Question

ABCD is a trapezium in which AB || BC, AD = 10m, BC = 18m and CD = 17m. Find the are of trapeziumm.​

Answer 1

Correct Question:

ABCD is a trapezium in which AB || BC, AD = 10m, BC = 18m and CD = 17m. Find the area of trapezium.

Answer :-

• The area of trapezium is 210m².

Step-by-step explanation:

To Find :-

• The area of trapezium.

Solution:

Given that,

• ABCD is a trapezium
• AB || BC
• CD = 17m
• AD = 10m
• BC = 18m

In the given figure, DE ⊥ BC, therefore, ∆DEC is a right triangle and the measure of EC is,

$\bigstar \: \: \bf{(EC = BC - BE)} \: \: \sf{and} \: \: \{ \bf{BE = AD = 10m} \}$

$\sf{ \mapsto \: EC = BC - BE} \\ \\ \sf{ \mapsto \: EC = 18m - 10m} \\ \\ \sf{ \mapsto \: EC = (18 - 10)m} \\ \\ \mapsto \: \sf{ \orange{ \: EC = 8m}}$

Hence, EC = 8m. Now, from ∆DEC we will find out DE,

$\boxed{\bf{DE = \sqrt{{DC}^{2} + {EC}^{2}}}}$

Where,

• DC = 17cm
• EC = 8cm

$\mapsto \: \bf{DE = \sqrt{{DC}^{2} - {EC}^{2}}} \\ \\ \\ \mapsto \: \bf{DE = \sqrt{{17}^{2} - {8}^{2}}} \\ \\ \\ \mapsto \: \bf{DE = \sqrt{289 - {EC}^{2}}} \\ \\ \\ \mapsto \: \bf{DE = \sqrt{289 - 64}} \\ \\ \\ \mapsto \: \bf{DE = \sqrt{225}} \\ \\ \\ \mapsto \: \bf{ \red{DE = 15m}}$

Hence, DE = 15m. Now, The area of trapezium :-

As we know that,

$\boxed{\bf{Area \: of \: trapezium = \dfrac{1}{2} \times (AD + BC) \times DE}}$

Where,

• AD = 10m
• BC = 18m
• DE = 15m.

According the question,

$\bf{\mapsto \: \dfrac{1}{2} \times (AD + BC) \times DE} \\ \\ \\ \bf{\mapsto \: \dfrac{1}{2} \times (10 + 18) \times 15} \\ \\ \\ \bf{\mapsto \: \dfrac{1}{2} \times 28 \times 15} \\ \\ \\ \bf{\mapsto \: \dfrac{1}{\cancel{2}} \times \cancel{28} \times 15} \\ \\ \\ \bf{\mapsto \: 1 \times 14 \times 15} \\ \\ \\ \bf{\mapsto \: 14 \times 15} \\ \\ \\ \bf{\mapsto \: \red{{210m}^{2}}}$

Therefore, The area of trapezium is 210m².