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Find the area of the trapezium ABCD in which AB//DC, AB = 18 cm, ZB = 18 cm, ZB = ZC = 90° CD = 12 cm and AD = 10 cm.
Find the area of the trapezium ABCD in which AB//DC, AB = 18 cm, ZB = 18 cm, ZB = ZC = 90° CD = 12 cm and AD = 10 cm.
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Answer:
Trapezium ABCD is drawn in figure.
AE=(18−12) cm =6 cm
Δ AED is a right-angled Δ
∴ (AD)² +(ED)² + (AE)²
10² = h² + 6²
[tex]∴ \: h \: = \sqrt{100 - 36} \\ \\ Area of trapezium \: = \: \frac{1}{2} \\ Sum of parallel sides×Distance \\ between them.[/tex]
[tex] = \frac{1}{2} (18 + 12) \times 8[/tex]
[tex] = \frac{1}{2} (30) \times 8[/tex]
= 30 ×4 = 120 cm².
Answer:
Area of trapezium=1/2× sum of parallel sides × distance
therefore,Area=1/2(18+12)×8
=15×8
=120cm^2 unit
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