9类NCERT解决方案-第10章圈子-练习10.6

问题1.证明两个相交圆的中心线在两个相交点处对角相等。

解决方案：

Given: Two circles with Centre A and B circle intersects at C and D.

To prove: ∠ACB=∠ADB

Construction: Join AD,BC and BD

Proof: In ∆ACB and ∆ADB

AC=AD            ——–[radii of the same circle]

BC=BD           ———[radii of the same circle]

AB=AB            ——–[common]

∴∆ACB≅∆ADB     ——— [by  S.S.S]

∠ACB=∠ADB          ——–[c.p.c.t.]

问题2：两个和弦AB和CD的长度分别为一个圆的5厘米和11厘米，彼此平行，并位于圆心的相反两侧。如果AB与CD之间的距离为6厘米，请找到圆的半径。

解决方案：

Let O be the centre of circle and be r cm

Given: AB=5cm, CD =11cm

Construction: Draw OM perpendicular AB  and  OL perpendicular CD.

Because OM perpendicular AB  and  OL perpendicular CD and AB||CD.

∴Points O,L, and  M are collinear, than ∠M=6cm  

Let OL=x  

Then OM=6=x

Join AO and CO

OA=OC =r

OL=1/2CD=1/2*11=5.5cm  —–[perpendicular from bisects the chord]

AM=1/2AB=1/2*5=2.3cm   —–[perpendicular from bisects the chord]

Now, In right ∆DLC

r²=(OL)²+(CL)²

r²=x²+(5.5)²

r²=x²+30.25       ———–1

Now in right ∆OMA

r²=(OM)²+(MA)²

r²=(6-x)²+(2.5)²

r²=36+x²=12x+6.25

r²=x²-12x+42.25             ———–2

Now equating equation 1 and 2

X²+30.25=x²-12x+30.25

12x=42.25-30.25

X=12/12=1

Putting value of x in equation 1

r²=x²+30.25

r²=(1)²+30.25

r²=31.25

r=√31.25=5.6  (approx.)

Radius of circle is 5.6cm.

问题3.圆的两个平行弦的长度分别为6厘米和8厘米。如果较小的和弦距离中心4厘米，那么另一个和弦距离中心的距离是多少?

解决方案：

Let AB and CD are|| chord of circle with centre O which AB=6cm and CD=8cm and radius of circle =r cm.

Construction: Draw OP perpendicular AB  and OM perpendicular CD.

Because AB||CD and OP perpendicular AB and OM perpendicular CD therefore. Point O, M and P are collinear.

Clearly, OP=4cm ———-[According to question]

OM=to find?

P is midpoint of AB.

∴AP=1/2 AB=1/2*6=3cm  

M is midpoint of AB.

CM=1/2 CD=1/2*8cm=4cm

Join AO and CO

Now in Right ∆OPA,

r²=AP²+PO²

r²=3²+4²

r²=9+16=25

Now in ∆OMC

r²=CM²+MO²

25=4²+MO²

25-16=MO²

9=MO²

√9=MO  

3=MO

∴Therefore distance of the other chord from the centre is 3cm  

问题4.让一个角度ABC的顶点位于圆的外部，并使该角度的边与该圆相交相等的弦AD和CE。证明∠ABC等于中心的AC和DE弦所夹的角度之差的一半。

解决方案：

Give: Vertex B of ∆ABC lie outside the circle,chord AD=CE

To prove: ∠ABC=1/2(∠DOE-∠AOC)

Construction: Join AE

Solution: Chord DE subtends ∠DOE at the center and ∠DAE at point A on the circle.

∴∠DAE=1/2∠DOE       ———-1

chords AC subtends ∠AOC at the centre and ∠AEC at point  

∴∠AEC=1/2∠AOC          ———2

In ∆ ABE,∠DAE is exterior angle  

∠DAE=∠ABC +∠AEC  

1/2∠DOE=∠ABC+1/2∠AOC

½(∠DOE-∠AOC)= ∠ABC

问题5.证明以菱形任意边作为直径绘制的圆通过其对角线的交点。

解决方案：

Given: A rhombus ABCD in which O is intersecting point of diagonals AC and BD.

A circle is drawn taking CD as diameter.

To prove: circle points through O or Lies on the circles.

Proof: In rhombus ABCD,

∠DOA=90°       ——–[diagonals of rhombus intersect at 90°] 1

In circle:

∠COD=90° ——–[angle made in segment  O is right angle] 2

From 1 and 2

O lies on the circle.

