[tex] \pink{\rule{190pt}{5pt}}[/tex][tex]\huge\color{cyan}\boxed{\colorbox{black}{♛Answer♛}}[/tex]
1. Draw EM and DN perpendicular on BC.
So, MN = ED = 6 cm.
Therefore BM + CN = 12 cm - 6cm = 6 cm.
But BEDC is a Isosceles Trapezium
BE = DC (GIVEN)
EM = DN ( GIVEN)
∠EMB = ∠DNC (ASSUMED)
So △BEM ≅△DNC.[by RHS rule of congruency]
BM = CN (cpctc)
so BM = CN = 3 cm.
EM = DN = 13 cm - 8 cm = 5 cm.
Area of the polygon = Area of triangle AED + 2 × Area of triangle BEM + Area of rectangle EMND.
= (1/2 × 6 cm × 8 cm)+(2 × 1/2 × 3 cm × 5 cm)+(6 cm × 5 cm).
= 24 cm² + 15 cm² + 30 cm²
= 69 cm²
[tex] \pink{\rule{190pt}{5pt}}[/tex]
Answer:
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[tex] \pink{\rule{190pt}{5pt}}[/tex][tex]\huge\color{cyan}\boxed{\colorbox{black}{♛Answer♛}}[/tex]
1. Draw EM and DN perpendicular on BC.
So, MN = ED = 6 cm.
Therefore BM + CN = 12 cm - 6cm = 6 cm.
But BEDC is a Isosceles Trapezium
BE = DC (GIVEN)
EM = DN ( GIVEN)
∠EMB = ∠DNC (ASSUMED)
So △BEM ≅△DNC.[by RHS rule of congruency]
BM = CN (cpctc)
so BM = CN = 3 cm.
EM = DN = 13 cm - 8 cm = 5 cm.
Area of the polygon = Area of triangle AED + 2 × Area of triangle BEM + Area of rectangle EMND.
= (1/2 × 6 cm × 8 cm)+(2 × 1/2 × 3 cm × 5 cm)+(6 cm × 5 cm).
= 24 cm² + 15 cm² + 30 cm²
= 69 cm²
[tex] \pink{\rule{190pt}{5pt}}[/tex]
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Answer:
1. Draw EM and DN perpendicular on BC.
So, MN = ED = 6 cm.
Therefore BM + CN = 12 cm - 6cm = 6 cm.
But BEDC is a Isosceles Trapezium
BE = DC (GIVEN)
EM = DN ( GIVEN)
∠EMB = ∠DNC (ASSUMED)
So △BEM ≅△DNC.[by RHS rule of congruency]
BM = CN (cpctc)
so BM = CN = 3 cm.
EM = DN = 13 cm - 8 cm = 5 cm.
Area of the polygon = Area of triangle AED + 2 × Area of triangle BEM + Area of rectangle EMND.
= (1/2 × 6 cm × 8 cm)+(2 × 1/2 × 3 cm × 5 cm)+(6 cm × 5 cm).
= 24 cm² + 15 cm² + 30 cm²
= 69 cm²