Question

Find the altitude of a Trapezium, the sum of the lengths of whose bases is 6.5 cm and whose area is 26 cm².​

Answer 1

Answer:

Altitudeofthe

trapezium(h)=8cm

Step-by-step explanation:

Let a and b are two parallel sides of a trapezium .

Given ,

a+b = 6.5 cm ,

Let altitude = h cm

Area of the trapezium (A)=26 cm²

$\frac{1}{2}\times (a+b)h=26$

2

1

×(a+b)h=26

$\implies \frac{1}{2}\times 6.5 \times h = 26⟹$

2

1

×6.5×h=26

$\begin{gathered}\implies h = \frac{26\times 2}{6.5}\\=4 \times 2 \\=8\:cm\end{gathered}$

⟹h=

6.5

26×2

=4×2

=8cm

Therefore,

$\begin{gathered} Altitude \: of \: the \\ trapezium (h) = 8 \:cm \end{gathered}$

Altitudeofthe

trapezium(h)=8cm

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Answer 2

Question :-

• Find the altitude of a Trapezium, the sum of the lengths of whose bases is 6.5 cm and whose area is 26 cm².

Given :-

• Sum of the bases = 6.5cm
• Area of Trapezium = 26cm²

To Find :-

• Altitude or Height of the Trapezium .

Solution :-

$\qquad \quad {:} \longrightarrow \sf{ \boxed{{\bf{ Area\:_{(Trapezium)}\: = \: \frac{1}{2} \: \times (Sum \: of \: bases) \: \times \: Altitude}}}}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ 26 \: = \: \frac{1}{2} \: \times \: 6.5 \: \times \: Altitude}}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{26}{6.5} \: = \: \frac{1}{2} \: \: \times \: Altitude}}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{26 \times \: 2 \: }{6.5} \: = \: Altitude}}$

$\qquad \quad {:} \longrightarrow \sf{\bf{ 8 \: = \: Altitude}}$

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Hence, the Altitude of the Trapezium = 8cm

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More For Knowledge :-

• Trapezium - A Trapezium is a quadrilateral whose two sides are parallel.
• Base - Each of two parallel sides of a Trapezium is called base of the trapezium.
• Height or Altitude - The distance between the two bases ( parallel sides ) is called the height or altitude of a Trapezium.

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