Question

$\sf\fbox\green{question = > }$

Find the area of the trapezium ABCD in which AB//DC, AB = 18 cm, B = ZC = 90°, CD = 12 cm and AD = 10 cm. ​

Answer 1

Answer:

${\huge{\underline{\small{\mathbb{\pink{REFER \ TO \ THE \ ATTACHMENT}}}}}}$

${10}^{2} = {6}^{2} + x_{o} ^{2} \\ \\ { x_{o}}^{2} = 100 - 36 \\ \\ { x_{o}}^{2} = 64 \\ \\ x_{o} = \sqrt{64} \\ \\ \boxed{ = 8cm}$

Now,

Area of ABCD

$\sf = \frac{1}{2} \times ( sum \: of \parallel \: sides) \times height \\ \\ = \frac{1}{2} \times (12 + 18) \times 8 \\ \\ = \frac{1}{ \cancel2} \times { 30} \: \times \cancel8 \: \: ^{4} \\ \\ =30 \times 4 \\ \\ \green{\boxed{ \red{= {120cm}^{2} }}}$

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Additional information:-

• A trapezium is a quadrilateral having two parallel sides of unequal length and the other two sides are non-parallel.
• All quadrilaterals' interior angles sum to 360°. In isosceles trapezoids, the two top angles are equal to each other. Similarly, the two bottom angles are equal to each other as well.
• The parallel sides of a trapezium are called bases and the non-parallel sides of a trapezium are called legs. It is also called a trapezoid.
• In a trapezoid, the parallel sides are known as the bases, while the pair of non-parallel sides are known as the legs. The perpendicular distance between the two parallel sides of a trapezium is known as a trapezoid height. 

Answer 2

Answer:

Step-by-step explanation:

rapezium ABCD is drawn in figure.

AE=(18−12) cm =6 cm

Δ AED is a right-angled Δ

∴(AD)

2

+(ED)

2

+(AE)

2

10

2

=h

2

+6

2

∴h=

100−36

​

h=8 cm

Area of trapezium=

2

1

​

×Sum of parallel sides×Distance between them.

=

2

1

​

(18+12)×8

=

2

1

​

(30)×8

=30×4 =120 cm

2

.