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given: ABCD is a trapezium with AB||CD.......(1)and AB = 2 CD......(2)

In the triangles AOB and COD;

∠DOC = ∠ BOA [vertically opposite angles are equal]∠ CDO = ∠ ABO [alternate interior angles ]∠ DCO = ∠ BAO

thus Δ AOB ≈ Δ COD.

by the similarity rule, the ratio of the areas of the similar triangles is the ratio of the square of corresponding sides.thereforearea(Δ AOB) : area(Δ COD)=AB^2:CD^2area(Δ AOB) : area(Δ COD)=(2CD)^2:CD^2area(Δ AOB) : area(Δ COD)=4CD^2:CD^2area(Δ AOB) : area(Δ COD) = 4 : 1.

it is of herons formula

1st day=15 m2nd day=15+10=25 m3rd day=25+10=35m4th day=35+10=45m5th day=45+10=55mTotal depth of well=15+25+35+45+55=175 m

15+25+35+45+55=175 m

15+10+10+10+10 = 55m

angle 5 = 55° ( 3 and 5 are alternate angles and alternate angles are equal; as angle 3 = 55°)

angle 1 = 55° ( 1 and 5 are corresponding angles and corresponding angles are equal; as angle 5 = 55°)

angle 7 = 55° ( 7 and 5 are alternate angles and alternate angles are equal; as angle 5 = 55°)

angle 4 = angle 6= 125° ( 4 and 3 are allied angles and they are supplementary so 4 = 125°)

angle 8 = angle 2 = 110° ( 8 and 5 are allied angles and are supplement to each other so 8 = 110°)

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It is 36,36/4=936/3=123+6=9

answer 36 and 72 hai

Nakula and Sahadeva were Madri's sons.

Madri's sons were nakula and sahadeva

nakula and shdev were Madri's sons

The eldest of Kunti'schildren wascalled Yudhishthir, the second Bhimasen, and the third Arjun, and of Madri'ssons, the first-born of the twinswascalledNakuland the nextSahadev. And thosesons wereborn at an interval of one year after one another. And king Pandu, looking at hischildren, became very happy.

Nakul and Sahdev were Nadri's son

Madri sons were the Nakula and Sahdev

(6^3)^86^24will be the answer

(8√15)/(2√3)=4√5

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4648

four thousand six hundred forty eight

Incomplete question

angle p+q+r = 180

angle r = 180 - 70 - 65

angle r = 45

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Here ABCDE is a pentagon and each interior angles A,B,C,D,E=108Consider quadrilateral ABQESum of angles in Quadrilateral =360108+108+108+Q=360Q=36Similarly all P=Q=R=S=T=36P+Q+R+S+T=180

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Ruksana ate 1-1/3=2/3 partsRuksana had the larger share as 2/3>1/3She had larger share by 2/3-1/3=1/3

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sorry the answer is wrong

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in ∆PRQ and ∆PSQPQ = PQRPQ = SPQRQP = SQPso ∆PRQ and ∆PSQ are congurentBy CPCT PR = PS and RQ = SQ

Current Assets = Rs 500000Quick Assets = Rs 300000

Value of inventory= 500000 + 300000= Rs 800000

50% is the correct answer of the given question

Multiply first equation with 3and subtract-6x+16x+6/x+16/x -6-26=010x+22/x -32=05x+11/x -16=05x^2-16x+11=05x^2-5x-11x+11=05x(x-1)-11(x-1)=0(x-1)(5x-11)=0x=1,11/5

Given :

In Δ XYZ with XY = 4 cm, YZ = 6 cm and XZ = 3 cm.

Steps for construction :

1.Draw a line segment XY = 4 cm

2. From vertex X draw an arc of radius 3 cm the help of compass.

3.From vertex Y draw an arc of radius 6 cm the help of compass intersecting the previous arc.

4. The point of intersection is Z.

5. Join XZ & YZ

∆XYZ is a required triangle.

thank you soo much

when x → 0- , then the value of f(x) = 1+(0) = 1when x→ 0+ then also the value of f(x) = 1+(0) = 1

so, the limit near x= 0 will always be 1.

gardener plant 16 trees in 4 hoursfor 1 hour he plants 16/4=4 treeasso in 6 hours he plants 4×6=24 trees

Let sum of ratio=2x+3x=5xSo 5x=5500X=5500/5=1100So 1st will get=2x=2*1100=2200rs2nd will get=3x=3*1100=3300rs

2:3 sum of the ratio 2+3=5ramanns share = 2/5×5500=Rs.2200Amans share= 3/5 ×5500=Rs.3300

√ 8 = 2√2√18 =3√ 2√32 =4√2common difference = 4√2 - 3√2=√2next term = 5√2 = √50

Please post a clear image with proper lighting. We cannot provide a solution without a clear image of question.

I got answer

If it is a solution x= 3 y = 02(3)+3(0)= kk= 6