Question

The parallel sides of a Trapezium are 58 cm and 42cm the other two sides are of equal length which is 17 cm find the area of trapezium​

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

The parallel sides of a trapezium are 58 cm and 42 cm the other two sides are of equal length which is 17 cm.

$\setlength{\unitlength}{1.2cm}\begin{picture}(8,2)\thicklines\put(8.6,3){\large{A}}\put(7.7,0.9){\large{B}}\put(10,0.7){\sf{\large{42cm}}}\put(10,3.1){\sf{\large{58cm}}}\put(13.1,0.9){\large{C}}\put(11.8,0.7){\large{R}}\put(8,1){\line(1,0){5}}\put(12,1){\line(0,2){2}}\put(9,3){\line(3,0){3}}\put(11,1.8){\sf{\large{h\:cm}}}\put(13,1){\line(-1,2){1}}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{D}}\end{picture}$

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The area of trapezium.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We have two sides are equal length AB = DC = 17cm

Difference in parallel sides are 58 cm - 42 cm = 16 cm

We have RC = 16/2 = 8 cm

$\underline{\underline{\bf{In\:\triangle DRC\::}}}$

$\underline{\bigstar\:{\boldsymbol{Using\:pythagoras\:theorem\::}}}$

$\mapsto\sf{(Hypotenuse)^{2} =(Base)^{2} +(Perpendicular)^{2} }\\\\\mapsto\sf{(17cm)^{2} =(8cm)^{2} +(h)^{2} }\\\\\mapsto\sf{(h)^{2} =(17cm)^{2} -(8cm)^{2}}\\\\\mapsto\sf{(h)^{2} =289cm^{2} -64cm^{2} }\\\\\mapsto\sf{(h)^{2} =225cm^{2} }\\\\\mapsto\sf{h=\sqrt{225cm^{2} } }\\\\\mapsto\sf{\pink{h=15\:cm}}$

$\underline{\bigstar\:{\boldsymbol{Area\:of\:trapezium\::}}}$

$\mapsto\sf{Area=\dfrac{1}{2} \times (sum\:of\:base)\times height}\\\\\\ \mapsto\sf{Area=\frac{1}{2} \times(AD+BC)\times DR}\\ \\\\ \mapsto\sf{Area=\dfrac{1}{2} \times (58+42)\times 15}\\\\\\\mapsto\sf{Area=\dfrac{1}{\cancel{2}} \times \cancel{100} \times 15}\\\\\\\mapsto\sf{Area=(50\times 15)cm^{2} }\\\\\\\mapsto\sf{\pink{Area=750cm^{2} }}$

Thus;

$\underbrace{\sf{The\:area\:of\:trapezium \:is\:750\:cm^{2} }}}}$

Answer 2

Question -

The parallel sides of a Trapezium are 58 cm and 42cm .

The other two sides are of equal length which is 17 cm .

Find the area of trapezium .

Solution -

This can be solved by two methods .

• Method 1 - .

See the above attached figure .

Here, we have the following information -

The parallel sides of a Trapezium are 58 cm and 42cm .

The other two sides are of equal length which is 17 cm .

Now, to find the area , we need to find the height of the trapezium .

AB = AD = 17 cm

So,

Let us find a Pythagorean triplet containing 17 as the largest number .

Here,

8, 15, 17 is the required Pythagorean triplet .It is an example of a primitive Pythagorean triplet .

AB = 58 cm .

DC = 42 cm .

Subtracting,

AE + FB = 58 - 42 = 16 cm.

Now, AE = FB .

Hence , AE = FB = 8 cm .

So, the third side = 17 ^ 2 - 8^2 = 15 cm

Hence , the required height of the Parallelogram is 15 cm.

Now we know that -

Area of A Parallelogram = ( 1 / 2) × { Sum Of \\ Sides } × { Distance Between Them }

=> { 1 / 2 } × 100 × 15

=> 50 × 15

=> 750 cm^2 .

____________________

• Method 2 -

Using the Brahmagupta's Theorem for area of Cyclic Quadrilaterals -

According to this theorem ,

Area = ( 1 / 2 ) × [ ( s - a ) × ( s - b ) × ( s - c ) × ( s - d ) ]

Where ,

S is the Semiperimeter .

a, b, c and d are the sides of the Quadrilateral .

S = { 42 + 58 + 17 + 17 } / 2

=> 67

Substuting these values into the formula above -

Area = ( 1 / 2 ) × [ ( 67 - 42 ) × ( 67 - 14 ) × ( 67 - 58 ) × ( 67 - 17 ) ]

=> Solving, we get the area of 750 square centimeters .

_________________________