Question

Can you plz help me with this question?

HOW MANY ISOSCELES TRIANGLES WITH INTEGER SIDES ARE POSSIBLE SUCH THAT THE SUM OF ITS TWO OF THE SIDE IS 18?



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Answer 1

Answer:

I am Giving A example

Step-by-step explanation:

There are 2 possibilities

1)equal sides add up to 12 .

2)unequal sides add up to 12.

Case 1

Sum of 2 equal side add up to 12.

Let x be the unequal side.

Equal sides must be 6 to add up to 12

So the triangle has sides 6,6,x.

x can vary from 1-11 for 6,6,x to be a triangle(sum of 2 sides must be greater than the third side)

So,we get 11 triangles from case 1

Case 2

2 unequal sides add up to 12

This can be

1+11 , 2+10 , 3+9 , 4+8 , 5+7

Possible triangles from above combinations

1,11,11

2,10,10

3,9,9

4,8,4

5,7,7

5,5,7

So,we get 6 triangles from case 2.

Total no. of triangles =17

Sorry,

It will be 4,8,8 in case 2 not 4,8,4

Answer will be 17

U will solve it by yourself because now you have example

Answer 2

Answer:

Let us name the three sides as x,y,z.

According to the triangle’s property, sum of two sides is always greater than the third side.

i.e x+y>z

Given info - sum of two sides is 18

x+y=18

one case where x=y

so x =9

now z<18

so z will have 17 values 1- 17

When the sum of two unequal sides is equal to 18

x+y=18 and z=x

so it will give the following 8 combination.

z x y

1) 17 , 1 , 17

2) 10 , 8 , 10

3) 15 , 3 , 15

4) 12 , 6 , 12

5) 16 , 2 , 16

6) 14 , 4 , 14

7) 13 , 5 , 13

8) 11 , 7 , 11

pls mark Brainliest.

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