Question

Hello ♥️
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.​

Answer 1

Answer:

• ello xd

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

$∴ \: h \: = \sqrt{100 - 36} \\ \\ Area of trapezium \: = \: \frac{1}{2} \\ Sum of parallel sides×Distance  \\ between them.$

$= \frac{1}{2} (18 + 12) \times 8$

$= \frac{1}{2} (30) \times 8$

= 30 ×4 = 120 cm².

Answer 2

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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