Question

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Answer 1

3)

Given:-

• Parallel sides of Trapezium = 19cm and 13cm
• Non-parallel sides are equal and each measure as 5cm

Find:-

• Area of Trapezium

Solution:-

In ∆ABC

▶H² = P² + B² [Pythogoras Theorem]

▶H² = 5² + 4²

▶H² = 25 + 16

▶H² = 41

▶H = √(41)

▶H = 6.40cm

we, know that

$\large{ \underline{\boxed{ \sf Area \: of \: Trapezium =area \: of \: rectangle + 2(area \: of \: triangle)}}}$

$\large{ \underline{\boxed{ \sf Area \: of \: rectangle = l \times b}}}$

$\large{ \underline{\boxed{ \sf Area \: of \: triangle = \dfrac{1}{2} \times b \times h}}}$

So,

$\dashrightarrow\sf Area \: of \: Trapezium =area \: of \: rectangle + 2(area \: of \: triangle)$

$\dashrightarrow\sf Area \: of \: Trapezium =l \times b+ 2 \bigg( \dfrac{1}{2} \times b \times h\bigg)$

where,

• Breadth, b = 6.40cm
• Length, l = 19cm
• Height, h = 6.40cm
• Base, b = 4cm

So,

$\dashrightarrow\sf Area \: of \: Trapezium =19\times6.40+ 2 \bigg( \dfrac{1}{2} \times 4 \times 6.40\bigg)$

$\dashrightarrow\sf Area \: of \: Trapezium =121.6+ 2 \bigg( \dfrac{1}{2} \times 4 \times 6.40\bigg)$

$\dashrightarrow\sf Area \: of \: Trapezium =121.6+ 2 \bigg( \dfrac{1}{2} \times25.6\bigg)$

$\dashrightarrow\sf Area \: of \: Trapezium =121.6+ 2 \times 12.8$

$\dashrightarrow\sf Area \: of \: Trapezium =121.6+ 25.6$

$\dashrightarrow\sf Area \: of \: Trapezium =147.2 {cm}^{2}$

Hence, Area of given Trapezium = 147.2cm²

____________________________

4)

Given:-

• Perimeter of Trapezium = 48cm
• Non-parallel sides = 10cm
• Height = 8cm

Find:-

• Area of Trapezium

Solution:-

AD = BC = 10cm

DE = 8cm

➟Perimeter = Sum of all sides

➟Perimeter = AB + BC + CD + DA

where,

Perimeter = 48cm

BC = 10cm = DA = 10cm

So,

➟ 48 = AB + 10 + CD + 10

➟ 48 = AB + CD + 10+10

➟ 48 = AB + CD + 20

➟ 48 - 20 = AB + CD

➟ AB + CD = 28cm

Now, we know that

$\large{ \underline{\boxed{ \sf Area \: of \: Trapezium = \dfrac{1}{2} \times (sum \: of \: parallel \: sides) \times h}}}$

where,

• AB + CD = 28cm
• DE = 8cm

So,

$\implies\sf Area \: of \: Trapezium = \dfrac{1}{2} \times (AB + CD) \times DE$

$\implies\sf Area \: of \: Trapezium = \dfrac{1}{2} \times (28) \times 8$

$\implies\sf Area \: of \: Trapezium = \dfrac{1}{2} \times 28 \times 8$

$\implies\sf Area \: of \: Trapezium = \dfrac{1}{2} \times224$

$\implies\sf Area \: of \: Trapezium = \dfrac{224}{2}$

$\implies\sf Area \: of \: Trapezium = 112 {cm}^{2}$

Hence, Area of Trapezium = 112cm²

____________________________

5)

Given:-

• Parallel Sides = 34m and 10m
• Distance between them = 5m

Find:-

• Area of Trapezium

Solution:-

we, know that

$\large{ \underline{\boxed{ \sf Area \: of \: Trapezium = \dfrac{1}{2} \times (sum \: of \: parallel \: sides) \times h}}}$

where,

• Sum of parallel sides = (34 + 10)m
• h = 5m

So,

$\dashrightarrow\sf Area \: of \: Trapezium = \dfrac{1}{2} \times (sum \: of \: parallel \: sides) \times h$

$\dashrightarrow\sf Area \: of \: Trapezium = \dfrac{1}{2} \times (34 + 10) \times 5$

$\dashrightarrow\sf Area \: of \: Trapezium = \dfrac{1}{2} \times (44) \times 5$

$\dashrightarrow\sf Area \: of \: Trapezium = \dfrac{1}{2} \times 44 \times 5$

$\dashrightarrow\sf Area \: of \: Trapezium = \dfrac{1}{2} \times 220$

$\dashrightarrow\sf Area \: of \: Trapezium = \dfrac{220}{2}$

$\dashrightarrow\sf Area \: of \: Trapezium = 110m^2$

Hence, Area of Trapezium = 110m²