Question

find the altitude of trapezium whose area is 105cm square and bases are 14cm and 7cm​

Answer 1

Given :-

• A trapezium whose area is 105cm square and bases are 14cm and 7cm.

To find :-

• Altitude of trapezium

Solution :-

• Area of trapezium = 105 cm²

• Bases of trapezium = 14cm and 7cm

As we know that

→ Area of trapezium = ½ (a + b) × h

Where " (a + b) " is sum of parallel sides and " h " is Altitude.

• According to the question

→ Bases of trapezium = parallel sides

→ Area of trapezium = 105 cm²

→ ½ (a + b) × h = 105

→ ½ (14 + 7) × h = 105

→ ½ × 21 × h = 105

→ 21/2 × h = 105

→ h = 105 × 2/21

→ h = 10 cm

Therefore,

• Height of trapezium is 10cm

More to know :-

• Area of circle = πr²

• Circumference of circle = 2πr

• Perimeter of rectangle =2(l + b)

• Perimeter of square = 4 × side

• Area of square = side × side

• Area of rhombus = ½ × product of diagonals

• Area of Parallelogram = base × height

Answer 2

Answer:

✯ Given :-

• Area of the trapezium is 105 cm² and base is 14 cm and 7 cm.

✯ To Find :-

• What is the altitude of trapezium.

✯ Formula Used :-

Area of trapezium = $\dfrac{1}{2}$(a + b) × h

✯ Solution :-

⋆ Given :-

• Area = 105 cm²
• Base = 14 cm and 7 cm

➣ According to the question by using the formula we get,

⇒ $\dfrac{1}{2}$(14 + 7) × h = 150

⇒ $\dfrac{1}{2}$ × 21 × h = 150

⇒ $\dfrac{21}{2}$ × h = 105

⇒ h = $\dfrac{2}{21}$ × 105

➠ h = 10 cm

$\therefore$ The height of the trapezium is $\boxed{\bold{\large{\rm{\green{10\: cm}}}}}$.

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