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As we know, if two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
= AC
2
/DF
2
= BC
2
/EF
2
∴ 64/121 = BC
2
/EF
2
⇒ (8/11)
2
= (BC/15.4)
2
⇒ 8/11 = BC/15.4
⇒ BC = 8×15.4/11
⇒ BC = 8 × 1.4
⇒ BC = 11.2 cm
In ΔAOB and ΔCOD, we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
∴ ΔAOB ~ ΔCOD [AAA similarity criterion]
As we know, if two triangles are similar, then the ratio of their areas is equal to the square of the ratio of their corresponding sides. Therefore,
Area of (ΔAOB)/Area of (ΔCOD) = AB
2
/CD
2
= (2CD)
2
/CD
2
[∴ AB = 2CD]
∴ Area of (ΔAOB)/Area of (ΔCOD)
= 4CD
2
/CD
2
= 4/1
Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1
We know that area of a triangle = 1/2 × Base × Height
In ΔAPO and ΔDMO,
∠APO = ∠DMO (Each 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ΔAPO ~ ΔDMO (AA similarity criterion)
∴ AP/DM = AO/DO
⇒ Area (ΔABC)/Area (ΔDBC) = AO/DO
Now, let us prove ΔABC ≅ ΔPQR
ΔABC ~ ΔPQR
∴ Area of (ΔABC)/Area of (ΔPQR) = BC
2
/QR
2
⇒ BC
2
/QR
2
=1 [Since, Area(ΔABC) = (ΔPQR)
⇒ BC
2
/QR
2
⇒ BC = QR
Similarly, we can prove that
AB = PQ and AC = PR
Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]
We have to prove: Area(ΔABC)/Area(ΔDEF) = AM
2
/DN
2
Since, ΔABC ~ ΔDEF (Given)
∴ Area(ΔABC)/Area(ΔDEF) = (AB
2
/DE
2
) ……………………………
(i)
and, AB/DE = BC/EF = CA/FD ………………………………………
(ii)
In ΔABM and ΔDEN,
Since ΔABC ~ ΔDEF
∴ ∠B = ∠E
AB/DE = BM/EN [Already proved in equation
(i)
]
∴ ΔABC ~ ΔDEF [SAS similarity criterion]
⇒ AB/DE = AM/DN …………………………………………………..
(iii)
∴ ΔABM ~ ΔDEN
The areas of two similar triangles are proportional to the squares of the corresponding sides.
∴ area(ΔABC)/area(ΔDEF) = AB
2
/DE
2
= AM
2
/DN
2
Hence, proved.
Given, ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.
Area(ΔBQC) = ½ Area(ΔAPC)
ΔAPC and ΔBQC are both equilateral triangles,
∴ ΔAPC ~ ΔBQC [AAA similarity criterion]
∴ area(ΔAPC)/area(ΔBQC) = (AC
2
/BC
2
) = AC
2
/BC
2
Since, Diagonal = √2 side = √2 BC = AC
⇒ area(ΔAPC) = 2 × area(ΔBQC)
⇒ area(ΔBQC) = 1/2area(ΔAPC)
Hence, proved.
∴ BD = DC = 1/2BC
Let each side of the triangle be 2
a
.
ΔABC ~ ΔBDE
∴ Area(ΔABC)/Area(ΔBDE) = AB
2
/BD
2
= (2
a
)
2
/(
a
)
2
= 4
a
2
/
a
2
= 4/1 = 4:1
Hence, the correct answer is (C).
Let ABC and DEF be two similar triangles, such that,
ΔABC ~ ΔDEF
And AB/DE = AC/DF = BC/EF = 4/9
The ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides.
∴ Area(ΔABC)/Area(ΔDEF) = AB
2
/DE
2
∴ Area(ΔABC)/Area(ΔDEF) = (4/9)
2
= 16/81 = 16:81
Hence, the correct answer is (D).
