Question

Please solve these questions except Question11.
Give only correct answer.
Full solution...​

Answer 1

Answer:

12) there is a circle having center O touches the sides AB and AC of the triangle at point E and F respectively.

Let the length of the line segment AE is x.

Now in △ABC,

CF=CD=6 (tangents on the circle from point C)

BE=BD=6 (tangents on the circle from point B)

AE=AF=x (tangents on the circle from point A)

Now AB=AE+EB

⟹AB=x+8=c

BC=BD+DC

⟹BC=8+6=14=a

CA=CF+FA

⟹CA=6+x=b

Now

Semi-perimeter, s=

2

(AB+BC+CA)

s=

2

(x+8+14+6+x)

s=

2

(2x+28)

⟹s=x+14

Area of the △ABC=

s(s−a)(s−b)(s−c)

=

(14+x)((14+x)−14)((14+x)−(6+x))((14+x)−(8+x))

= (14+x)(x)(8)(6)

= (14+x)(x)(2×4)(2×3)

Area of the △ABC=4

3x(14+x

Area of △OBC=

2

1

×OD×BC

=

2

1

×4×14=28

Area of △OBC=

×OF×AC

=

2

1

×4×(6+x)

=12+2x

Area of ×OAB=

2

1

×OE×AB

=

2

1

×4×(8+x)

=16+2x

Now, Area of the △ABC=Area of △OBC+Area of △OBC+Area of △OAB

4

3x(14+x)

=28+12+2x+16+2x

4

3x(14+x)

=56+4x=4(14+x)

3x(14+x)

=14+x

On squaring both sides, we get

3x(14+x)=(14+x)

2

3x=14+x ------------- (14+x=0⟹x=−14 is not possible)

3x−x=14

2x=14

x=

2

14

x=7

Hence

AB=x+8

AB=7+8

AB=15

AC=6+x

AC=6+7

AC=13

So, the value of AB is 15 cm and that of AC is 13 cm.

13)Let ABCD be a quadrilateral circumscribing a circle with centre O.

Now join AO, BO, CO, DO.

From the figure, ∠DAO=∠BAO [Since, AB and AD are tangents]

Let ∠DAO=∠BAO=1

Also ∠ABO=∠CBO [Since, BA and BC are tangents]

Let ∠ABO=∠CBO=2

Similarly we take the same way for vertices C and d

Sum of the angles at the centre is 360

Recall that sum of the angles in quadrilateral, ABCD = 360

=2(1+2+3+4)=360

=1+2+3+4=180

In ΔAOB,∠BOA=180−(1+2)

In ΔCOD,∠COD=180−(3+4)

∠BOA+∠COD=360−(1+2+3+4)

=360

–180

=180

Since AB and CD subtend supplementary angles at O.

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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Answer 2

Step-by-step explanation:

here, AD = AE (tangent)

DC = CF (tangent)

BE= BF (tangent)

<DOF+ <EOF + <DOE = 360 ° (COMPLETE ANGLE)

You get <DOE = 120

triangle doe is isosceles.

therefore, angle ode and angle oed are equal. which gives angle ode as 30°.

in quadrilateral adoe, angle ado is 90° (radius on tangent)

we can find angle ade as 60.

similarly angle aed is 60 and angle a is 60

which makes triangle ade equilateral.

now in triangle abc,

de is parallel to bc

therefore, de is half of bc

there fore de is 7 cm.

which gives ad = ae = 7 cm.

this gives ac = 7+6=13.

gives ab = 7+8=15