If the side lengths of ABC and def are 6,5,8 and 30,25,40 respectively, So this triangle - yes, or no - is similar to a triangle, according to-

If the side lengths of ABC and def are 6,5,8 and 30,25,40 respectively, So this triangle - yes, or no - is similar to a triangle, according to-

My heart flies_ 815：
ABC and def are similar triangles
Reason: the three sides 30, 25 and 40 of △ def are 5 times of the three sides 6, 5 and 8 of △ ABC
A similar triangle is proportional to its corresponding edge. Conversely, a similar triangle is proportional to its corresponding edge
Good luck and goodbye

If we know that the triangle ABC is equal to the triangle def, and ab: de = 2:3, the area of the triangle ABC + the area of the triangle def = 75, then the surface of the triangle def
Given that the triangle ABC is equal to the triangle def, and ab: de = 2:3, the area of triangle ABC + the area of triangle def = 75, what is the area of triangle def?

First of all, it should be similar
The area ratio of similar triangles is the square of the length ratio of each side
So if the area of DEF is 9x, the area of abd is 4x
9x+4x=75
X = 75 / 13
So the area of DEF is 9 * 75 / 13

As shown in the figure, triangle ABC and triangle def are isosceles right triangles respectively. Given DF = 6, ab = 5, EB = 2.6, the area of the shadow part is______ ．

S △ AGH = 12 × 2.4 × 2.4 × 12 = 1.44; BF = ef-eb = 3.4, so FC = bc-bf = 5-3.4 = 1.6; s △ FCI = 12 × 1.6 × 1.6 = 1.28; so the area of shadow is: s shadow = s △ abc-s △ agh-s △ FCI = 12 × 5 × 5-1.44-1.28 = 12.5-1.44-1.28 = 9.78; answer: the area of shadow is 9.78

It is known that the circumference of a triangle is 15cm, and both sides of the triangle are equal to twice of the third side

It is known that the perimeter of a triangle is 15cm, and both sides of the triangle are equal to twice of the third side, then the shortest side of the triangle is ()
A. 1cm B. 2cm C. 3cm D. 4cm

Let the shortest side of the triangle be x cm, which means x + 2x + 2x = 15, 5x = 15, x = 3. So the shortest side of the triangle is 3 cm

If the perimeter of a triangle is 15cm, and both sides are equal to twice of the third side, then the shortest side length of the triangle is?

3cm

It is known that the perimeter of a triangle is 15cm, and the length of both sides is equal to twice the length of the third side, then the longest side of the triangle is 15cm______ ．

Let the length of the third side of the triangle be xcm, and the length of the two longer sides be 2xcm. According to the meaning of the question, there is 2 × 2x + x = 15. If x = 3, then 2x = 6. So the longest side of the triangle is 6cm. So the answer is 6cm

Given that point n (3,1), point a and point B are on the straight line y = x and y = 0 respectively, then the minimum value of the circumference of △ ABN is___ ．

According to the meaning of the question, make the symmetrical point m of n about X axis and the symmetrical point D of n about y = x, connect the intersection X axis of MD to a and the intersection line y = x to B, then the perimeter of △ ABN is the minimum, that is, the length of DM is the minimum perimeter of △ ABN, and m (3, - 1), D (1,3) can be obtained from the distance formula between two points, so | DM | = (3-1) 2 + (- 1-3) 2 = 25, so the answer is: 25

If the point R (3,5) is known and the moving point P.Q is on the X-Y + 1 + = 0 and Y axis respectively, then the minimum perimeter of the triangle PQR =?

Let R (3,5) be the symmetric point R '(- 3,5) about the y-axis, and the symmetric point R' '(4,4) about the line X-Y + 1 + = 0
From the shortest line between two points, the
The minimum perimeter of triangle PQR = r'r '' = 5 √ 2

1. If n (3, l) is known and points a and B are on the line y = x and y = 0 respectively, then the minimum perimeter of triangle ABN is?
I've asked this question many times. Many people have got it wrong. Some of them answered correctly but didn't explain it. I'm really distressed. It doesn't count. I'll explain my ideas

First, make the symmetric point n '(1,3) of N point with respect to the straight line y = x, and then make the symmetric point n' '(3, - 1) of N point with respect to the x-axis to connect n'n' ', and the intersection points with y = x and y = 0 are a and B respectively, then an = an', BN = BN '', the minimum perimeter of triangle ABN = | n'n ''; = √ [(1-3) & sup2; + (3 + 1) & sup2;] = 2 √ 5 (∵ the distance between two points is linear