Question

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Find the area of the following polygon if AB = 12 cm, AC = 2.4 cm, CE = 6 cm, AD = 4.8 cm, CF = GE = 3.6 cm, DH = 2.4 cm.

Note:
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Answer 1

Answer:

GIVEN:

AP = 4 cm, BQ = 6 cm and AC = 9 cm. Lengths of the tangent from an exterior point to a circle are equal.

AR = AP = 4 cm [ From A]..(1)

BQ = BP = 6 cm [From B]...

(2)

CR = CO = 5 cm [From C]..(3)

[AC = AR + RC, 9 = 4 cm + RC, RC = 9 -

= 5 cm]

Adding equations 1, 2 & 3. AR + BQ + CR = AP + BP + CQ

Perimeter of AABC = AB + BC + AC

Perimeter of AABC = (AP +PB) + ( CQ +

BQ) + (AR + RC)

Perimeter of AABC = (AP + AR) + (PB +

BQ )+ (CQ + RC)

Perimeter of AABC = (AP + AP) + (PB +

PB)+ (CQ + CQ)[FROM EQUATION 1, 2 AND 3]

Perimeter of AABC = 2AP + 2PB +2CQ Perimeter of AABC = 2(AP + PB +CQ) AP + PB +CQ = 2(Perimeter of AABC) 4 + 6 +5 = 2(Perimeter of AABC)

[FROM EQUATION 1, 2 AND 3] 15 = 2(Perimeter of AABC)

Hence, semi perimeter of AABC is 15 cm

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Answer 2

Given

• A polygon
• AB = 12 cm
• AC = 2.4 cm
• CE = 6 cm
• AD = 4.8 cm
• CF = GE = 3.6 cm
• DH= 2.4 cm

_____________________________

To Find

• The Area

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Solution

We can observe that the above diagram consists of triangles and rectangles, so we will find their area separately and add them up.

Let's first find the area for the triangle GEB.

GE = 3.6 cm

BE = AB - AE

GB = x

Let's find BE.

BE = AB - AE

BE = 12 - (AC + CE)

BE = 12 - (2.4 + 6)

BE = 12 - 8.4

BE = 3.6 cm

∴ BE = 3.6 cm

Area of triangle = $\dfrac{1}{2} \times Base \times Height$

⇒ $\dfrac{1}{2} \times 3.6 \times 3.6$

⇒ $\dfrac{1}{2} \times 12.96$

⇒ $6.48$

∴ The area of ΔGEB is 6.48 cm²

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Now let's find the area of rectangle GECF.

GE = CF = 3.6 cm

GF = CE = 6 cm

Area of rectangle = $Length \times Breadth$

⇒ 3.6 × 6

⇒ 21.6

∴ The area of rectangle GECF is 21.6 cm²

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Now let's find the area of triangle CAF.

CF = 3.6 cm

FA = x

CA = AB - CB

CA = 12 - CB

CA = 12 - (CE + EB)

CA = 12 - (6 + 3.6)

CA = 12 - 9.6

CA = 2.4 cm

∴ CA = 2.4 cm

Area of triangle = $\dfrac{1}{2} \times Base \times Height$

⇒ $\dfrac{1}{2} \times 2.4 \times 3.6$

⇒ $\dfrac{1}{2}\times 8.64$

⇒ $4.32$

∴ The area of ΔCAF is 4.32 cm²

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Let's find the area of triangle HDB

HD = 2.4 cm

BH = x

DB = AB - DA

DB = 12 - 4.8

DB = 7.2 cm

∴ DB = 7.2 cm

Area of triangle = $\dfrac{1}{2} \times Base \times Height$

⇒ $\dfrac{1}{2}\times 7.2 \times 2.4$

⇒ $\dfrac{1}{2} \times 17.28$

⇒ $8.64$

∴ The area of ΔHDB is 8.64 cm²

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Let's find the area of triangle AHD

AH = x

AD = 4.8 cm

HD = 2.4 cm

Area of triangle = $\dfrac{1}{2} \times Base \times Height$

⇒ $\dfrac{1}{2} \times 4.8 \times 2.4$

⇒ $\dfrac{1}{2} \times 11.52$

⇒ $5.76$

∴ The area of ΔAHD is 5.76 cm²

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Now that we have got the area of all the figures in the diagram we will add the area of all of them to find the final answer.

6.48 cm² + 21.6 cm² + 4.32 cm² + 8.64 cm² + 5.76 cm² = 46.8 cm²

∴ The area of the given polygon is 46.8 cm²

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