Area and perimeter

Advanced Math 4 · Unit 11 · area-perimeter-rectilinear · Teacher edition

area-perimeter-rectilinear

Name ______________________________Date ______________

1.
A rectangle is $9$ cm long and $4$ cm wide. What is its perimeter, in cm?
Answer: ______________
2.
An L-shaped patio has six sides, and every corner is a right angle. Its bottom edge is $12$ m long and its left edge is $7$ m tall. Its top edge is $5$ m long and its right edge is $3$ m tall. What is the area of the patio, in m²?
Answer: ______________
3.
A rectangle is $6$ cm by $4$ cm. A second rectangle has double the length and double the width. What is the area of the second rectangle, in cm²?
1. (A)
  $96$
2. (B)
  $48$
3. (C)
  $40$
4. (D)
  $24$

Answer key — Area and perimeter

1.
26area-perimeter-rectilinear-01
$P = 2(9 + 4) = 2 \times 13 = \mathbf{26}$ cm. (The area, $9 \times 4 = 36$ cm², is a different question.)
2.
56area-perimeter-rectilinear-02
Whole rectangle: $12 \times 7 = 84$ m². Cut-out corner: $(12 - 5) \times (7 - 3) = 7 \times 4 = 28$ m². Patio: $84 - 28 = \mathbf{56}$ m². Check by splitting instead: a $12 \times 3$ strip along the bottom ( $36$ ) plus a $5 \times 4$ block above it ( $20$ ) gives $56$ .
3.
(A)
$96$
area-perimeter-rectilinear-03
The new rectangle is $12$ cm by $8$ cm, so its area is $12 \times 8 = \mathbf{96}$ cm² — four times the original $24$ . ( $48$ doubles the old area, which is the trap; $40$ is the new perimeter, not an area; $24$ is the original area.)