Englishmathematicspart-2(geometry)set_3_ans.pdf

Mathematics Part II STD 10th Question Paper No. 3 Answersheet

Q. 1 (A) (1) d(A, B) = 4 - (-8) = 4 + 8 = 12 (2) ∠ RHG = ∠ DHP .................(Opposite angles) = 85˚ ∠ HGS = ∠ DHP ................ (Corresponding angles) = 85˚ (3) ∠ ACD = ∠ B + ∠ A .............. (theorem of remote interior angle) = 40 + 70 = 110˚ (4) WY = 2 OY = 2×5 = 10 cm (Diagonals of parallelogram bisect each other) (5) Point A(-3, 2) is in second quadrant and point B(12, 0) is on X- axis. (6) Curved surface area of sphere = 4πr2 = 4 × 3.14 × 12 ( r = 1 cm) = 4 × 3.14 × 1 = 4 × 3.14 = 12.56 sq. cm Q . 1 (B) (1) 2.sin30 + 3.tan45 1 =2× 2 +3×1 =1+3 ∴

=4 1 1 (2) MB = 2 × AB = 2 × 12 = 6 cm (perpendicular drawn from the centre of the circle to the chord bisects the chord) OB2 = OM2 + MB2 .....................(Pythagoras thearem) = 82 + 62 = 64 + 36 = 100 ∴ OB = 10 cm 1

(3) In ∆ PQR 12 cm > 10 cm > 8 cm ∴ QR > PQ > PR ∴ ∠P>∠R>∠Q The biggest angle is ∠P and the smallest angle is ∠Q. Q 2 (A) (1) A (2) C (3) A Q. 2 (B) (1) ∆ ABC ∼ ∆ DEF 2 A( ∆ ABC) = AB2 A( ∆ DEF)

(4) B

DE



1 2

=



1 2

=

42 DE 2 16 DE 2

∴ DE2 = 16 × 2 ∴ DE = 4 2 (2) Chords EN and FS intersect each other externally. ∴ ∠ NMS =

1 2 1 2 1 2

× [m(arc NS) - m(arc EF)] × (125 - 37)



=



=



= 440

× 88

(3) P(0, 6) Q(12, 20) (x1, y1) (x2, y2) Let co-ordinates of midpoint be (x, y) By formula for midpoint., x +x x= 1 2 2 0 + 12 = 2

=

12 2

y= y=

y1 + y 2 2 6 + 20 2 26 2

= = 13

=6

∴ PQ co-ordinates of midpoint of segment PQ are(6, 13) 2

Q. 3 (A) (1) AB = BC ∠ BAC = ∠ BCA = 45˚



1

AB = BC =

2



=

1



=

1



=2



× AC

× 8 = 2 2



∠ EFG = ∠ FGH =

2

× 4×2

×2 2

(2) Proof : ∠ EFG = ∠FGH



1

.......... Alternate angles

........(Inscribed angle theorem) (II)

1 2

[m(arc EG)]

1 2

[m(arc FH)] ........ (Inscribed angle theorem) (III)

∴ m(arc EG) = m(arc FH)

.........

=

θ × πr2 360 22 90 × × 72 7 360 22 1 × ×7×7 7 4

=

154 4

Sector

D-AXC =

=

= 38.5 cm2

[(I), (II), (III) ]

∴ chord EG ≅ chord FH..... (corresponding chords of congruent arcs)

(3) Area of square ABCD = side2 = 72 = 49 cm2

(I)

∴ Area of shaded portion = 49 - 38.5



 = 10.5 cm2 3

Q 3 (B) (1) NQ2 = MQ × QP ................. (Theorem of Geometric mean) =9×4 = 36 ∴ NQ = 6

(2) secθ + tanθ

=

1 cos θ

+

sin θ cos θ

=

1 sin  cos 

=

(1  sin  )(1  sin  ) cos  (1  sin  )

=

12  sin 2  cos  (1  sin  )

=

cos 2  cos  (1  sin  )

∴ secθ + tanθ = cos  1 sin  (3) r1 = 5 cm, r2 = 2 cm, h = 9 cm 1 Area of frustum= 3 πh (r12 + r22 + r1 × r2) Q 4 (1) P

