Class 10 WBBSE Math Solution In English Chapter 7 Theorems Related To Angles In A Circle Exercise 7.5
Question 1. In AABC, ZB is a right angle. A circle drawn taking AC as diameter intersects AB at the point P; let us write the correct information from the followings:
1. AB > AD
2. AB = AD
3. AB < AD.
Solution: We know the angle in a semicircle angle = 90°
∴ ∠ADC = 90° (given)
But ∠ABC 90°
It is possible that points B & point D are the same points.
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∴ AB = AD
Question 2. Let us prove that the circle drawn with any one of the equal sides of an isosceles triangle as diameter bisects the unequal side.
Solution: In Δ ABC, AB = AC.
A circle with center O and AB as the diameter is drawn.
The circle cuts BC at D.
To prove D bisects BC.
Proof: Join A, B.
As AB is the diameter,
.. ∠ADB is one right angle.
i.e., AD ⊥ BC & ∠ADC = 90°
Now, in two right-angled ΔABD & ΔACD, Hypotenuse AB Hypotenuse AC (given), And AD are common.
∴ΔABD ≅ ΔACD
∴ BD = CD
∴ D bisects the unequal side of the isosceles triangle ABC.
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Question 3. Sahana drew two circles that intersect each other at the points P and Q. If the diameters of the two circles are PA and PB respectively, then let us prove that A, Q, and B are collinear.
Solution: Two circles cut each other at P & Q.
If PA & PB are the diameters of the two circles,
prove that points A, Q, and B are collinear.
Proof: Join Q, A; Q, B & P, Q.
As AP is the diameter & AQP is a semi-circle angle,
∴ ∠AQP = 1 rt. angle
Similarly, BP is the diameter.
∴∠BQP = 1 rt. angle
∴ ∠AQP+∠BQP = 2 rt. angles
Now, AQ & BQ meet at Q & the sum of the adjacent angles = 2 rt. angles
∴ AQ and BQ are on the same straight line.
∴ A Q & B are collinear.
Question 4. Rajat drew a line segment PQ whose midpoint is R and two circles are drawn with PR and PQ as diameters. I drew a straight line through point P which inter- sects the first circle at point S and the second circle at point T. Let us prove with the reason that PS = ST.
Solution: PQ is a straight line whose midpoint is R. Now two circles are drawn with PR & PQ as diameter.
A straight line passing through P is drawn which cuts the 1st circle at S & cuts the 2nd circle at T.
To prove PS = ST.
Join R, S & Q, T.
As PR is the diameter of the 1st circle,
∴ Semicircle angle PSR = 90°
∴ SR ⊥ PT
Similarly, PTQ = 90°
∴ QT I PT
∴ SR & QT are both perpendiculars on PT.
∴ SR || QT
In APQT, the midpoint PQ is R
and SR || OT.
∴ S is the midpoint of PT.
∴PS = ST.
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Question 5. Three points P, Q, and R lie in a circle. The two perpendiculars PQ and PR at point P intersect the circle at points S and T respectively. Let us prove that RQ = ST.
Solution: Let P, Q & R are the three points on a circle.
Perpendiculars drawn from P on the chords PQ & PR are PS & PT, which cut the circle
at S & T respectively.
To prove RQ ST.
Proof Join R, Q; S, T; S, Q; & T, R.
Let the straight lines RT & SQ intersect at O.
∠SPQ = 1 rt. angle [as PS ⊥ PQ (given)]
∴SQ is a diameter.
Again, RPT 1 rt. angle [as PT ⊥ PR (given)]
∴RT is a diameter.
∴O is the center as the diameters SQ & RT intersect each other.
∴ OR= OQ =OT
∴In triangles ORQ & OST,
OR=OS (Radius of the same circle)
OQ= OT (Radius of the same circle)
and ∠ROQ=∠SPT (vertically opposite)
∴ΔORQ ≅ ΔOST
∴RQ = ST. Proved.
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Question 6. ABC is an acute-angled triangle. AP is the diameter of the circumcircle of the triangle ABC; BE and CF are perpendiculars on AC and AB respectively and they intersect each other at point Q. Let us prove that BPQC is a parallelogram.
Solution: ABC is an acute-angled triangle. AP is the diameter of the circumcircle of ΔABC.
BE & CF are the perpendiculars on the sides AC & AB, respectively.
They meet at Q.
To prove BPCQ is a parallelogram..
Proof: CF AB (given)
and ∠ABP is a semicircle angle.
∴ BP ⊥ AB
CF || BP or CQ || BP
Again, BE 1 AC (given)
and ∠ACP is an angle in a semi-circle.
∴CP ⊥ AC
∴CP || BE or CP || BQ.
∴BPCQ is a parallelogram. Proved.
Maths WBBSE Class 10 Solutions
Question 7. The internal and external bisectors of the vertical angle of a triangle intersect the circumcircle of the triangle at the points P and Q. Let us prove that PQ is the diameter of the circle.
Solution: ABC is a triangle, and internal & external bisectors of ∠A of ΔABC are AP & AQ, respectively, which cut the circumcircle of ΔABC at P & Q respectively.
To prove PQ is the diameter of the circle.
Proof: As AP & AQ are the internal & external bisectors of∠A.
∴ ∠PAQ = 90°
∴ PAQ is a semi-circle angle.
∴ PQ is the diameter. Proved.
Maths WBBSE Class 10 Solutions Question 8. AB and CD are two diameters of a circle. Let us prove that ABCD is a rectangular
Solution: Let AB & CD be two diameters of the circle with center O
Join AC, BD, AD & BC.
