisible by 3 or 4

A perfect square leaves a remainder 0 when it's divisible by 3 Or 4

a perfect square leaves a reminder 0 when it's divisible by 3 or 4

a perfect square leaves a reminder 0 when it's divisible by 3 or 0

cosA +sinA= √2cosA

=> sinA= √2cosA- cosA

=> sinA = cosA(√2 -1)

Now multiplying both side by ( √2+1)

=>sinA(√2+1) = cosA(√2–1)(√2+1)

=>√2sinA + sinA=cosA(2–1)

=> cosA-sinA= √2sinA (Answer)

1)area of parallelogram is base into height and AB is equal to DC o area of the parallelogram is AE X DC = 16X8 and area of para. is also base x height = AD x FC = 10 X AD. now by comparing both the results we get AD is equal to 16 X 8 / 10 that is 12.8

calculate the area of triangle agar zero aagaya to points collinear hai

1) slope AB = (-0.5-2)/(1-0) = -2.5/1 slope BC = (-3+0.5)/(3-1) = -2.5/1 since they are equal , so collinear

2) slope PQ = (8/5-2)/(2-1) = -2/5 slope QR = (6/5-8/5)/(3-1) = -2/5 since they are equal , so collinear..

area (∆ABC)=1/2{0(-0.5+3)+1(-3-2)+2(2+0.5)} =1/2{0-5+5} =1/2×0 =0So this point is collinear

100 - 0.60 - 0.25100 - 0.850.15

100- 0.60- 0.25100 - 0.850.15

Distance covered in one round = circumference of circle=2×pi×r=2×22/7×24.5=154 mthen,Distance run by a boy in 4 rounds=154×4=616 m

you are ginious

Radius of circular field = 24.5 m

In one complete round distance covered = 2 × pi × r

= 2 × 22/7 × 24.5 m = 154 m

Distance covered in 4 rounds = 4 × 154 m= 616 m

Area of APB= Ap= √10^2-8^2= 6now area= 1/2*6*8= 24area if PQBC= 8*6= 48total area= area of PQBC + area of 2 triangle= 48+48= 96cm^2

Area of APB=1/2 *10*8=40sqcm

triangle APB is congruent to triangle DQC40sqcm=40sqcmArea of PBQC8*648sqcm

Area ofAPQDCB40+40+48=128sqcm

Given:- ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on BD (Maine angle ko ese likha h <)

PROOF (1):- In ∆APB and ∆ CQD <ABP =<CDQ ( Alternate angle) AB = CD (opposite side is llgm) <APB = <CQD (each 90 angle) ∆APB =~ ∆ CQD (ASA Rule) (2.) so, AP = CQ ----------------Prove that----------------

By converse of mid point if two sides are parallel then the ratio of the corresponding sides are equal

so ,

ad/db = ae/ec ( ad = ab - ad)

4/2 = ae/3

cross multiplying '

ae = 6 cm

so ac = ae + ec = 6+3 = 9

ac = 9 cm

hence, option (d) is correct.

LHS =cos²(2x) - cos²(6x) Use the formula, a² - b² = (a - b)(a + b) = (cos2x - cos6x)(cos2x + cos6x)Use the formulacosC + cosD = 2cos(C+D)/2.cos(C-D)/2cosC - cosD = 2sin(C+D)/2.sin(D-C)/2

= {2sin(2x + 6x)/2.sin(6x-2x)/2}{2cos(2x+6x)/2.cos(6x -2x)/2}={2sin2x.sin4x}{2cos4x.cos2x}={2sin2x.cos2x}{2sin4x.cos4x}

Now, Use the formula, sin2A = 2sinA.cosA

= sin2(2x).sin2(4x)= sin4x.sin8x = RHS

Like if you find it useful

total area of square park=side^2=100^2=10000m^2flower bed in form of quadranthencearea of 4 quadrants will be 4(πr^2/4)=22/7*14*14=616m^2hence remaining area is 10000-616=9384m^2

In 5 rounds = 1 km = 1000 min 1 round = 1000/5 = 200 mso,perimeter of square = 200 m4×side = 200side = 200/4side = 50 m

In 5 rounds = 1 km = 1000 min 1 round = 1000/5 = 200 mso,perimeter of square = 200 m4×side = 200side = 200/4side = 50 m

Length of park = 600 mWidth of park = 300 m

Perimeter of park= 2(length + breadth)= 2(600+300) = 1800 m

Total distance covered = 3*1800= 5400 m