Question

Find area of the following polygons​

Answer 1

Answer:

answer of your question is 39.5 cm²

like if you need explaination

Answer 2

GIVEN :-

• FD = 50 cm.
• EJ = 2 cm.
• DI = 4 cm.
• FC = 7 cm.
• FB = 6.5 cm.
• HB = 4 cm.
• AG = 2 cm.

TO FIND :-

• The area of the polygon ABCDEF.

SOLUTION :-

In ∆ ABF we have,

• Altitude , AG = 2 cm.
• Base , FB = 6.5 cm.

Now , as we know that the area of triangle is given by,

$\\ : \implies \displaystyle \sf \: Area_{(ABF)} = \frac{1}{2} \times base \times height$

$: \implies \displaystyle \sf \: Area_{(ABF)} = \frac{1}{2} \times 6.5 \times 2$

$: \implies \displaystyle \sf \: \underline{ Area_{(ABF)} = 6.5 \: cm ^{2} }$

Now in ∆ BCF we have,

• Altitude , BH = 4 cm.
• Base , CD = 7 cm.

$\\ : \implies \displaystyle \sf \: Area_{(BCF)} = \frac{1}{2} \times base \times height$

$: \implies \displaystyle \sf \: Area_{(BCF)} = \frac{1}{2} \times 7 \times 4$

$: \implies \displaystyle \sf \: Area_{(BCF)} = 7 \times 2$

$: \implies \underline{\displaystyle \sf \: Area_{(BCF)} = 14 \: cm ^{2} }$

Now , in ∆ CDF we have,

• Height , DI = 4 cm.
• Base , FC = 7 cm.

$\\ : \implies \displaystyle \sf \: Area_{(CDF)} = \frac{1}{2} \times base \times height$

$: \implies \displaystyle \sf \: Area_{(CDF)} = \frac{1}{2} \times 7 \times 4$

$: \implies \displaystyle \sf \: Area_{(CDF)} = 7 \times 2$

$: \implies \underline{\displaystyle \sf \: Area_{(CDF)} = 14 \: cm ^{2} }$

Now in ∆ DEF we have,

• Height , EJ = 2 cm.
• Base , FD = 5 cm.

➳ Area of DEF = ½ × Base × height

➳ Area of DEF = ½ × 5 × 2

➳ Area of DEF = 5 cm²

Now , according to question we have to find the area of given Polygon. So,

➳ Area of polygon = Sum of Areas of figures inside it i.e, (DEF + CDF + BCF + ABF)

➳ Area of polygon = 5 + 14 + 14 + 6.5

➳ Area of polygon = 39.5 cm²

Hence the area of the given polygon ABCDEF is 39.5 cm².