To prove ACBD is a rectangle.
Proof: ∠ADB = ∠ACB= 1 rt. angle
& ∠CAD = DBC = 1 rt. angle
In ACBD quadrilateral,
∠A = ∠C = ∠B = <D = 1 rt. angle
∴ACBD is a rectangle.
Question 9. Let us prove that if the circles are drawn having sides of a rhombus as diameter then the circles pass through a fixed point.
Solution: If circles are drawn with diameters of the sides of a rhombus, they will pass a fixed point.
Let ABCD is a rhombus. The circle with AB as diameter cuts BC or produced BC at D. Join A, D.
∠ADB= 1 rt. angle (as a semicircle angle)
∴∠AOC = 1 rt. angle
Again the circle as AC diameter will pass through point D.
∴ Two circles with AB & AC as diameters intersect each other at D.
∴Circles with diameters of the sides of a rhombus will pass through a fixed point. Proved.
Maths WBBSE Class 10 Solutions Chapter 7 Theorems Related To Angles In A Circle Exercise 7.5 Multiple Choice Question
Question 1. PQ is a diameter of a circle with center O, and PR RQ; the value of <RPQ is
1. 30°
2. 90°
3. 60°
4. 45°
Solution: As PR RQ,
∴ PRQ 90° (semi-circle angle)
∴RPQ = RQP = 180°-90° / 2
=90° /2
Answer. 4. 45°
Question 2. QR is a chord of a circle and POR is a diameter of a circle. OD is perpendicular on QR. If OD = 4 cm, the length of PQ is
1. 4 cm
2. 2 cm
3. 8 cm
4. none of these
Solution: PQ = 2 x OD = 2 x 4 = 8 cm.
Answer. 3. 8 cm.
WBBSE Solutions Guide Class 10 Question 3. AOB is a diameter of a circle. The two chords AC and BD when extended meet at point E. If ∠COD = 40°, the value of CED is
1. 40°
2. 80°
3. 20°
4. 70°
Answer. 3. 20°
Question 4. AOB is the diameter of a circle. If AC = 3 cm, BC = 4 cm, then the length of AB is
1. 3 cm
2. 4 cm
3. 5. cm.
4. 8 cm.
Solution: AB2 = √AC2+BC2
= √32+42
=√25
= 5
Answer. 3. 5. cm
WBBSE Solutions Guide Class 10 Question 5. In the adjoining figure, O is centre of circle & AB is a diameter, if BCE = 20°, ZCAE= 25°, the value of ZAEC is
1. 50°
2. 90°
3. 45°
4. 20°
Solution: ACE = 90° + 20° = 110°; AEC = 180°- (110° +25°) = 45° A
Answer. 3. 45°
Chapter 7 Theorems Related To Angles In A Circle Exercise 7.5 True Or False
1. The angle in the segment of a circle that is greater than a semi-circle is an obtuse angle.
False
2. O is the midpoint of the side AB of the triangle ABC, and OA = OB = OC; if we draw a circle with side AB as diameter, then the circle passes through point C.
True
Class 10 WBBSE Math Solution In English Chapter 7 Theorems Related To Angles In A Circle Exercise 7.5 Fill In The Blanks
1. Semicircular angle is a Right angle.
2. The angle in the segment of a circle that is less than a semicircle is an Obtuse angle.
3. The circle is drawn with the hypotenuse of a right-angled triangle as the diameter passes through the Vertices.
Class 10 WBBSE Math Solution In English Chapter 7 Theorems Related To Angles In A Circle Exercise 7.5 Short Answer
Question 1. In isosceles triangle ABC, AB = AC; a circle drawn taking AB as diameter meets the side BC at point D. If BD = 4 cm, let us find the value of CD.
Solution. ABC is an isosceles triangle, where AB AC. A circle is drawn with AB as the diameter, the circle will cut BC at D.
If BD = 4 cm.
Find CD.
BD= CD = 4 cm.
Question 2. Two chords AB and AC of a circle are mutually perpendicular to each other. If AB = 4 cm, AC = 3, let us find the length of the radius of the circle.
Solution. AB & AC the two chords of a circle are perpendicular to each other.
AB = 4 cm, AC 3 cm,
the radius of the circle =?
Diameter (BC)=√32+42
= √25
= 5,
radius = 5/2
= 2.5 cm.
Question 3. Two chords PQ and PR of a circle are mutually perpendicular to each other. If the length of the radius of the circle is r cm, let us find the length of the chord QR.
Solution. PQ & PR are two chords of a circle perpendicular to each other;
if radius = r cm. find the length of the chord QR.
QR-Diameter – 2r cm [.. QPR = 1 rt. Angle]
Question 4. AOB is a diameter of a circle. Point C lies on the circle. If ZOBC = 60°, let us find the value of ZOCA.
Solution. AOB is a diameter of a circle. C is any point on the circle; if ZOBC= 60°, find ZOCA.
∴ OB OC
ZOBC= ZOCB = 60°
∴ ZOCA = 90° – 60° 30°
Question 4. In the picture beside, O is the center of the circle and AB is the diameter. The length of the chord CD is equal to the length of the radius of the circle. AC and BD produced meet at the point P, let A us find the value of ∠APB.
Solution. OA = OC = OB = OD = CD
∴ ΔOCD is an equilateral triangle.
∴ ∠COD 60°
∠APB = 60°
