Math, asked by harinand8479, 7 months ago

given triangle Abc similar to triangle def find the area of triangle Abc divided by area of triangle def.if A and d = 90 degree,ab=3cm,DF=12 and EF =13cm.​

Answers

Answered by Anonymous
9

Answer:

i think it will be helpful for you

Step-by-step explanation:

ANSWER

(i) We know that the ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides.

area(△DEF)

area(△ABC)

=(

EF

BC

)

2

25

16

=(

EF

2.3

)

2

5

4

=

EF

2.3

⇒ EF=

4

2.3×5

∴ EF=2.875cm

(ii) We know that the ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides.

area(△DEF)

area(△ABC)

=(

DE

AB

)

2

64

9

=(

DE

AB

)

2

8

3

=

5.1

AB

∴ AB=1.91cm

(iii) We know that the ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides.

area(△DEF)

area(△ABC)

=(

DF

AC

)

2

area(△DEF)

area(△ABC)

=(

8

19

)

2

area(△DEF)

area(△ABC)

=(

64

361

)

(iv) We know that the ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides

area(△DEF)

area(△ABC)

=(

DE

AB

)

2

64

36

=(

DE

AB

)

2

8

6

=

6.2

AB

∴ AB=4.65cm

(v) We know that the ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides.

area(△DEF)

area(△ABC)

=(

DE

AB

)

2

area(△DEF)

area(△ABC)

=(

1.4

1.2

)

2

area(△DEF)

area(△ABC)

=

49

36

Answered by PoojaBurra
0

Given: Triangle ABC similar to triangle DEF. A and D = 90 degree, AB = 3 cm, DF = 12 and EF = 13 cm.​

To find: The area of triangle ABC divided by area of triangle DEF.

Solution:

In the triangles ABC and DEF, ∠A and ∠D are 90°. So, they are right-angled triangles. In triangle DEF, the perpendicular is 12 cm and the hypotenuse is 13 cm. Hence, the base can be found using the Pythagoras theorem.

base = \sqrt{(13)^{2} - (12)^{2}}

       = \sqrt{25}

       = 5 cm

Since the two triangles are similar, their sides are in proportion. The proportion can be represented as follows.

\frac{AC}{DF} = \frac{AB}{DE} = \frac{BC}{EF}

\frac{AC}{12 cm} = \frac{3 cm}{5 cm} = \frac{BC}{13 cm}

AC = 7.2 cm

In the two triangles, AC and DF form the perpendicular and the length of the perpendicular is the height of the triangle. The area of a triangle is given by the formula,

Area = \frac{1}{2} * base * height

The area of triangle ABC divided by the area of triangle DEF can be calculated as,

\frac{Ar. ABC}{Ar. DE F } = \frac{1/2 * 3 * 7.2}{1/2 * 5 * 12}

            = \frac{9}{25}

            = 0.36

Therefore, the area of triangle ABC divided by area of triangle DEF is 0.36.

Similar questions