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156=2×2×3×1313=1×13hcf=13is your answer

4 + 3/2 + 4x = - 4(3x + 2)

4x + 12x = - 4 - 3/2 - 8

16x = (-12*2 - 3)/2

32x = - 27

x = - 27/32

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I joined o and cIn ΔOPA and ΔOCA, OP = OC (Radii of the same circle) AP = AC (Tangents from point A) AO = AO (Common side)∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)⇒ ∠POA = ∠COA … (i)Similarly, ΔOQB ≅ ΔOCB ∠QOB = ∠COB … (ii)Since POQ is a diameter of the circle, it is a straight line.∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 ºFrom equations (i) and (ii),2∠COA + 2∠COB = 180º⇒ ∠COA + ∠COB = 90º⇒ ∠AOB = 90°

thanks

Volume will be the same as that of cube that is side^3=6^3=216cm^3

Given: ABC is a triangle; AD is the exterior bisector of 4A and meets BCproduced at D; BA is produced to F.To prove: AB/AC = BD/DCConstruction: Draw CE||DA to meet AB at E.Proof: In A ABC. CE||AD cut by AC.

∠CAD = ∠ACE (Alternate angles)Similarly CE || AD cut by AB

∠FAD = ∠AEC (corresponding angles)AC = AE (by isosceles △ theorem)

Now in △BAD, CE || DAAE=DC(BPT)AB BD

But AC = AE (proved above)AC=DCAB=BD orAB/AC = BD/DC (proved).

2*6/5 = 2.4 cm

3*6/5 = 3.6 cm

please post a complete question

Length of road = 10 mNo. Of flags placed = 15Distance between two flags = 10/14 =.714

No. Of flags can be placed on 1.5 km= 1500/.714 = 2101

x is equal to 62degree

tan(5theta) = 1/√3tan(5 theta) = tan 30°5theta= 30°theta = (30°/5)6°

Join A & C. Proof: BC extends upto point Q. ∴ CQ is also parallel to AD. △ ACQ and △ DQC are on the same base CQ and between the same parallels CQ ∥ AD, so they are equal in area ,i.e., Area (△ ACQ) = Area (△ DQC) Subtracting Area (△ CPQ) from both the triangles, Area (△ ACQ) - Area (△ CPQ) = Area (△ DQC) - Area (△ CPQ) ⇒ Area (△ APC) = Area (△ DPQ) .....(i) Since, AB ∥ DC (∵ Opposite sides of a parallelogram are equal) ∴ AB ∥ PC (∵ PC is a part of DC) △ APC and △ BCP are on the same base PC and between the same parallels PC ∥ AB, so they are equal in area ,i.e., Area (△ APC) = Area (△ BCP) .....(ii) From (i) and (ii), Area (△ BCP) = Area (△ DPQ)

thanks a lot Abhijit

Volume will be same as that of cube that is side^3=6^3=216cm^3

Here we have joined " AP " and " CQ " and assume that QP and BD intersect at ' O ' .

Given : P and Q are mid points of ' BC ' and ' AD ' respectively , So

BP = PC =12BC and DQ = QA =12DA and we know BC = DA ( As we know ABCD is a parallelogram and opposite sides are equal to each other )

Then,

BP = PC = DQ = QA --- ( 1 )

i ) We know ABCD is a parallelogram and opposite sides are parallel to each other , So

BC | | DA , So

PC | | QA --- ( 2 ) ( As here PC is part of line BC and QA is a part of line DA and we know BC | | DA )

And

PC = QA ---- ( 3 ) ( From equation 1 )

We know is a pair of opposite sides of any quadrilateral are equal and parallel to each other then that quadrilateral is a parallelogram . Thus from equation 2 and 3 we get :

APCQ is a parallelogram . ( Hence proved )

ii ) We know : BC | | DA , So

BP | | QA --- ( 4 ) ( As here BP is part of line BC and QA is a part of line DA and we know BC | | DA )

And

BP = QA ---- ( 5 ) ( From equation 1 )

We know is a pair of opposite sides of any quadrilateral are equal and parallel to each other then that quadrilateral is a parallelogram . Thus from equation 4 and 5 we get :

ABPQ is a parallelogram .So

AB | | PQ as we know opposite sides are parallel to each other , So

AB | | OQ --- ( 6 ) ( As here OQ is part of line PQ and we know AB | | PQ )

Here we also know ' Q ' is mid point of AD and we consider information " AB | | OQ " from equation 6

Then , In triangle ABD we get from converse of mid point theorem :

' O ' is mid point of BD , So we can say that :

QP is bisect line BD . ( Hence proved )

First arrange terms in ascending order11, 27, 36, 58, 65

As number of terms are 5. Then median is (5 + 1)/2 th term.

So, 3rd term = 36 is median

1)26 = 2 x 1391 =7 x 13HCF = 13LCM =2 x 7 x 13 =182Product of two value 26 x 91 = 2366Product of HCF and LCM 13 x 182 =2366

2) 510 = 2 x 3 x 5 x 1792 =2 x 2 x 23HCF =2LCM =2 x 2 x3 x 5 x 17 x 23 = 23460Product of both values 510x92 = 46920Product of HCF and LCM 2x23460=46920

18+21+17+28+16+30+15/7=145/7=2.7

1)rotational motion→ rotated with 90°

2) and translational Motion.

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thank you dear

3m = 300 cm

hence,(6/300) x 1006/3= 2%

3m= 300cmhence 6\300×1006/32%

20*2πr=66r=66/20*2πr=66/20*2*22/7r=6*7/20*2*22r=21/440m

loss=200*100/1200= =16.67%

loss 200*100/1200=16.67

Solution

Loss = 45%=>0.55∗Cost price=3,7400.55∗Cost price=3,740

Cost price =37400.5537400.55=6800

loss=45°\•; cost price=374005537400.55= 6800

3740×45/100=374×4.5= 1683.00

1683 is correct answer

1683 is the correct answer

In given ∆ABP. & ∆CDP

AP/CP = BP/DP

angleBPA=angle DPC. (opposite angle)

from S.A.S similarity∆ABP ~ ∆CDP

proved...