Question

7. Two parallel sides of a trapezium are 60 cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of trapezium.

Answer 1

Answer

1644cm2

Explanation

area ABEC

s=225+26+17=34

area ΔBEC=34×9×8×17

=17×3×2×2=204cm2

area ΔBEC=21×17×h=204

h=17204×2=24

area trap. ABCD =21(60+77)×24=137×12=1644cm2

Answer 2

QUESTION :- Two parallel sides of a trapezium are 60 cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of trapezium.

ANSWER :-

Draw a line BC parallel to AD.

Draw a perpendicular line BE on DF.

ABCD is a parallelogram.

$\sf BC = AD = 25 \: cm$

$\sf CD = AB = 60 \: cm$

$\sf CF = 77 - CD = 17 \: cm$

We know that,

$\sf \fbox{s = \frac{Perimeter}{2}}$

Here, s = half perimeter

For △BCF

Semi - Perimeter =

$\sf \frac{(25 + 26 + 17)}{2}$

$\sf = \frac{68}{2}$

$\sf = 34$

By using Heron's formula,

$\sf \fbox{\sqrt{s(s - a)(s - b)(s - c)}}$

$\sf = \sqrt{34(34 - 25)(34 - 26)(34 - 17)}$

$= \sqrt{34 \times 9 \times 8 \times 17}$

$\sf = 17 \times 3 \times 2 \times 2$

$\sf = {204 \: cm}^{2}$

Area of Trapezium ABFD

$\sf = \frac{1}{2} (AB+DF)×BE$

$\sf = \frac{1}{2} (60 + 77) \times 24$

$\sf = \frac{1}{2} \times 137 \times 24$

$\sf = 134 \times 12$

$\sf \fbox{ = {1644 \: cm}^{2} }$

∴ The area of trapezium is 1644 cm².

[NOTE :- Refer the attachment for figure⤴️]

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