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Proof: in tri. BDL and tri. CDMDL=DM (given)BD=CD (given)Angle BLD=Angle CMD (DL per. to AB and DM per. to AC)Therefore tri. BDL = tri. CDM (RHS)Angle B =angle C (by c.p.c.t.)Hence proved, AB = AC ( side opp. to equal angles are equal)

SP = ₹ 1872

gain% = 4%

Let CP be x ,

x + 4% of x = 1872

x + 4/100 × x = 1872

x + x/25 = 1872

26x/25 = 1872

x = (1872 × 25)/26

x = 1800

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brass weight is 28.8 gram0.6 cooper is mix in brassif we cut brass into 3 pieces= 0.6÷ 3 = 0.2there will be 0.2 copper In one brass piece

Diameter is the longest chord if a circle.

diameter is the longest chord of the circle.

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Heron's formula= √s(s-a)(s-b)(s-c)longest chord=diameter

The longest chord of any circle is its diameter. Here the radius of the circle being 5 cm, the longest chord is 10 cm.

length of the longest chord=diameter of the circle=2(2.9cm)=5.8cm

answer jg =2617, g=1257

Amount (A1) = Rs. 4840

Amount (A2) = Rs. 5324

Let the principal = Rs. P

Required Rate %

=(5324−4840)/4840×100=10

Solution of this part 2

Thanks

Considering 24 hour clock time:

From 6:15 to 8: 40, there are 2 hours 25 minutes.

quarter past seven: 7:45 quarter to 10 = 9:45

Hence, the difference between two times is 2 hours

AOC + COB = 180 ( ANGLES IN A STRAIGHT LINE )AOC = 180 - COBAOC = 180 - 130AOC = 50AO = CO ( RADIUS OF A CIRCLE)OAC = ACO = X ( BY LAW OF ISOSCELES TRIANGLE)OAC + ACO + COA = 180 ( SUM OF ANGLE OF TRIANGLE)X + X + 50 = 1802X + 50 = 1802X = 180 - 502X = 130X = 65THEREFORE , ACO = 65

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given: angle boc=90°proof: angle BOC+ angle COA=180°130°+ ANGLE COA=180°ANGLE COA=180°-130°THEREFORE, ANGLE COA =50°ANGLE OAC=90° by thereorm 10.1Angle AOC+COA+OAC=180°( Triangle sum property)AOC+50°+90°=180°AOC+140°=180°AOC=40°

sorry one mistakeAOC+ACO+OAC=180°50°+ACO+90°=180°ACO=180°-140°angle ACO=40°

join OT that is radius let it be rj ,PT is tangent , so PT is perpendicular to OT .Now considering Right triangle OPT ,apply Pythagoras theoremso OT² + PT² = OP² r² + 8² = 10² r² = 100-64 = 36 so r = 6 which radius of circle.

It is given that Principle + Simple Interest for 3.5 years = 873 Rs-----(1)Principle + Simple Interest for 2 years = 756 Rs -------(2)Subtracting (1) – (2) we getSimple Interest for 1.5 years = 117 Rs.Therefore, S.I = P x R xT.Simple Interest for 2 years = Rs. 117/1.5 x2 =156 Rs.Therefore P = 756 - 156 = 600 RsAnd rate of interest =100 x 156/600 x2 = 13% per annum

thank you so much

In the Circle inscribed in square ;

Diameter = Side of square = 10 cm

Radius = 10/2 = 5 cm

Area of the circle = πr^2

= (22/7) × (5×5)

= 550/7

= 78.57 cm^2

As tangents drawn from an external point to a circleare equal in length

So, therefore, we get, AP=AQ (tangents from A) 1)

BP=BR (tangents from B) 2)

CQ=CR(tangents from C) 3)

As it is given that ABC is an isosceles triangle with sides AB=AC

Subtracting AP from both sides, we have,

AB-AP=AC-AP

implies AB-AP=AC-AQ (from 1)

BP=BQ

implies BR=CQ (from 2)

implies BR=CR(from 3)

So therefore BR=CR that shows that BC is bisected at the point of contact.

Let the circle touches the side AB at P and side AC at Q and side BC at RWe know that Tangents drawn from external points are equal.Then we have Tangents from point A i.e AP = AQ , Tangents from point B gives BP = BR , Tangents from point C gives RC = CQ.We have AB=AC ⇒ AP+PB=AQ +QC as AP= AQ ⇒ PB = QC ⇒ BR = RCThis gives that BC is bisected at point of contact.

Ans :- Let the circle touches the side AB at P and side AC at Q and side BC at R

We know that Tangents drawn from external points are equal.

Then we have Tangents from point A i.e AP = AQ , Tangents from point B

gives BP = BR , Tangents from point C

gives RC = CQ.

We have AB=AC ⇒ AP+PB=AQ +QC

as AP= AQ ⇒ PB = QC ⇒ BR = RC

This gives that BC is bisected at point of contact.

thanks for the answer . I appreciate it

-105 is the answer........

-105 is the right answer

-105 the correct answer of the given question

105 is the best answer

-105 is the right answer

plz like my answer

-105 is the answer for 3× (-5)× 7

-105is correct answer

3*(-5)*7=21*(-5)= -105

-105 is the correct answer

When it is circumscribed

the diameter will be square's side

therefore r = 5cm

area of circle = πr²= 22/7 × 5 × 5 = 78.57cm²

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when it is inscribed

the diameter will be it's diagonal

r = 5√2

area of circle = πr²= 22/7 × 5√2 × 5√2 = 157.14cm²

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Let us first convert all years in months.

8yrs. = 96 months, 168yrs. = 2016 months.

Therefore A.P. = 96,100,104....

Now, a = 96

S = 2016

d = 4

n=?

S = n/2( 2a + (n-1)d)

2016 = n/2( 192 + n-1)4)

4032 = n( 192 + 4n-4)

4032 = 4n2+188n

4n2+188n - 4032 = 0

n = [47 +/- ?(2209+4032)]/ (? means under root)

n = [47+/- ?(6241)] / 2

n = (47+/- 79) / 2

n = (47+ 79) / 2 or n = (47-79) / 2

Now the latter option is not possible as no. of students can never be negative.

Therefore n = (47+79) / 2

n = 169 / 2

Therefore n = 63

Tanθ+sinθ=mtanθ-sinθ=n∴, m+n=tanθ+sinθ+tanθ-sinθ=2tanθm-n=tanθ+sinθ-tanθ+sinθ=2sinθmn=(tanθ+sinθ)(tanθ-sinθ) =tan²θ-sin²θ∴(m²-n²)²=((m+n)(m-n))²=(2tanθ.2sinθ)²=16sin²θtan²θ

16mn=16(tan²θ-sin²θ)=16(sin²θ/cos²θ-sin²θ)=16sin²θ(1/cos²θ-1)=16sin²θ(1-cos²θ)/cos²θ=4sin²θ/cos²θ.sin²θ [∵,sin²θ+cos²θ=1]=16sin²θtan²θ

Hence proved.