Question

the ratio of area triangle lies above the diagonal ac of the quadrilateral on the figure is 42:14 3:1mc=7cmthen am=?what is the area of triangle amb?if dm =5cmthen bm=? ji​

Answer 1

Answer:

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Answer 2

Given :-

• The ratio of area triangle lies above the diagonal AC of the quadrilateral on the figure is 42 : 14 = 3 : 1.
• MC = 7 cm .
• DM = 5 cm .
• Area CMB = 28 cm² .

To Find :-

• AM = ?
• Area of ∆AMB = ?
• BM = ?

Answer :-

→ Area ∆AMD : Area ∆AMC = 42 : 14 : 3 : 1

→ (1/2) * AM * MD * sin θ : (1/2) * MC * MD * sin (180 - θ) = 3 : 1

→ AM * sin θ : MC * sin θ = 3 : 1

→ AM : MC = 3 : 1

→ AM/MC = 3/1

→ AM/7 = 3/1

→ AM = 21 cm (Ans.)

similarly,

→ Area AMB : Area CMB = AM : CM

→ Area AMB : 28 = 21 : 7

→ Area AMB / 28 = 3/1

→ Area AMB = 28 * 3

→ Area AMB = 84 cm² .

similarly,

→ Area DMC : Area BMC = DM : BM

→ 14 : 28 = 5 : BM

→ (1/2) = 5/BM

→ BM = 10 cm (Ans.)

Conclusion :- Ratio of areas of ∆'s lies on either side of a straight line is equal to ratio of their base .

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