Answer:
397.7 m²
Step-by-step Explanation:
Step 1: find m < W
W = 180 - (33+113) (sum of ∆)
W = 34°
Step 2: find side UV using the law of sines
Multiply both sides by sin(34)
(approximated)
Step 3: find the area using the formula, ½*UV*VW*sin(V)
area = ½*29.8*29*sin(113)
Area = 397.7 m² (rounded to the nearest tenth.
Answer:
the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.
Step-by-step explanation:
From the given question.
Let p be the length of the of the printed material
Let q be the width of the of the printed material
Therefore pq = 2400 cm ²
q =
To find the dimensions of the poster; we have:
the length of the poster to be p+30 and the width to be
The area of the printed material can now be:
=
Let differentiate with respect to p; we have
Also;
For the smallest area
p² = 3600
p =√3600
p = 60
Since p = 60 ; replace p = 60 in the expression q = to solve for q;
q =
q =
q = 40
Thus; the printed material has the length of 60 cm and the width of 40cm
the length of the poster = p+30 = 60 +30 = 90 cm
the width of the poster = =
= 40 + 20 = 60
Hence; the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.
