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3 years ago

Are the two triangles below similar?

Mathematics

2 answers:

gizmo_the_mogwai [7]3 years ago

8 0

Answer:

Yes, because there are two pairs of congruent corresponding angles.

Step-by-step explanation:

Two triangles ABC and DEF are given .

The measure of angles in triangle ABC are:

m<A=30°,m< C=65°,

m<B= 180-(30+65)=85°(measure of sum of angles in any triangle is 180°)

In triangle DEF the measure of the angles are :

m<D=30°,m<E=85°,m<F=180-(30+85)=65°(sum of angles in a triangle add to 180°)

In triangles ABC and DEF,

< A=<D=30°

<B=<E=85°

<C=<F=65°

The two triangles are similar by AAA property.

Luda [366]3 years ago

3 0

Answer :- Yes the two triangles are similar because there are two pairs of congruent corresponding angles.

Explanation:-

In Δ ABC

∠A=30° , ∠C=65 °

By angle sum property of triangle

∠A + ∠B + ∠C= 180°

⇒∠B= 180°-∠A-∠C=180°-30°-65°=85°

⇒∠B=85°

Now in ΔABC and ΔDEF

∠A=∠D=30° and ∠B=∠E=85°

⇒ there are two pairs of congruent corresponding angles.

So by AA-similarity criteria

ΔABC ≈ ΔDEF

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Read 2 more answers

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

eduard

The Lagrangian is

$L(x_1,\ldots,x_n,\lambda_1,\ldots,\lambda_n)=x_1+\cdots+x_n+\lambda_1({x_1}^2+\cdots+{x_n}^2)+\cdots+\lambda_n({x_1}^2+\cdots+{x_n}^2)$

with partial derivatives (set equal to 0)

$\dfrac{\partial L}{\partial x_i}=1+2x_i(\lambda_1+\cdots+\lambda_n)=0$

$\dfrac{\partial L}{\partial\lambda_i}={x_1}^2+\cdots+{x_n}^2-36=0$

for each $1\le i\le n$ .

Let $\Lambda$ be the sum of all the multipliers $\lambda_i$ ,

$\Lambda=\displaystyle\sum_{k=1}^n\lambda_k=\lambda_1+\cdots+\lambda_n$

We notice that

$x_i\dfrac{\partial L}{\partial x_i}=x_i+2{x_i}^2\Lambda=0$

so that

$\displaystyle\sum_{i=1}^nx_i\dfrac{\partial L}{\partial x_i}=\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0$

We know that $\sum\limits_{i=1}^n{x_i}^2=36$ , so

$\displaystyle\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0\implies\sum_{i=1}^nx_i=-72\Lambda$

Solving the first $n$ equations for $x_i$ gives

$1+2\Lambda x_i=0\implies x_i=-\dfrac1{2\Lambda}$

and in particular

$\displaystyle\sum_{i=1}^nx_i=-\dfrac n{2\Lambda}$

It follows that

$-\dfrac n{2\Lambda}+72\Lambda=0\implies\Lambda^2=\dfrac n{144}\implies\Lambda=\pm\dfrac{\sqrt n}{12}$

which gives us

$x_i=-\dfrac1{2\left(\pm\frac{\sqrt n}{12}\right)}=\pm\dfrac6{\sqrt n}$

That is, we've found two critical points,

$\pm\left(\dfrac6{\sqrt n},\ldots,\dfrac6{\sqrt n}\right)$

At the critical point with positive signs, $f(x_1,\ldots,x_n)$ attains a maximum value of

$\displaystyle\sum_{i=1}^nx_i=\dfrac{6n}{\sqrt n}=6\sqrt n$

and at the other, a minimum value of

$\displaystyle\sum_{i=1}^nx_i=-\dfrac{6n}{\sqrt n}=-6\sqrt n$

4 0

4 years ago