CBSE 10th Standard Maths Subject Circles Ncert Exemplar 3 Marks Questions 2021
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CBSE 10th Standard Maths Subject Circles Ncert Exemplar 3 Marks Questions 2021
10th Standard CBSE
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Reg.No. :
Maths
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The tangent at a point C of a circle and a diameter AB when extended intersect at P.If \(\angle PCA=110^0\), find \(\angle CBA\) [see figure] Join C with centre O
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In figure, AB and CD are common tangents to two circles of unequal radii.Prove that AB=CD.
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From an external point P, two tangents, PA and PB are drawn to a circle with centre O.At one point E on the circle tangent is drawn which intersect PA and PB at C and D, respectively.If PA = 10cm, find the perimeter of the triangle PCD.
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If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure.Prove that \(\angle BAT=\angle ACB\)
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AB is a diameter of a circle and AC is its chord such that \(\angle BAC=30°\). If the tangent at C intersects AB extended at D, then prove that BC = BD.
3 Marks
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CBSE 10th Standard Maths Subject Circles Ncert Exemplar 3 Marks Questions 2021 Answer Keys
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\(\angle \)PCA = 110°
\(\angle \)CBA = ?
\(\angle \)ACB = 90°e [Angle in a semicircle is right agle]
\(\angle \)PCB = 110° - 90° = 20°
\(\angle \)PCB = \(\angle \)CAB = 20° [Alternate segment theorm]
In \(\triangle\)ABC
\(\angle \)ABC + \(\angle \)CAB + \(\angle \)BCA = 180° [∵ Sum of angles of a is 180°]
⇒ \(\angle \)ABC = 180° - 110° = 70° -
Construction: Join AD and BC
Proof: The tangent drawn from an internal point to a circle are equal in length.
If A is external point for circle hving centre O.
AB = AD .....(i)
If C is external point then
BC = CD .....(ii)
Now, B is external point for circle having centre O
AB = BC .....(iii)
So, from (i), (ii) and (iii), we get
AB = BC = CD
So, AB = CD Hence proved. -
PA= 10 cm.
PA = PB [If P is external point] .....(i)
[ ∵ From an external point tangents drawn to a circle are equal in length]
If C is external point, then CA = CE
If D is external point, then
DB = DE ......(ii)
Perimeter of triangle \(\triangle\)PCD
= PC + CD + PD
= PC + CE + ED + PD
= pc + CA + DB + PD
= PA + PB
-PA + PA = 2 PA
= 2 x 10 = 20 cm.[From (ii)] -
In \(\triangle\)ABC
AC is diameter, B = 90° ......(i)
[In a semicircle there is always a right angle]
So, ACB + CAB = 90°
[∵ sum of angles of a\(\triangle\) is 180°]
OA ⊥ AT [Radius and tangent are ⊥ to each other at the point of contact]
\(\angle \)OAT = 90°
\(\angle \)OAB + \(\angle \)BAT = 90° ......(ii)
From (i) and (ii),
\(\angle \)ACB = \(\angle \)BAT Hence proved. -
\(50\sqrt { 3 } cm^{ 2 }\)
3 Marks
