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given DC = 2 ABDraw BH parallel to AD. H will be the middle point of DC as AB = DH.Sin EF || AB, FI = AB

TheΔBHC andΔBIE are similar as the correspondingsides are parallel.So IE / HC = BE / BC IE = HC * BE / BC = y * 4x/7x = 4 y /7

EF = FI + IE= y + 4 y /7 = 11 y / 7

so 7 EF = 11 AB

if sin theta is equals to cos theta and theta is acute then theta is equals to 45° so 2tan²θ+sin²θ-1=(2*1)+(1/2)-1=2+(1/2)-1=(5/2)-1=3/2

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Given :DE||BC and D is midpoint of AB.

So, By converse of midpoint theorem E is midpoint of AC, Similarly F and E are midpoints of BC and AC respectively so, by midpoint theorem FE || AB, and EF = 1/2 AB So,EF = BD

Let BD = 3y , CD = y, and thus BC = 4y.

Applying Pythagoras theorem,

In ∆ABD , AB² = AD²+BD²

In ∆ACD, AC² = AD²+CD²,

from equations above equate for AD, we get,

AB²–BD² = AC²–CD²

AB² = AC² + 9y²–y²= AC² + 8y²multiply by 2,

2AB² = 2AC²+16y²which is,

2AB² = 2AC² + BC²

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Given : AB || CD

In ∆ OBA and ∆ OCD

angle AOB=angle DOC(vertically opposite)

angle OAB= angle ODC ( Alternate angle)

angle OCD = angle OBA ( Alternate angle)

Though AAA rule not exist, no we can not say certainly but if a corresponding side are equal we can use a criterion.

thank you lots

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Given : AB || CD

In ∆ OBA and ∆ OCD

angle AOB=angle DOC(vertically opposite)

angle OAB= angle ODC ( Alternate angle)

angle OCD = angle OBA ( Alternate angle)

Though AAA rule not exist, no we can not say certainly but if a corresponding side are equal we can use a criterion.

Given :A circle C (0,r) and a tangentlat point A.

To prove :OA⊥l

Construction :Take a point B, other than A, on the tangentl. Join OB. Suppose OB meets the circle in C.

Proof:We know that, among all line segment joining the point O to a point onl, the perpendicular is shortest tol.

OA = OC (Radius of the same circle)

Now, OB = OC + BC.

∴ OB > OC

⇒ OB > OA

⇒ OA < OB

B is an arbitrary point on the tangentl. Thus, OA is shorter than any other line segment joining O to any point onl.

Here, OA ⊥l

In ∆ABCangle ABC = angle ACB ......(1)In ∆BCDangle DBC = angle DCB.......(2)

adding (1) and (2) ABC + DBC = ACB + DCBangle ABD = angle ACD

it is equal because AD equal to AC and CB and BD so it is equal

yes it is equal because AB are Angle bisecter of angle DBC and angle bisecter bisects the angle in equal parts

perimeter of a figure is sum of all sidessob) 8+16+16 = 8+32 = 40cmc) 5+4+3 = 12cm

in 15 min he runs one lapso 16/3 laps in16/3*1516*580 min0r 1 hour 20 min

in 15 min he runs one lapso 16/3 laps in16/3*1516*580 min0r 1 hour 20 min

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4+4+(-4)=8-4=4this right answer

4+4+-4=8-4=4the correct answer

4+4-48-4=0this answer is correct for this question

4+4+-48-4=4 is correct answer.

4 is a correct answer

4 is the correct answer

4+4+(4)=8-4=4the correct answer

in 15 min he runs one lapso 16/3 laps in16/3*1516*580 min0r 1 hour 20 min

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Surface area of sphere is 4πr^2=4*22/7*7*7=4*22*7=88*7=616cm^2

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Speed of motorcycle = 45 km/h

Length of Reactangular track = 800 m = .8 km

Width of Reactangular track = 400 m = .4 km

Perimeter of Reactangular track = 2(length + width) = 2(.8 +. 4) = 2*1.2 = 2.4 km

Total distance covered= 12*2.4 = 28.8 km

Time taken to cover a distance by motorcycle = Distance / speed= 28.8/45 = .64 hours = .64*60= 38.4 minutes

thanks

Given:Speed=45km/heNo of laps=12length =800mwidth=400m

Here total distance of track = 2(800+400)=2400 m

Total distance he has to travel = 12*2400=28800 m = 28.8 kmSince he takes 12 laps around

Therefore time taken =Distance/speed= 28.8/45=0.6 hours (apprx)

Just multiply 1800m with 5

Dipti run 9 km around the park

Let the radius of the pond be 'r'.Since the man takes exactly 600 steps with each step of 44 cm to go around the pond.Hence, the circumference of the pond = 44*600 = 26400 cm or 26400/100 = 264 mCircumference of the pond is 264 mCircumference of the circular pond = 2πr264 = 2*22/7*rr = (264*7)/44r = 42 mHence, the radius of the pond is 42 m.

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Area of a square = a² = 6² = 36 m

Area of rectangular part = 4*3*1 = 12 m

Remaining part = 36-12 = 24 m

I would give it to you, but I'm Bush getting deported. Sorry

meena share=1-(2/5+1/3)=1-(11/15)=15-11/15=4/15

Meena =1-(2/5+1/3)15a-(11/15)=4/15

1 - ( 2/5 + 1/3 )= 1 - ( 2×3 /5×3 + 1×5/3×5 )= 1 - ( 6/15 + 5/15 )= 1 - ( 6+5 /15 )= 1 - ( 11/15 )= 15 -11 /15= 4/15

meena=1-(2/5+1/3)15a-(11/15) =4/15like my answer

4/15 is correct answer