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Gennadij [26K]
3 years ago
13

What else would need to be congruent to show that ABC = XYZ by ASA?

Mathematics
1 answer:
Rom4ik [11]3 years ago
8 0

Answer:

Option B AC≅XZ

Step-by-step explanation:

we know that

ASA (angle, side, angle) means that we have two congruent triangles where we know two angles and the included side are equal

In this problem we have

∠X≅∠A

∠Z≅∠C

In the triangle ABC the included side between the angles ∠A and ∠C is the side AC

and

In the triangle XYZ the included side between the angles ∠X and ∠Z is the side XZ

therefore

AC≅XZ

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Step-by-step explanation:

3

Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

Therefore, by mid-point formula:

D = ( \frac{2 + 5}{2},  \:  \:  \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \:  \:  \frac{ 1}{2} ) \\ \therefore D= (3.5, \:  \: 0.5) \\  \& \: A=(-1,\:\:4)...(given) \\\\  now \: by \: distance \: formula \\  \\ Length  \: of \:  segment  \: AD \\  =  \sqrt{( - 1 - 3.5)^{2}  +  {(4 - 0.5)}^{2} }  \\ =  \sqrt{(4.5)^{2}  +  {(3.5)}^{2} }  \\ =  \sqrt{20.25 + 12.25 }  \\  =  \sqrt{32.5}  \\    \red{ \boxed{\therefore Length  \: of \:  segment  \: AD  = 5.7 \: units}}

4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

LHS = 3x - y-5\\=3\times 4-7-5\\= 12-12\\=0\\= RHS

Hence, point C (4, 7) lie on the straight line AB.

4(b)

Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

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