Question

(a)The drawing above shows how the sun, moon, and earth are aligned for a solar eclipse.Identify the tangents lines which partition an area on the earth that experiences a total solar eclipse.


(b)From the above drawing, identify the common external tangents between the sun and the moon.

(c)From the above drawing, identify the common internal tangents between the sun and the moon.

(d)What is the length of DM in the given picture, if EM = y units, HM = 10 units, GM = (y + 10) units.

(i) 15 units
(ii) 20 units
(iii) 10.5 units
(iv) 10 units

(e) What is the measure of ∠DOG in the given picture, if ∠DMG = x° .

(i) 180 °
(ii) ( 180 – x )°
(iii) ( 180 + x )°
(iv) 90°​

Answer 1

Answer:

Option A is the correct answer

Answer 2

Question A) :- (a)The drawing above shows how the sun, moon, and earth are aligned for a solar eclipse.Identify the tangents lines which partition an area on the earth that experiences a total solar eclipse.

Answer :-

• The Pink area, the area between the tangents EF and HI is the area that experiences a total solar eclipse.

Question B) :- From the above drawing, identify the common external tangents between the sun and the moon.

Answer :-

we know that, If a tangent touches the upper surface of both circles, it’s a common external tangent.

So,

• The common external tangent are DE and GH as they do not intersect the red dotted line.

Question c) :- From the above drawing, identify the common internal tangents between the sun and the moon.

Answer :-

we know that, if a tangent crosses from the upper surface of one circle to the lower surface of the other, then that tangent is a common internal tangent.

So,

• The common external tangent are DH and GE as they do intersect the red dotted line.

Question D) :- What is the length of DM in the given picture, if EM = y units, HM = 10 units, GM = (y + 10) units.

(i) 15 units

(ii) 20 units

(iii) 10.5 units

(iv) 10 units

Answer :-

we know that, if two tangent segments are drawn to one circle from the same external point, then they are congruent.

So,

→ EM = HM (same External point M from Moon.)

→ y = 10 units.

now,

→ EG = HG (same External point G from Moon.)

→ EM + GM = HG

→ y + (y + 10) = HG

→ 2y + 10 = HG

→ 2 * 10 + 10 = HG

→ 20 + 10 = HG

→ HG = 30 units.

then,

→ HD = HG (same External point H from Sun.)

→ HM + MD = 30

→ 10 + MD = 30

→ MD = 30 - 10

→ MD = 20 units.

therefore, The length of DM is 20 units. (Option ii) .

Question E) :- What is the measure of ∠DOG in the given picture, if ∠DMG = x° .

(i) 180 °

(ii) ( 180 – x )°

(iii) ( 180 + x )°

(iv) 90°.

Answer :-

→ DO = GO = radius of sun.

→ DM = GM = tangents from sun to external point M .

we know that,

• A radius that connects to the tangent is 90°.
• Sum of all four angles of a quadrilateral is 360° .

So, in quadrilateral DMGO we have,

• ∠DMG = x°
• ∠ODM = ∠OGM = 90° . (DO and GO are radius.)

then,

→ ∠DMG + ∠ODM + ∠OGM + ∠DOG = 360° (Angle sum Property of a quadrilateral.)

→ x° + 90° + 90° + ∠DOG = 360°

→ (x + 180)° + ∠DOG = 360°

→ ∠DOG = 360° - (x + 180)°

→ ∠DOG = 360° - 180° - x°

→ ∠DOG = (180 - x)° Option (ii) (Ans.)

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