Question

find the area of the trapezium given below.​

Answer 1

Answer :-

$\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 12\ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 15\ cm }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 10\ cm }\end{picture}$

Given :-

Trapezium ABCD with sides of length :-

• AB = 15 cm
• CD = 10 cm
• CB = 13 cm

To find :-

Area of trapezium

Formula used :-

Area of trapezium = ½ × Sum of parallel sides × Distance between the parallel sides

Solution :-

According to the question :-

Area of trapezium ABCD = ½ × ( AB + CD ) × CE

Applying Pythagoras theorem in ∆ CEB

EB = 15 - 10 = 5 cm

CB = 13 cm

$\implies\sf (EB)^2 + (CE)^2 = (CB)^2$

$\implies\sf 5^2 + (CE)^2 = 13^2$

$\implies\sf 25 + (CE)^2 = 169$

$\implies\sf (CE)^2 = 144$

$\implies\sf CE = \sqrt{144}$

$\implies\sf CE = 12 cm$

Now we have -

• AB = 15 cm
• CD = 10 cm
• CE = 12 cm

Substituting the value in formula :-

Area of trapezium ABCD = ½ × ( AB + CD ) × CE

$\sf = \frac{1}{2} \times (15 + 10) \times 12$

$\sf = \frac{1}{ \cancel 2} \times 25 \times 1\cancel 2$

$\sf = 25 \times 6$

$\sf = 150 cm^2$

$\boxed{\rm Area \:of \:trapezium = 150 cm^2 }$

Answer 2

Answer:

Answer:

अ + न् + ध + क + आ + र

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