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Exercise 6.4
Q: 1
Let
ABC ~
DEF their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
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Answer
Q: 2
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio
of the areas of triangles AOB and COD.
Answer
Since AB || CD
OCD
(Alternate interior angles)
OBA =
ODC
(Alternate interior angles)
AOB =
COD
(Vertically opposite angles)
Therefore
AOB ~
COD
in
OAB =
(By AAA rule)
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MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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Q: 3
In figure 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that
Answer
Since ABC and DBC are one same base,
Therefore ratio between their areas will be as ratio of their heights.
Let us draw two perpendiculars AP and DM on line BC.
APO and
DMO,
in
In
APO =
DMO = 90°
AOP =
DOM
(vertically opposite angles)
OAP =
ODM
(remaining angle)
Therefore
APO ~
DMO
(By AAA rule)
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MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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Q: 4
If the areas of two similar triangles are equal, prove that they are congruent.
Answer
Let us assume two similar triangles as
ABC ~
PQR
Q: 5
D, E and F are respectively the mid-points of sides AB, BC and CA of
DEF and
ABC. Find the ratio of the area of
ABC.
Answer
in
Since D and E are mid points of
ABC
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MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
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Q: 6
Prove that the ratio of the areas of two similar triangles is equal to the square
of the ratio of their corresponding medians.
Answer
ABC ~
in
Let us assume two similar triangles as
PQR. Let AD and PS be the medians of these triangles.
A = P, B = Q, C = R
Since, AD and PS are medians
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Q: 7
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the
equilateral triangle described on one of its diagonals.
Answer
in
Let ABCD be a square of side a.
Therefore its diagonal
Two desired equilateral triangles are formed as
Side of an equilateral triangle
ABE and
DBF
ABE described on one of its side = a
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Side of an equilateral triangle DBF described on one of its diagonal
We know that equilateral triangles are having all its angles as 60º and all its sides of same length. So, all
equilateral triangles are similar to each other. So, ratio between areas of these triangles will be equal to the
square of the ratio between sides of these triangles.
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Q: 7
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the
equilateral triangle described on one of its diagonals.
Answer
Let ABCD be a square of side a.
Therefore its diagonal
Two desired equilateral triangles are formed as
Side of an equilateral triangle
ABE and
DBF
ABE described on one of its side = a
Side of an equilateral triangle DBF described on one of its diagonal
We know that equilateral triangles are having all its angles as 60º and all its sides of same length. So, all
equilateral triangles are similar to each other. So, ratio between areas of these triangles will be equal to the
square of the ratio between sides of these triangles.
in
Q: 8
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles
ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
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(D) 1 : 4
Answer
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We know that equilateral triangles are having all its angles as 60º and all its sides of same length. So, all
equilateral triangles are similar to each other. So, ratio between areas of these triangles will be equal to the
square of the ratio between sides of these triangles.
Let side of
ABC = x
Hence, (c)
Q: 9
Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A)
(B)
(C)
(D)
2:3
4:9
81 : 16
16 : 81
in
Answer
If, two triangles are similar to each other, ratio between areas of these triangles will be equal to the square of the
ratio between sides of these triangles.
Given that sides are in the ratio 4:9.
Hence, (d).
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in
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CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
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