1

= 3 × 3.14 × 9 (52 + 22 + 5 × 2) = 3.14 ×3(25 + 4 + 10) = 3.14 ×3× 39 = 367.38 cm2 A Q

Given :In ∆ABC line l || Side BC line l intersects side AB and side AC in P and Q respectively.

l

To prove : B





C

Construction : Draw seg PC and seg QB. 4

AP PB

=

AQ QC

A( ∆ APQ)

Proof : A( ∆ PQB) = A( ∆ APQ) A( ∆ PQB)



AP PB AQ QC

=

...... (I) (Areas are in proportion to the bases) ....... (II) (Areas are in proportion to the bases)

∆ PQB and ∆ PQC have the same base PQ and PQ || BC, their height is also same. ∴ A(∆ PQB) = A(∆ PQC) ..... (III) ∴

A( ∆ APQ) A( ∆ PQB) AP

∴ PB =

AQ QC

=

A( ∆ APQ) A( ∆ PQC)



........ from ((I), (II) and (III) ........ from (I) , (II)

(2) A

O

P

M B

(3) slope of the line = P(2, 4), Q(3, 6)

y 2 − y1 x2 − x1

slope of the line PQ =

6−4 3− 2

=

2 1

k −1 5−3

=

k −1 2

=2

R (3, 1), S (5, k)

slope of the line RS =

But line PQ || line RS ∴ slope of line PQ = slope of line RS k −1

∴ 2= 2 ∴ 4=k-1 ∴ 4+1=k ∴ k = 5

5

A

Let AB be the light house.

M

o

60

The boat is at C and observer is at A.

∠ MAC is the angle of depression. 90 m

∠ MAC = ∠ ACB = 60o .....(Alternate angle)

AB = 90 m.

o

60

B



C

From the figure, tan60o =



3

=



BC =

90 BC 90 3

=

AB BC

90 × 3 3× 3

=

90 3 3

= 30

3

∴ BC = 30 × 1.73 ∴ BC = 51.90 ∴ The boat is at a distance of 51.90m from the light house. Q. 5 (1)

A

P B

D

Q

C

Draw Seg PQ. APQD is a cyclic qudrilateral. ∠ ADQ + ∠ APQ = 180o ....... (1) PBCQ is a cyclic qudrilateral. ∴ ∠ BCQ + ∠ BPQ = 180o ...... (2) o o ∴ ∠ ADQ + ∠ APQ + ∠ BCQ + ∠ BPQ = 180 +180 .... [from (1),(2) ] ∴ ∠ ADQ + ∠ BCQ + ∠ APQ + ∠ BPQ = 180o+ 180o ...... (3) But∠ APQ + ∠ BPQ = 180o ............ (4) (angles in linear pair) o o o ∴ ∠ ADQ + ∠ BCQ + 180 = 180 + 180 ............ [from (3) , (4) ] ∴ ∠ ADQ + ∠ BCQ = 180o ∴ ∠ ADC + ∠ BCD = 180o 6

(2)

R

5. 1

cm

N

Y

600

4 cm Y1 Y2 Y3 Y4

Q. 6

P

X

Y5 Y6

A F

E

× ×

B

O D

C

(1) In ∆ AOB, OF is bisector of ∠ AOB ∴

OA OB

=

AF ....... (1) (by angle bisector theoerm) BF

In ∆ BOC, OD is bisector of angle ∠ BOC .

∴

OB OC

=

BD ....... (2)(by angle bisector theoerm) CD

In ∆ AOC , OE is bisector of angle∠ AOC. CE ....... (3)(by angle bisector theoerm) AE

∴

OC OA

∴

OB OA BD CE OC AF × × = × × OC OB CD AE OA BF

∴

OA× OC× OB OB× OA× O C

=

∴

=

AF× CE × BD BF× AE × CD

1 =

AF× CE × BD BF× AE × CD

∴ BF × AE × CD = AF × CE × BD 7



from (1), (2) and (3)

(2) Volume of hemisphere = volume of cone =





2 πR3 3

1 πr2 × h 3

By the given condition ;

2 × volume of cone = volume of hemisphere

∴2×

1 2 2 3 3 πr h = 3 πR

∴ r h = R3

2

∴ if r = h = R .......then both sides will be equal.



∴ if radius of base of the cone is R and its height is R, which is equal



to radius of the bowl, then a cone satisfying the given condition can be made.

8