Question

8) Find the area of a quadrilateral ABCD in which AD = 24 cm, <BAD = 90°
and BCD forms an equilateral triangle whose each side is equal to 26 cm .​

Answer 1

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❇ Required Answer:

✏ GiveN:

• ABCD is any quadrilateral.
• AD = 24 cm
• < BAD = 90°
• BCD forms an equilateral triangle of side 26 cm

✏ To FinD:

• Area of the quadrilateral.

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❇ How to solve?

The quadrilateral is form of two kind of triangles. One is triangle ABD, a right angled triangle and another one is BCD, an equilateral triangle. To Find the area of the quadrilateral, we need to find the areas of this triangles.

✒We all know that,

$\large{ \boxed{ \rm{Area \: of \: \triangle = \frac{1}{2} \times base \times height}}}$

In a right angled triangle, we have the base as well as height, so we can calculate its area by this method.

✒ For a triangle whose all sides are given, the area is calculated by heron's formula. But, in a equilateral triangle, all the sides are equal. So, by simplifying the heron's formula, we get

$\large{ \boxed{ \rm{Area \: of \: eq. \: \triangle = \frac{ \sqrt{3} }{4} \times {(side)}^{2} }}}$

By applying these formula, we can find the areas of triangle and eventually the area of the quadrilateral.

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❇ Solution:

According to question,

Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle ACD.

So, let's find the area of triangles first,

In triangle ABD We have,

• AD = 24 cm
• BD = 26 cm

By applying pythagoras theoram

$\large{ \rm{ \longrightarrow {AD}^{2} + {AB}^{2} = {BD}^{2} } } \\ \\ \large{ \rm{ \longrightarrow \: {24}^{2} + {AB}^{2} = {26}^{2} }} \\ \\ \large{ \rm{ \longrightarrow \: AB= \sqrt{ {26}^{2} - {24}^{2} } \: cm }} \\ \\ \large{ \rm{ \longrightarrow \: \boxed{ \rm{AB= 10 \: cm}}}}$

Finding the area of triangle ABD,

By using formula,

$\large{ \rm{ \longrightarrow \: Area \: of \: \triangle \: ABD = \frac{1}{2} \times AD \times AB}} \\ \\ \large{ \rm{ \longrightarrow \: Area \: of \: \triangle \: ABD= \frac{1}{2} \times 24 \times 10 \: {cm}^{2} }} \\ \\ \large{ \rm{ \longrightarrow \: \boxed{ \rm{Area \: of \: \triangle \: ABD = 120 \: {cm}^{2} }}}}$

Now, For triangle BCD, side = 26 cm

By using formula,

$\large{ \rm{ \longrightarrow \: Area \: of \: \triangle \: BCD = \frac{ \sqrt{3} }{4} \times {side}^{2} }} \\ \\ \large{ \rm{ \longrightarrow \: Area \: of \: \triangle \: BCD = \frac{ \sqrt{3} }{4} \times {26}^{2} {cm}^{2} }} \\ \\ \large{ \rm{ \longrightarrow \: Area \: of \: \triangle \: BCD = 169 \sqrt{3} \: {cm}^{2} }} \\ \\ \large{ \rm{ \longrightarrow \: \boxed{ \rm{Area \: of \: \triangle \: BCD \approx 292.71 \: {cm}^{2} }}}}$

So, Area of quadrilateral ABCD,

$\large{ \rm{ \longrightarrow \: Area \: of \: ABCD = Area \: of \: ABD + Area \: of \: BCD}} \\ \\ \large{ \rm{ \longrightarrow \: Area \: of \: ABCD = 120 {cm}^{2} + 292.71 {cm}^{2} }} \\ \\ \large{ \rm{ \longrightarrow \: \boxed{ \rm{ \red{ Area \: of \: ABCD \approx \: 412.71 \: {cm}^{2} }}}}}$

Hence, the quadrilateral is having an area of 412.71 cm².

$\large{ \therefore{ \underline{ \underline{ \rm{ \green{Hence \: solved \: \dag}}}}}}$

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