Question

The perimeter of a quadrilateral is 40 cm . if the first three sides of quadrilateral taken in order are 13cm , 10 cm , and 12cm respectively , and the angled between fourth side and the third side is a right angle . find the area of the quadrilateral.​

Answer 1

Answer:

90 cm²

Step-by-step explanation:

Consider the provided figure in the attachment.

As per provided question,

• First three sides taken in order are 13 cm , 10 cm and 12 cm.
• Perimeter of the quadrilateral is 40 cm.

Before going further, we need to calculate the length of the fourth side first.

$\longmapsto \rm { Perimeter_{(Quadrilateral)} = Sum_{(All \; sides)} }$

According to the question,

$\longmapsto \rm { 40 = 13 + 10 + 12 + 4th \; side}$

Performing the addition in R.H.S.

$\longmapsto \rm { 40 = 35 + 4th \; side}$

Transposing 35 from R.H.S to L.H.S, changing its sign.

$\longmapsto \rm { 40 -35= 4th \; side}$

Subtracting 35 from 40.

$\longmapsto \bf { 5 \; cm = 4th \; side}$

∴ Fourth side of the quadrilateral is 5 cm.

Now, let us split the quadrilateral in two triangles.So, the sum of the area of these two triangles is the area of the quadrilateral. ABC and ADC. AC is the diagonal of the quadrilateral.

In order to calculate the area of quadrilateral, we need to find the length of AC. Since ADC is a right-angled triangle, so by using pythagoras property,

$\longmapsto \rm { (AC)^2 = (AD)^2 + (DC)^2}$

Substituting the values.

$\longmapsto \rm { (AC)^2 = (5\; cm)^2 + (12\; cm)^2}$

$\longmapsto \rm { (AC)^2 = 25 \; cm^2 + 144\; cm^2}$

Performing addition.

$\longmapsto \rm { (AC)^2 = 169 \; cm^2}$

$\longmapsto \rm { AC = \sqrt{169 \; cm^2}}$

$\longmapsto \rm { AC = 13 \; cm}$

• Length of AC = 13 cm

Calculating the area of ∆ADC :

Since it is a right angled triangle, so perpendicular of the triangle is the height of the triangle.

• Height (AD) = 5 cm
• Base (DC) = 12 cm

By using the formula of area of triangle,

$\longmapsto \rm { Area_{(Triangle)} = \dfrac{1}{2} \times Base \times Height}$

Substituting values.

$\longmapsto \rm { Area_{(ADC)} =\Bigg ( \dfrac{1}{2} \times 12 \times 5 \Bigg ) \; cm^2}$

Performing multiplication.

$\longmapsto \rm { Area_{(ADC)} =\Bigg ( \dfrac{60}{2} \Bigg ) \; cm^2}$

Dividing 60 by 2.

$\longmapsto \bf \underline { Area_{(ADC)} = 30 \; cm^2}$

Finding area of ∆ABC :

By using heron's formula,

$\longmapsto \rm { Area_{(Triangle)} = \sqrt{s(s - a)(s -b )(s - c)} }$

Here,

• s is the Semi-perimeter of half of the perimeter.
• a, b and c are sides.

$\longmapsto \rm { s= \dfrac{Perimeter_{(ABC)}}{2} }$

Perimeter of triangle is the sum of all sides.

$\longmapsto \rm { s= \dfrac{(13+10+13)}{2} cm}$

Performing addition in the numerator.

$\longmapsto \rm { s= \dfrac{36}{2} cm}$

$\longmapsto \rm { s=18 \; cm}$

So, semi-perimeter of ABC is 18 cm. Now, substitute the values in the heron's formula.

$\longmapsto \rm { Area_{(ABC)} = \sqrt{18(18-13)(18-10 )(18-13)} \; cm^2 }$

Performing subtraction in the brackets.

$\longmapsto \rm { Area_{(ABC)} = \sqrt{18(5)(8 )(5)}\; cm^2 }$

Performing multiplication.

$\longmapsto \rm { Area_{(ABC)} = \sqrt{3600}\; cm^2 }$

Finding the square root of 3600.

$\longmapsto\bf \underline { Area_{(ABC)} = 60 \; cm^2 }$

Finding the area of the quadrilateral,

$\longmapsto \rm { Area_{(ABCD)} = Area_{(ABC)} + Area_{(ADC)} }$

Substituting the values of area we found above.

$\longmapsto \rm { Area_{(ABCD)} = (60+30) \; cm^2}$

$\longmapsto \bf { Area_{(ABCD)} = 90 \; cm^2}$

∴ Area of the quadrilateral is 90 cm².