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Ksenya-84 [330]
3 years ago
15

What other information do you need to prove triangle INF=CAT by the ASA Postulate?

Mathematics
1 answer:
tekilochka [14]3 years ago
7 0
Label the angles on the given figure an x and y and label the unknown angles as z (see the figure below).
Because the sum of angles in a triangle = 180°, angle z is correct for all three triangles.

In order to prove that ΔINF = ΔCAT by the ASA Postulate, 
1.  We should show that AC = IN, or
2. We should show that AT = NF.
Proving either of the two options means that we have congruent isosceles triangles.

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The prime numbers which come between <br> 5 and 15.
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There are 7 boys in a class of 15 students ( a ) what is the ratioo of all students in the class to girls? (b) what is the ratio
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Find the other two vertices of a square with one vertex (0, 0) and another vertex (4, 2). Can you find another answer?
Margarita [4]

It is given that two vertices of square are (0,0) and (4,2).

Now the problem is that you haven't given that whether these two vertices are adjacent vertices or opposite vertices of the square.

1. By Supposing that these two are adjacent vertices of Square

The third vertex will be at (-4,2) which lies in third quadrant.

Suppose the coordinate of fourth vertex be (x,y).

Mid point of line joining (4,2) and (-4,2) is{ [4+(-4)]/2,(2+2)/2} is (0,2).

Mid point of line joining (x,y) and (0,0) is (x/2,y/2).

Since diagonals of square bisect each other,

∵ x/2=0

⇒x=0

and

y/2=2

⇒y=4

So, The Coordinate of  fourth vertex is (0,4).

Now coming back to second condition if these are two opposite vertex of Square.

Let the third coordinate be (a,b).

Length of diagonal=\sqrt{(4-0)^2+(2-0)^2}=\sqrt20=2\sqrt5

Now,let side of Square be A.

Then length of Diagonal of square =√2 A

⇒√2 A=2√5

⇒A =√10

As third vertex is (a,b).

Using distance formula

a² + b²=10  -------------(1)

(a-4)²+ (b-2)²=10  --------------(2)

Solving expression (1) and (2), we get

⇒a²+ b²=(a-4)² +(b-2)²

⇒2a + b =5

⇒b=5-2a

Putting the value of b in (1),we get

⇒a² +(5-2a)²=10

⇒a²+25+4a²-20a =10

⇒5a²-20a+15=0

⇒a² - 4a + 3=0

Splitting the middle term,we get

⇒(a-3)(a-1)=0

⇒a=3  ∧  a=1

we get b=5-2×1=3 and b=5-2×3=5-6=-1

So,the other vertex are (1,3) and(3,-1).





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