Question

$\huge \sf {QUESTION:}$
In the attacchment given above:
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Answer 1

Required Answer:-

A quadrilateral can be divided into two triangles and if you know how to calculate the areas of traingles, we can add them to get the area of the quadrilateral.

Here:

• Join the diagonal BD.
• We will get ∆DAB and ∆DBC
• ∆DAB is a right angled triangle whose perpendicular and base are given. So, we can find the area by 1/2 × base × height formula.
• ∆DBC is a scalene triangle whose all sides are given. We can use the √s(s-a)(s-b)(s-c) formula.

Let's start..

Area of quadrilateral ABCD = Area of ∆DAB + Area of ∆DBC$.$

Area of ∆DAB:

= 1/2 × b × p

= 1/2 × 12 cm × 5 cm

= 30 cm²

Area of ∆DBC:

Here, s = (13 + 14 + 15)/2 cm = 21 cm

Putting the values in formula,

= √{s(s-a)(s-b)(s-c)}

= √{21(21-13)(21-14)(21-15) cm²

= √(21 × 8 × 7 × 6) cm²

= √(3 × 7 × 2³ × 7 × 2 × 3) cm²

= 3 × 7 × 2² cm²

= 84 cm²

Hence,

Area of quadrilateral ABCD

= 30 cm² + 84 cm²

= 114 cm²

Note:-

• The diagonal BD is the hypotenuse of ∆DAB. Find the length using Pythagoras theorem.
• s is the semiperimeter i.e. perimeter/2.
• a,b,c in √{s(s-a)(s-b)(s-c)} are the sides of the scalene triangle.

Answer 2

$\large\underline{\underline{\maltese{\red{\pmb{\sf{ \: Given :-}}}}}}$

• ➙ Figure in the attachment .

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$\large\underline{\underline{\maltese{\gray{\pmb{\sf{ \: To \: Find :-}}}}}}$

• ➙ Area the the given figure .(i.e, Quadrilateral)

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$\large\underline{\underline{\maltese{\purple{\pmb{\sf{ \: Solution :-}}}}}}$

✴ How to Solve :

We will apply a diagonal BD .After the the figure will be divided into 2 adjacent triangles namely DAB and DCB . We will find the area of the both triangles and add them .We will get the Area of the Quadrilateral.

$\qquad{━━━━━━━━━━━━━━━━━━━━━━━━━━}$

✴ Formula Used :

Area :

$\large\blue{\bigstar{\underline{\boxed{\red{\sf{Area {\small_{(Triangle)}} = \dfrac{1}{2} \times b \times h}}}}}}$

Area :

$\large\blue{\bigstar{\underline{\boxed{\red{\sf{Area{\small_{(Triangle)}} = \sqrt{s ( s - a)(s - b)(s - c)}}}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━━━━━━━}$

✴ Area of Triangle DAB :

$\qquad{\twoheadrightarrow{\sf{Area = \dfrac{1}{2} \times b \times h}}}$

$\qquad{\twoheadrightarrow{\sf{ \: \: \: \: \: \dfrac{1}{2} \times 12 \times 5}}}$

$\qquad{\twoheadrightarrow{\sf{ \: \: \: \: \: \dfrac{1}{\cancel2} \times \cancel{60}}}}$

$\large{\red{:\longmapsto{\green{\underline{\sf{ Triangle \: DAB = 30 cm}}}}}}$

✴ Area of Triangle DCB :

Semi - Perimeter :

$\qquad{\twoheadrightarrow{\sf{S = \dfrac{a + b + c}{2} }}}$

$\qquad{\twoheadrightarrow{\sf{S = \dfrac{13 + 14+ 15}{2} }}}$

$\qquad{\twoheadrightarrow{\sf{S = \dfrac{42}{2} }}}$

$\qquad{\twoheadrightarrow{\sf{S = \cancel\dfrac{42}{2} }}}$

$\qquad{\twoheadrightarrow\orange{\sf{S =21 \: cm }}}$

Area :

$\qquad{\twoheadrightarrow{\sf{Area = \sqrt{s ( s - a)(s - b)(s - c)}}}}$

$\qquad{\twoheadrightarrow{\sf{Area = \sqrt{21( 21- 13)(21 - 14)(21- 15)}}}}$

$\qquad{\twoheadrightarrow{\sf{Area = \sqrt{21 \times 8 \times 7 \times 6}}}}$

$\large{\red{:\longmapsto{\green{\underline{\sf{ Triangle \: DCB = 84 \: {cm}^{2} }}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━━━━━━━}$

✴ Area of Quadrilateral :

$\qquad{\twoheadrightarrow{\sf{Area {\small_{(Quadrilateral)}} = Sum \: of \: both \: triangles}}}$

$\qquad{\twoheadrightarrow{\sf{Area {\small_{(Quadrilateral)}} = {84 \: cm}^{2} + {</u><u>3</u><u>0</u><u> \: cm}^{2} }}}$

$\large{\red{:\longmapsto{\green{\underline{\sf{ Area {\small_{(Quadrilateral)}} = 105 cm²}}}}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

✴ Therefore :

❝ Area of Quadrilateral is 114 cm² . ❞

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