Question

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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

Given:-

• Area of trapezium = 105cm²
• base = 14cm and 7cm

To find:-

• altitude of this trapezium

Formula to be used:-

$\underline{\boxed{\sf area\: of \: trapezium =\dfrac{1}{2} × (sum\: of \: parallel\: sides)×h }}$

Solution:-

$\longrightarrow$ $\bf 105 = \dfrac{1}{2} × (14+7) × h$

$\longrightarrow$ $\bf 210 = 21×h$

$\longrightarrow$ $\bf h = 10cm$

hence, the altitude of this trapezium is 10cm

Extra information:-

• Area of rectangle :-

$\longrightarrow$ $\underline{\boxed{\sf area\: of \: rectangle = 2 × (length + breadth)}}$

• Area of parallelogram:-

$\longrightarrow$ $\underline{\boxed{\sf area\: of \: parallelogram = base×height }}$

• area of square :-

$\longrightarrow$ $\underline{\boxed{\sf area\: of \: square = side² }}$

• area of triangle :-

$\longrightarrow$ $\underline{\boxed{\sf area\: of \: triangle =\dfrac{1}{2} × base × height }}$

• Area of rhombus:-

$\longrightarrow$ $\underline{\boxed{\sf area\: of \: rhombus =\dfrac{1}{2} × (product\: of \: diagonals)}}$