Answer:
174 square centimetres
Step-by-step explanation:
The formula for finding the surface area of a rectangular prism is
SA = 2LW + 2LH + 2HW
Therefore surface area will be
= 2(11*5) + 2(11*2) +2(2*5)
= 2 * 55 + 2 * 22 + 2 * 10
= (110 + 44 + 20) square cm
Which gives you 174 square centimetres.
5. A. (4, -2)
6. C. (x, y) — (x, -y + 5)
Step-by-step explanation:
5. For the formula y = x, the x and y coordinates get swapped.
M = (-2, 4) — M’ = (4, -2)
6. If the coordinates get reflected across the x-axis, the y coordinates become negative.
(x, y) — (x, -y)
Now that the coordinates are reflected, you go 5 units up (+ 5) to get to the reflection of the coordinates if it was 5 units down before it reflected across the x-axis (- 5).
Ex. 1, 6 gets reflected across the x-axis and moved 5 units up. It’s reflection would be equivalent to (1, -1) because it moved 5 units down (1, 1) then reflected across the x-axis (1, -1).
(x, y - 5) reflected across the x-axis is equivalent to (x, -y + 5)
Answer:
C
Step-by-step explanation:
To make it easy let's start by organizing our information :
- AC=12 AND BD=8
- ABCD is a rhombus
- K and L are the midpoints of sides AD and CD
- we notice that the rhombus ABCD is divided into four right triangles
What do you think of when you hear a right triangle ?
- The pythagorian theorem !
AC and BD are khown so let's focus on them .
If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?
Simply because they are the diagonals of a rhombus .
ow let's apply the pythagorian theorem :
- (AC/2)² + (BD/2)² = BC²
- 6²+4²=52
- BC²= 52⇒
=BC
Now we khow that : AB=BC=CD=AD=
This isn't enough . Let's try to figure out a way to calculate the length of KL wich is the base of the triangle
- KL is parallel to AC
- k is the midpoint of AD and L of DC
I smell something . yes! Thales theorem
- KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE
- KL/12=
/2*
- KL/12=1/2⇒ KL=6
Now we have the length of the base kl
Now the big boss the height :
- notice that you khow the length of KL
- BD crosses kl from its midpoint and DL =
/2
What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D
- DL²=(KL/2)²+D²
- 52/4= 9+ D²
- D² = 52/4-9 +4 SO D=2
now the height of the trigle is H= BD-D= 8-2=6
NOw the area of the triangle is :
- A=(KL*H)/2 ⇒ A= (6*6)/2=18
THE ANSWER IS 18 SQ.UN