Given that - ABC - DEF-BC - 3 cm-EF - 4 cm ans the area of - ABC is 54 sq-cm-Then find the area of - DEF-

Solution REF. Image Area ABCArea DEF= ( BCEF )^2 Area DEF = 54×4^23^2 Area ( DEF)= 6× 4^2 = 96 cm^2 ...

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If Two Sides of a Triangle Are Unequal, Then the Angle Opposite to the Greater Side Is Greater Than Angle Opposite to the Smaller Side

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Question

Given that $Δ A B C \sim Δ D E F, B C = 3 c m, E F = 4 c m$ ans the area of $Δ A B C$ is $54 s q . c m .$ Then find the area of $Δ D E F .$

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Solution

Solution -
REF. Image

$\frac{A r e a △ A B C}{A r e a △ D E F} = {(\frac{B C}{E F})}^{2}$

Area $△ D E F = 54 \times \frac{4^{2}}{3^{2}}$

Area $(△ D E F) = 6 \times 4^{2}$

$= 96 c m^{2}$

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Similar questions

Q. If

Δ A B C \sim Δ D E F

B C = 3 c m

E F = 4 c m

and area of

Δ A B C = 54 c m^{2}

then find the area of

Δ D E F

If $Δ A B C$ is similar to $Δ D E F$ such that BC = 3 cm, EF = 4 cm and area of $Δ A B C = 54 {c m}^{2}$ . Determine the area of $Δ D E F$ . [4 MARKS]

Δ A B C \sim Δ D E F

, area of

Δ A B C = 64 c m^{2}

and area of

Δ D E F = 121 c m^{2}

. If

E F = 15.4 c m

, find

B C

If $Δ A B C \sim Δ D E F$ such that area of $Δ A B C$ is $9 c m^{2}$ and the area of $Δ D E F$ is $16 c m^{2}$ and BC = 2.1 cm. Find the length of EF. [2 MARKS]

Q. ∆ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If ∆DEF ∼ ∆ABC and EF = 4 cm, then perimeter of ∆DEF is

(a) 7.5 cm
(b) 15 cm
(c) 22.5 cm
(d) 30 cm

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Given that ΔABC∼ΔDEF,BC=3cm,EF=4cm ans the area of ΔABC is 54sq.cm.Then find the area of ΔDEF.

Solution -REF. ImageArea△ABCArea△DEF=(BCEF)2Area △DEF=54×4232Area (△DEF)=6×42=96cm2

Given that $Δ A B C \sim Δ D E F, B C = 3 c m, E F = 4 c m$ ans the area of $Δ A B C$ is $54 s q . c m .$ Then find the area of $Δ D E F .$

Solution -
REF. Image

$\frac{A r e a △ A B C}{A r e a △ D E F} = {(\frac{B C}{E F})}^{2}$

Area $△ D E F = 54 \times \frac{4^{2}}{3^{2}}$

Area $(△ D E F) = 6 \times 4^{2}$

$= 96 c m^{2}$