1. Find the areas of given polygons
Answers
in trapezium BGFC;
parallel sides, a & b = 6 cm and 10 cm respectively
height, h = GF = 5 cm
so, ar(trapezium)
in parallelogram EFCD;
base, b = CF = 6 cm
height, h = 4.5 cm
so, ar(parallelogram)
= b x h = 6 x 4.5
= 27 cm² ... (ii)
in triangle BAG;
base, b = BG = 10 cm
height, h = AH = 3 cm
so, ar(triangle)
now,
area of the given polygon
= ar(trapezium) + ar(parallelogram) + ar(triangle)
= 40 cm² + 27 cm² + 15 cm² ... (from i, ii and iii)
= 82 cm².
Hence, area of the given polygon is 82 cm².
Step-by-step explanation:
in trapezium BGFC;
parallel sides, a & b = 6 cm and 10 cm respectively
height, h = GF = 5 cm
so, ar(trapezium)
\begin{gathered} = \frac{h}{2} (a + b) \\ = \frac{5}{2} (6 + 10) = \frac{5}{2} \times 16 = 5 \times 8 \\ = 40 \: {cm}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (i)\end{gathered}=2h(a+b)=25(6+10)=25×16=5×8=40cm2...(i)
in parallelogram EFCD;
base, b = CF = 6 cm
height, h = 4.5 cm
so, ar(parallelogram)
= b x h = 6 x 4.5
= 27 cm² ... (ii)
in triangle BAG;
base, b = BG = 10 cm
height, h = AH = 3 cm
so, ar(triangle)
\begin{gathered} = \frac{1}{2} \times b \times h \\ = \frac{1}{2} \times 10 \times 3 = \frac{1}{2} \times 30 \\ = 15 \: {cm}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (iii)\end{gathered}=21×b×h=21×10×3=21×30=15cm2...(iii)
now,
area of the given polygon
= ar(trapezium) + ar(parallelogram) + ar(triangle)
= 40 cm² + 27 cm² + 15 cm² ... (from i, ii and iii)
= 82 cm².
Hence, area of the given polygon is 82 cm².