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2 years ago

Based only on the information given in the diagram, it is guaranteed that ABC~DEF

Mathematics

2 answers:

slamgirl [31]2 years ago

4 0

False because it doesn’t meet any triangle congruence theoreme

Varvara68 [4.7K]2 years ago

4 0

Answer: False.

Step-by-step explanation:

To prove any pair of triangles similar , some necessary things are needed :-

1) At least two angles :-Two prove a pair of angles are similar , we need at-least two corresponding angles are congruent for AA-similarity criteria.

2) Two sides and one included angle :- To apply SAS similarity criteria , we need two sides of one triangle proportional two two sides of another traingle and one pair of congruent corresponding angle.

3) Three sides :-To apply SSS similarity criteria , we need all the three sides of triangle proportional to the sides of the another triangle.

In the given figure we have given only one angle and one side in each triangle.

To apply any similarity criteria to prove then similar , we need at least an angle or side CB and EF or all the sides.

Hence, on the basis of the given diagram it is not guaranteed that ABC~DEF.

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if the volume of the prism is represented by 15x2 + x + 2 and the height is x2, which expression represents B, the area of the b

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1 year ago

Please answer all of them need this

VikaD [51]

First Question

For a better understanding of the solution provided here please find the first attached file which has the diagram of the the isosceles trapezoid.

We dropped perpendiculars from C and D to intersect AB at Q and P respectively.

As can be seen in $\Delta BCQ$ , we can easily find the values of CQ and BQ.

Since, $Sin(75^0)=\frac{CQ}{8}$

$\therefore CQ=8\times Sin(75^0)\approx 7.73$ ft

In a similar manner we can find BQ as:

$Cos(75^0)=\frac{BQ}{8}$

$BQ\approx2.07$ ft

All these values can be found in the diagram attached.

Thus, because of the inherent symmetry of the isosceles trapezoid, PQ can be found as:

$PQ=22-(AP+QB)=22-(2.07+2.07)=17.86$

Let us now consider $\Delta AQC$

We can apply the Pythagorean Theorem here to find the length of the diagonal AC which is the hypotenuse of $\Delta AQC$ .

$AC=\sqrt{(AQ)^2+(QC)^2}=\sqrt{(AP+PQ)^2+(QC)^2}=\sqrt{(2.07+17.86)^2+(7.73)^2}\approx21.38$ feet.

Thus, out of the given options, Option B is the closest and hence is the answer.

Second Question

For this question we can directly apply the formula for the area of a triangle using sines which is as:

Area= $\frac{1}{2}$ (First Side)(Second Side)(Sine of the angle between the two sides)

Thus, from the given data,

Area= $\frac{1}{2}\times 218.5\times 224.5\times sin(58.2^0)\approx20845$ $m^2$

Therefore, Option D is the correct option.

Third Question

For this question we will apply the Sine Rule to the $\Delta ABC$ given to us.

Thus, from the triangle we will have:

$\frac{AB}{Sin(\angle C)}=\frac{BC}{Sin(\angle A)}$

$\frac{c}{Sin(\angle C)}=\frac{a}{Sin(\angle A)}$

$\frac{17}{Sin(25^0)}=\frac{a}{Sin(45^0)}$

This gives $a$ to be:

$a\approx28.44$

Which is not close to any of the given options.

Fourth Question

Please find the second attachment for a better understanding of the solution provided her.

As can be clearly seen from the attached diagram, we can apply the Cosine Rule here to find the return distance of the plane which is CA.

$AC=\sqrt{(AB)^2+(BC)^2-2(AB)(BC)\times Cos(\angle B)}$

$\therefore AC=\sqrt{(172.20)^2+(111.64)^2-2(172.20)(111.64)\times Cos(177.29^0)}\approx283.8$ miles.

Thus, Option D is the answer.

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