A triangle ABC with sides BC -7 cm - - B -45- - A -105- is given- The image of constructing a similar triangle of - ABC whose sides are 3-1 times the corresponding sides of the - ABC is given below- Then A- C - AC is equal to -

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Byju's Answer

Standard X

Mathematics

Construction of a Similar Triangle, When One Vertex Is Common

A triangle AB...

Question

A triangle ABC with sides BC =7cm, $∠ B = 45^{\circ}, ∠ A = 105^{\circ}$ is given. The image of constructing a similar triangle of $Δ A B C$ whose sides are $\frac{3}{4}$ times the corresponding sides of the $Δ A B C$ is given below. Then $\frac{A^{'} C^{'}}{A C}$ is equal to :

\frac{3}{4}

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\frac{4}{3}

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\frac{3}{7}

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\frac{4}{7}

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Solution

The correct option is A $\frac{3}{4}$
As we draw $Δ A^{'} C^{'} B$ similar to $Δ A C B$ with corresponding side ratio is $\frac{3}{4}$ :
$\frac{B A^{'}}{B A} = \frac{B C^{'}}{B C} = \frac{A^{'} C^{'}}{A C} = \frac{3}{4}$

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Similar questions

Q. A triangle ABC with sides BC =7cm,

∠ B = 45^{\circ}, ∠ A = 105^{\circ}

is given. The image of constructing a similar triangle of

Δ A B C

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\frac{3}{4}

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Δ A B C

is given below.

The condition used to draw A'C' parallel to AC:

Q. A triangle ABC with sides BC =7cm,

∠ B = 45^{\circ}, ∠ A = 105^{\circ}

is given. The image of constructing a similar triangle of

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whose sides are

\frac{3}{4}

times the corresponding sides of the

Δ A B C

is given below.

The condition used to draw A'C' parallel to AC:

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and

∠ A B C = 60^{\circ}

The image of constructing a similar triangle of

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times the corresponding sides of the

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is a/an :

Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are times the corresponding side of ΔABC. Give the justification of the construction.

Question 6
Draw a triangle ABC with side $B C = 7 c m, ∠ B = 45^{\circ}, ∠ A = 105^{\circ}$ . Then, construct a triangle whose sides are 4/3 times the corresponding sides of $Δ A B C$ .

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A triangle ABC with sides BC =7cm, ∠B=45∘,∠A=105∘ is given. The image of constructing a similar triangle of ΔABC whose sides are 34 times the corresponding sides of the ΔABC is given below. Then A′C′AC is equal to :

The correct option is A 34As we draw ΔA′C′B similar to ΔACB with corresponding side ratio is 34: BA′BA=BC′BC=A′C′AC=34

A triangle ABC with sides BC =7cm, $∠ B = 45^{\circ}, ∠ A = 105^{\circ}$ is given. The image of constructing a similar triangle of $Δ A B C$ whose sides are $\frac{3}{4}$ times the corresponding sides of the $Δ A B C$ is given below. Then $\frac{A^{'} C^{'}}{A C}$ is equal to :

The correct option is A $\frac{3}{4}$
As we draw $Δ A^{'} C^{'} B$ similar to $Δ A C B$ with corresponding side ratio is $\frac{3}{4}$ :
$\frac{B A^{'}}{B A} = \frac{B C^{'}}{B C} = \frac{A^{'} C^{'}}{A C} = \frac{3}{4}$