问题6. ABCD是平行四边形。穿过A，B和C的圆与E处的CD相交(如果需要，可以产生)。证明AE = AD。

解决方案：

ABCD is a ||gm. The  circle through A,B and C intersect at E.

To prove: AE=AD

Proof: Here ABCE is a cyclic quadrilateral

∠2+∠4=180°      —–[sum of opposite is of a cyclic quadrilateral is 180°]

∠4=180°-∠1     ——-1

Now ∠4+∠6=180°-∠6         ———2

From 1 and 2

180°-∠2=∠180°-∠6

∠2=∠6          ———–3

Also ∠2=∠5       ———[opposite angles of ||gm are equal]        —–4

From 3 and 4

∠5=∠6

Now, In ∆ADE,

∠5=∠6

∴AE=AD        ——[sides apposite to equal angles in  a∆ are equal]

问题7：AC和BD是彼此一分为二的圆的和弦。证明(i)AC和BD是直径，(ii)ABCD是矩形。

解决方案：

Given: Two chords AC and BD bisects each other i.e OA=OC,OB=OD

To prove: In ∆AOB and COB

AO=CO        ——-[given]

∠AOB=∠COD    ——[vertically opposite angle]

OB=OD         ——–[give]

∴∆AOB≅∆COB   ——[s.s.s]

AB=CD   ——[C.P.C.T.]         1

similarly ∆AOD≅∆COB   (S.A.S)

AD=CD           (C.P.C.T.)   2

From 1 and 2     ABCD is a ||gm

Since, ABCD  is cyclic quadrilateral

∴∠A+∠C=180°  

∠B+∠B=180°

2∠B=180°

∠B=180°/2

∠B=90°

∴ ∠A  and ∠B lies in a semicircle

→ AC and BD are diameter of circle.

ii) Since ABCD is a ||gm and ∠A=90°

∴ ABCD is a rectangle.

问题8.三角形ABC的角度A，B和C的等分线分别在D，E和F处与其外接圆相交。证明三角形DEF的角度为90°– 1 2 A，90°– 1 2 B和90°– 1 2C。

解决方案：

Given:∆ABC and it circum-circle AD,BE and CF are bisectors of ∠A,∠B and ∠C  

Respectively.

To proof:∠ D=90°-1/2∠A ,    ∠E=90°-1/2∠B    ,  ∠F=90°-1/2∠C

Construction: Join AE and AF.

Solution: ∠ADE=∠ABE       ———-1  [angle in the same segment are equal]

              ∠ADF=∠ACF      ———–2 [angle in the same segment are equal]

Adding 1 and 2

∠ADE+∠ABF=∠ABE+∠ACF

∠D=1/2∠B+1/2∠C  ——[BC and CF are bisector of ∠B & ∠c]

∠D=1/2(∠B+∠C)

∠D=1/2(180°-∠A)

∠D=1/2(180°-∠A)

∠D=90°-1/2∠AC

问题9：两个全等圆在点A和B处相交。通过A绘制任意线段PAQ，使得P，Q位于两个圆上。证明BP = BQ。

解决方案：

Given: two congruent circles which intersect at A and B.  

PAB is a line segment

To prove: BA=BQ

Construction: join AB

Proof: AB is a common chord of both the congruent circle.

Segment of both circles will be equal

∠P=∠Q

Now, in ∆ BPQ,

∠P=∠Q

BP=BQ       ——[sides opposite to equal angles are equal]

问题10.在任何三角形ABC中，如果∠A的角平分线和BC的垂直平分线相交，则证明它们在三角形ABC的外接圆上相交。

解决方案：

Given: A ∆ABC, in which AD is angle bisector of ∠A and OD is ⊥ bisector of BC.

To prove: D lies on circumcircle.

Construction: Join OB and OC

Proof: Since BC subtends ∠BAC at A on the remaining of the circle.

∠BOC=2∠BAC               ——-1

Now, In ∆BOE and ∆ COE

BO=OE      ——–(radii of the same circle)

BE=CE        —–(give)

∴∆BOE≅COE       ——-(S.S.S)

∠1=∠2                     ——-(c.p.c.t)

Now,

∠1+∠2=∠BOC

2∠1=∠BOC

2∠1=2∠BAC         ———- (from 1)

∠1=∠BAC

∠BOE=∠BAF

∠BOD=∠BAC

∠BOD=2∠BAD        [AD is bisector of ∠BAC]

This is possible only if BD is chord of the circle.

D lies on the circle.