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Let us consider a rectangle ABCD.We know that the diagonals of a rectangle are equal. Therefore, AC = DBIn triangle ABC and triangle DCB,AB = DC (Opposite sides of a parallelogram)AC = DBBC = CB (Common side)Therefore, triangle ABC is congruent to triangle DCB.And, angle ABC = angle DCB (CPCT)And, angle ABC + angle DCB = 180° (Adjacent angles of a parallelogram are supplementary/Co interior angles)Therefore, angle ABC = angle DCB = 90°And, angle ABC = angle ADC = 90° and angle DCB = angle BAC = 90° (Opposite angles of a parallelogram are equal.)Hence, each angle of a rectangle measures 90°Hence proved.

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Area of the small segment.(Green Area) = 56 Square cm.

Area of the big segment = 260 Square cm.

Step-by-step explanation:

Please see the attached diagram for the problem description.

We need to measure the green colored area for small segment.

The remaining area in circle is area of bigger segment.

AB is the Chord. Radii drawn from points A & B are making 90 degrees at the center of the circle.

Thus ∆AOB is a right angl triangle with height and base both equal to radius ‘r’. = 14 cm

Area of the small segment.(Green Area) = Total are covered by sector AoB – Area of the triangle AOB.

Area of the sector AOB = (ᶿ/360) ∏*r^2

Since ᶿ = 90

Area of the sector AOB = (90/360) 22*14*14/7 = 154 Square cm

Area of the Triangle AOB = height * base / 2 = 14 * 14 / 2 = 98 Square cm.

Area of the small segment.(Green Area) = 154 – 98 = 56 Square cm.

Area of the big segment = Area of the circle - Area of the small segment.(Green Area)

= 22*14*14/7 – 56 = 316 – 56 = 260 Square cm.

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260 is the correct answer

260 CM square is the right answer

The extra 2.5 kg is obtained because of 20 reduction in price i.e. 20% of 38 = 7.6 rs.Reduced price 1 kg = 7.6/2.5 = 3.04 rs.

14+14= 28

. .

wait let me tell

C.P to Prem = Rs. 500Gain = 10%S.P = C.P*(100+G)/100 = 500*110/100 = Rs.550

C.P to Sudhir = Rs. 550Gain = 15%S.P = C.P*(100+G)/100 = 550*115/100 = 632.5

C.P to Hari = Rs. 632.5Hope it helps.

C.P. of 3 toffees is 1 rs.C.P. of 1 toffee=1/3 rsTo gain 50%,1/3*150/1001/2.2 toffees for 1 rs

Iftwoangles of a triangleare equal;then sides opposite to them are also equal.Ifthe altitude from one vertex of atriangle bisectsthe opposite side; then thetriangle is isosceles. ...If the bisectorof thevertical angle of a triangle bisects the baseof thetriangle;then thetriangle is isosceles.

Given:In ∆ABC , AD bisects ∠BAC, & BD= CD

To Prove:AB=AC

Construction:Produce AD to E such that AD=DE & then join E to C.

Proof:

In ∆ADB & ∆EDCAD= ED ( by construction)∠ADB= ∠EDC. (vertically opposite angles (

BD= CD (given)

∆ADB congruent ∆EDC (by SAS)

Hence, ∠BAD=∠CED......(1) (CPCT)

∠BAD=∠CAD......(2). (given)

From eq.1 &2 ∠CED =∠CAD......(3)

AB=CE (CPCT).......(4)

From eq 3 as proved that

∠CED=∠CAD

So we can say CA=CE......(5)

[SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL]

Hence, from eq 4 & 5

AB = AC

HENCE THE ∆ IS ISOSCELES..

Given:In ∆ABC , AD bisects ∠BAC, & BD= CDTo Prove:AB=ACConstruction:Produce AD to E such that AD=DE & then join E to C.Proof:In ∆ADB & ∆EDCAD= ED ( by construction)∠ADB= ∠EDC. (vertically opposite angles (BD= CD (given)∆ADB congruent ∆EDC (by SAS)Hence, ∠BAD=∠CED......(1) (CPCT)∠BAD=∠CAD......(2). (given)From eq.1 &2 ∠CED =∠CAD......(3) AB=CE (CPCT).......(4)From eq 3 as proved that ∠CED=∠CADSo we can say CA=CE......(5)

[SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL]Hence, from eq 4 & 5 AB = ACHENCE THE ∆ IS ISOSCELES..

PLEASE like the solution

4230 = p*(5/2)*7/100

p = 4230*200/5*7

p = 24171.4 rupees

thanks

Via right triangles and 'sohcahtoa':t = sin⁻¹(12/13) <==> sin t = 12/13 and cos t = 5/13s = cos⁻¹(4/5) <==> cos s = 4/5 and sin s = 3/5w = tan⁻¹(63/16) <==> sin w = 63/65 and cos w = 16/65.--------------Now, repeatedly use the sum of angles identitiescos(x+y) = cos x cos y - sin x sin y and sin(x+y) = sin x cos y + cos x sin y.

So, cos(sin⁻¹(12/13) + cos⁻¹(4/5) + tan⁻¹(63/16))= cos(sin⁻¹(12/13) + cos⁻¹(4/5)) cos⁻¹(tan⁻¹(63/16)) - sin(sin⁻¹(12/13) + cos⁻¹(4/5)) sin(tan⁻¹(63/16))

= [cos(sin⁻¹ 12/13) cos(cos⁻¹ 4/5)) - sin(sin⁻¹ 12/13) sin(cos⁻¹ 4/5))] cos⁻¹(tan⁻¹(63/16))- [sin(sin⁻¹ 12/13) cos(cos⁻¹ 4/5) + cos(sin⁻¹ 12/13) sin(cos⁻¹ 4/5)] sin(tan⁻¹(63/16))

= [(5/13)(4/5) - (12/13)(3/5)] (16/65) - [(12/13)(4/5) + (5/13)(3/5)] (63/65)= -1.

Since cos(sin⁻¹(12/13) + cos⁻¹(4/5) + tan⁻¹(63/16)) = -1,we conclude that sin⁻¹(12/13) + cos⁻¹(4/5) + tan⁻¹(63/16) = cos⁻¹(-1) = π.

yes

Since interest is compounded annually interest will be counted for 2 yrs only

P= 36000, R=10%, n = 2

A = P(1+R/100)^n = 36000(1 +.1)^2 = 36000(1.1x1.1) = 36000(1.21) = 43560

Rs. 43560 ans

Given,P = 219000, T = 292/365, R = 6/5 %

SI = P*R*T/100= 219000*292/365*6/5*1/100= 6*292*6/5= Rs 2102.4

a=3

d=8-3=5

An=302

An=a+(n-1)d

302=3+(n-1)5

302-3=(n-1)5

299=5n-5

299+5=5n

304=5n

n=304/5

So it is not the term of the A.P because it is in the fraction form.

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5/63 is the required answer

let the required number is X.

ER. RAVI KUMAR ROY..

A/QX - (-8/21) = 29/63

X = 29/63 +(-8/21)

X = 29/63 - 8/21

X = (29 - 24)/63

X = 5/63 ANS...

On Tuesdayit -5°-2° = -7°C

on Wednesday=-7°+4° -3°C

can you tell in equation s

m=-(-a) /b=a/bNegative, Positive, m=-a/b

public class ReverseNumber { public static void main(String[] args) { int num = 1234, reversed = 0; while(num != 0) { int digit = num % 10; reversed = reversed * 10 + digit; num /= 10; } System.out.println("Reversed Number: " + reversed); } }

outputs

output isReversed number 4321

thanks

thank you