Question

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

Answer 1

Answer:

The Area of quadrilateral ABCD equals the sum of the areas of triangles BAD and BCD.

the Area of an equilateral triangle is A = √(3)s²/4

Area of triangle BCD = √(3)26²/4 cm² = 169√3 cm²

The area of a right angle triangle A = (b*h)/2

the height of BAD is AD, and the base is AB

BD² = AD² + AB²

∴ AB = √(BD² − AD²)

∴ AB = √(26² − 24²) cm

∴ AB = √(100) cm

∴ AB = 10 cm

Area of triangle BAD = (10*24)/2 cm² = 120 cm²

The Area of quadrilateral ABCD = 120 + 169√3 cm² ≈ 412.7 cm² (1 dp)

hope it's helpful for you mate

Please follow me

and please mark it as a brainlist

Answer 2

$\large\underline\pink{Diagram:-}$

$\large\underline\pink{Given:-}$

• ABCD is any quadrilateral
• AD is 24 cm and ∆BAD ⇢90°
• ∆BCD forms an equilateral triangle of side 26 cm

$\large\underline\pink{To find:-}$

• find Area of the Quadrilateral ABCD ...?

$\large\underline\pink{Solutions:-}$

✰ In ∆ABD.

• AB ⇢ 24 cm
• BD ⇢ 26 cm

»★ By using Pythagoras theorem:

$\: \: \: \: \: \pink{\star \: \: \: {perpendicular}^{2} \: + \: {Base}^{2} \: \: = \: \: {Hypothesis}^{2}}$

$\: \: \: \: \: \leadsto \: \: {AD}^{2} \: + \: {AB}^{2} \: \: = \: \: {BD}^{2}$

$\: \: \: \: \: \leadsto \: \: {24}^{2} \: + \: {AB}^{2} \: \: = \: \: {26}^{2}$

$\: \: \: \: \: \leadsto \: \: {AB}^{2} \: \: = \: \: {26}^{2} \: - \: {24}^{2}$

$\: \: \: \: \: \leadsto \: \: {AB}^{2} \: \: = \: \: {676} \: - \: {576}$

$\: \: \: \: \: \leadsto \: \: {AB}^{2} \: \: = \: \: {100}$

$\: \: \: \: \: \leadsto \: \: {AB} \: \: = \: \: {\sqrt{100}}$

$\: \: \: \: \: \leadsto \: \: {AB} \: \: = \: \: {10}$

✰ The area of triangle ABD.

$\: \: \: \: \: \pink{\star \: \: \: \frac{1}{2} \: \times \: base \: \times \: height}$

$\: \: \: \: \: \leadsto \: \: \frac{1}{2} \: \times \: AD \: \times \: AB$

$\: \: \: \: \: \leadsto \: \: \frac{1}{2} \: \times \: {24} \: \times \: {10}$

$\: \: \: \: \: \leadsto \: \: {12} \: \times \: {10}$

$\: \: \: \: \: \leadsto \: \: {120} \: {cm}^{2}$

$\: \: \: \: \: \pink{\star \: \: The \: \: area \: \: of \: \: triangle \: \: ABD \: \: is \: \: {120} \: {cm}^{2}}.$

»★ Area of triangle BCD.

• Side ⇢ 26 cm.

$\: \: \: \: \: \pink{\star \: \: \: Area \: \: of \: \: \triangle \: \: BCD \: \: \leadsto \: \: \frac{\sqrt{3}}{4} \: \times \: {sides}^{2}}$

$\: \: \: \: \: \leadsto \: \: \frac{\sqrt{3}}{4} \: \times \: {26}^{2}$

$\: \: \: \: \: \leadsto \: \: \frac{\sqrt{3}}{4} \: \times \: {676}$

$\: \: \: \: \: \leadsto \: \: {\sqrt{3}} \: \times \: {169}$

$\: \: \: \: \: \leadsto \: \: {169}{\sqrt{3}}$

$\: \: \: \: \: \leadsto \: \: {292.71}$

✰ Area of Quadrilateral ABCD.

$\: \: \: \: \: \pink{\star \: \: Area \: \: of \: \: ABCD \: \: = \: \: Area \: \: of \: \: ABD \: + \: Area \: \: of \: \: BCD}.$

$\: \: \: \: \: \leadsto \: \: {120} \: + \: {292.71}$

$\: \: \: \: \: \leadsto \: \: {412.71} \: {cm}^{2}$

$\: \: \: \: \: \star \: \: \: Hence,$

$\: \: \: \: \: \pink{\star \: \: \:Area \: \: of \: \: Quadrilateral \: \: ABCD \: \: is \: \: {412.71} \: {cm}^{2}}.$

$\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star\star$