Answer: ΔABC, ΔDEF, and ΔGHI
Step-by-step explanation:
Given: In ΔABC, ΔDEF, and ΔGHI:
AB = DF AB = GI
BC = HI DE = HI
m∠B = m∠D = m∠I
In ΔABC and ΔGHI
AB = GI [given]
BC = HI [given]
m∠B = m∠I [given]
[ here m∠B and m∠I are the included angle of ΔABC and ΔGHI]
∴ ΔABC ≅ ΔGHI [by SAS congruence postulate]
In ΔABC and ΔDEF
AB = DF [given]
BC = DE [ Since BC = HI and DE = HI so by transitive property BC = DE]
m∠B = m∠D [given]
[ here m∠B and m∠D are the included angle of ΔABC and ΔDEF]
∴ ΔABC ≅ ΔDEF [by SAS congruence postulate]
Now, since ΔABC ≅ ΔGHI and ΔABC ≅ ΔDEF
⇒ ΔGHI ≅ ΔDEF [transitive property]
Hence, all the given triangles ΔABC, ΔDEF, and ΔGHI are con gruent to each other.
Answer:
side: 4 metres
perimeter: 16 metres
Step-by-step explanation:
Let's first find the area of this rectangle.
The area of a rectangle is denoted by A = lw, where l is the length and w is the width. Here, the length is l = 6.4 and the width is w = 2.5. Plug these in:
A = lw
A = 6.4 * 2.5 = 16 metres squared
We want to find the side of a square with area 16. Suppose the side length is x. The area of a square is denoted by A = x * x = x², so set this equal to 16:
x² = 16
x = √16 = 4
Thus, the side length is 4 metres.
The perimeter of a square is denoted by P = 4s, where s is the side length.
Here the side length is 4 metres, as we found, so:
P = 4s = 4 * 4 = 16
Hence the perimeter is 16 metres.