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Posted by Rehana Khan 6 years, 9 months ago
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Ishu Bhadrawle 6 years, 9 months ago
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Sia ? 6 years, 4 months ago
Suppose the number of 25 - paisa coins be x and the number of 50 - paisa coins be y.
Then,
{tex}x + y = 50{/tex}............(i)
She has a total of Rs 19.50,
{tex}25x + 50y = 19.50\times(100){/tex}
{tex}\Rightarrow{/tex} {tex}25x + 50y = 1950{/tex}
{tex}\Rightarrow{/tex} {tex}x + 2y = 78{/tex} ........(ii)
Subtracting equation (i) from (ii),
{tex}\Rightarrow y = 28{/tex}
Substituting y = 28 in (i), we get x = 22.
{tex}\therefore{/tex} the number of 25 - paisa coins = 22
the number of 50 - paisa coins = 28
Posted by Jitender Jangid 6 years, 9 months ago
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Posted by Akash Singh 6 years, 9 months ago
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Posted by Jay Tulsiyani 6 years, 9 months ago
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Posted by Utkarsh Gupta 6 years, 9 months ago
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Shikhar Singh 6 years, 9 months ago
Posted by Harnit Gautam 6 years, 9 months ago
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Gaurav Seth 6 years, 9 months ago
6=0
a+5d=0
a=-5d----(i)
now
33rd term=a+32d
putting the value of a from equation (i) we get
t33=-5d+32d
=27d----(ii)
now,
t15=a+14d
=-5d+14d
=9d-----(iii)
from equation (ii) and (iii)
3(t15)=(t33)
proved.
Posted by Utkarsh Gupta 6 years, 9 months ago
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Gaurav Seth 6 years, 9 months ago
Let triangles ABC and PQR have equal areas.
we have the formula for area of similar triangles.
area of ABC/area of PQR= AB2/PQ2 = BC2/QR2= AC2/PR2
but given that ar(ABC)=ar(PQR)
implies that AB=PQ, BC=QR, AC=PR.
therefore triangle ABC≅ triangle PQR
Harsimran Singh ? 6 years, 9 months ago
Posted by Ramesh Rekharamesh 6 years, 9 months ago
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Dia Khurana@1608 6 years, 9 months ago
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Harsimran Singh ? 6 years, 9 months ago
Posted by Mahesh Nagar 6 years, 9 months ago
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Gaurav Seth 6 years, 9 months ago
Given: A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.
To prove: PA = PB
Construction: Join OA, OB, and OP.
It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OA
PA and OBPB ... (1)
In
OPA andOPB:
OAP =OBP (Using (1))
OA = OB (Radii of the same circle)
OP = OP (Common side)
Therefore,OPA
OPB (RHS congruency criterion)
PA = PB
(Corresponding parts of congruent triangles are equal)
Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.
Harsimran Singh ? 6 years, 9 months ago
Posted by Faisal Pasha 6 years, 9 months ago
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Gaurav Seth 6 years, 9 months ago
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Given : PA and PB arc two tangents drawn from an external point P to a circle with centre O.
To prove : ∠AOB + ∠APB = 180°
Const : Join OA and OB.
Proof : ∵ The tangent at any point of circle is perpendicular to the radius through the point of contact.
∴ ∠OAP = 90° .....(i)
and ∠OBP = 90° .....(ii)
Adding (i) and (ii), we get
∠OAP + ∠OBP = 180°
Now in quadrilateral AOBP,
∠OAP + ∠OBP + ∠APB + ∠AOB = 360°
⇒ 180° + ∠APB + ∠AOB = 360°
∴ ∠APB + ∠AOB = 180°.
Posted by Rohit Verma 6 years, 9 months ago
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Shikhar Singh 6 years, 9 months ago
Gaurav Seth 6 years, 9 months ago
Equilateral triangles are drawn on the sides of a right-angle triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
<hr />Given: A right-angle triangle right angled at B. Three equilateral triangles ABE, BCF and ACD are described on sides AB, BC and AC respectively.
To Prove: ar(∆ACD) = ar(∆ABE) + (∆BCF)
Proof: ∵ All triangle are equilateral therefore, by using AAA condition.
When 
then
[Taking reciprocal of both sides]
When 
, then
...(ii)
Adding (i) and (ii), we get
Posted by Yashika Chauhan 6 years, 9 months ago
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Posted by Raushan Lucky Raj 6 years, 9 months ago
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Shikhar Singh 6 years, 9 months ago
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