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AP = 3{radius when perpendicular to chord bisects it }QC = 4{radius when perpendicular to chord bisects it }Applying pythagoras theoremIn OAPOP=4In OCQOQ=3OP=PQ+OQ4=PQ+3therefore PQ = 1

explain it please

Given: PS = QR = 25 cm, PQ = 18 cm and SR = 32 cm.

Drawing perpendiculars from P and Q to meet SR at X and Y respectively.

Let AB be the diameter of the circle, where AB = PX.

We see that XY = PQ = 18.

Also SX = YR which gives us SX = YR = (32 – 18)/ 2 = 7 cm

By applying the Pythagorean Theorem in ∆ PXS, we get

PX2+ 72= 252

PX2= 576

PX = 24.

As PX = AB and AB is the diameter of the circle, therefore the diameter of the circle is 24 cm.

x = sum of money to be lenty = % rate per annuma/qwhen time= 2 years, Amount = 3224so, 3224 = x + x*y*2/100. --------- (I)

and when time= 5 years, Amount = 4160so, 4160 = x + x*y*5/100. --------- (II)

solving equation I and ii we get..

x =2600and y = 12%

let the principle be P and rate be R S.I = A - P

ATQ

744-P = PR*2/100 .....(1) => 744-P = PR/50 => PR = 50(744-P)

and also 816-P = 3PR/100...... (2)

so, 816-P = 3*[50(744-P)]/100=> 81600-100P = 111600-150P=> 50P = 30000=> P = 30000/50 = 600so, R = 50*(744-600)/600 = 12%

Thank you

Mean deviations = sum of deviations/number of observations

= 100/20=5

100/20=5 is the answer

Thank u so much

Th

Side of a triangle are. 6, 10, x.Area = 1/2 * 6 * 10 * sin∠BAC.Area is maximum, when ∠BAC = 90◦x = √(100+36) = √136

The question is"Sides of a triangle are 6, 10 and x for what value of x is the area of the △ the maximum?"

Hence, the answer is of these.

Choice D is the correct answer.

D is the correct answer of the given question

D is the correct answer

Yes D is the right answee

(D) of these is the answer.

none of these is correct answer

If you find this solution helpful, Please like it.

12 is the L.C.M of.

ab²=1/a²+1/2b²100=36+1/2b²100-36=1/2b2√64=1/2b²16=Barea = 12*16= 192sq cms

Thank you

d) 45 × (-1) = -450 sahi hai.

Kyoki - times + = - hota hai

x²+px-4= 0(-4) ²+P(-4) -4=016-4p=4-4p=4-16-4p= -12P= 3

Two Triangles are said to be similar if their i)corresponding angles are equal and ii)corresponding sides are proportional.(the ratio between the lengths of corresponding sides are equal)

GIVEN:AK = 10cm, BC = 3.5cm & HK = 7cm∆AHK ~ ∆ABC

AK/AC = HK/BC [corresponding sides of similar triangles are proportional]

10/AC = 7/3.510/AC = 70/3510/AC = 10/510AC = 10 × 5AC = (10 × 5) /10

AC = 5 cm

Hence, the value of AC is 5 cm.

Radius = 0.5*Diameter = 0.5 * 10= 5cmCircumference = 2*(22/7)*5= 220/7 cm

Length of arc = 36/360 * circumference= 36/360 * (220/7)= 22/7 cm

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Statement : A diagonal of a parallelogram divides it into two congruent triangles.

Given : A parallelogram ABCD.

To prove : ΔBAC ≅ ΔDCA

Construction : Draw a diagonal AC.

Proof :

In ΔBAC and ΔDCA,

∠1 = ∠2 [alternate interior angles]∠3 = ∠4 [alternate interior angles]AC = AC [common]

ΔBAC ≅ ΔDCA [ASA]

Hence, it is proved.

Therefore, diagonals of a parallelogram divide it into two congruent triangles.

Given: A parallelogram ABCD and AC is its diagonal .

To prove :△ABC≅△CDA

Proof : In△ABC and△CDA, we have

∠DAC =∠BCA [alt. int. angles, since AD| | BC]

AC = AC [common side]

and∠BAC=∠DAC[alt. int. angles, since AB| | DC]

∴ By ASA congruence axiom, we have

△ABC≅△CDA

∆ABC and ∆ABD; AC= AD /CAB=/DAB AB=AB • S-A-S;• • ∆ABC~∆ABD

Given AP is: √7, √28, √63,.......

= √7, 2√7, 3√7,......

>> Next term = 4th term = a + 3d= √7 + 3*√7

(first term a =√7 and common difference d = 2*√7-√7 = √7)

= 4*√7

= √(7*4*4)

= √112

2

3

4

5

tankyou very much

Area= 1/2πr^2

= 824 .72cm^2

Given,For upper cylinderHeight h1 = 10 cmDiameter = 10 cm, Radius r1 = 5 cm

For lower cylinderHeight h2 = 10 cmDiameter = 20 cm, Radius r2 = 10 cm

Then,Volume of upper cylinder= pi*r1^2*h1= 22/7*(5*5)*10= 22*250/7= 785.7 cm^2

Volume of lower cylinder= pi*r2^2*h2= 22/7*(10*10)*10= 22000/7= 3142.85 cm^2

Explanation:(29 + 13)/3 = 14,(76 + 26)/3 = 34,(74 + 14)/3 = 26.