Question
Find the area of the polygon in the given picture, cm,
Hint:
The figure consists of 4 Right angled triangles and 2 trapeziums find the
areas of them individually and add their areas and add them up to get the area of polygon
So, Area of polygon = area of ΔMAN + area of ANOD+ area of ΔDOP +
area of ΔMBR + area of RBCQ + area of ΔQCP
The correct answer is: 31.75cm2
Ans :- 31.75cm2.
Explanation :-
Given ,MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm and MA = 2 cm.
OD =3 cm , NA = 2.5 cm , RB = 2.5 cm and QC = 2cm
Step 1:- Find the area of ΔMAN
Consider ΔMAN ,
we have MA = 2 cm and NA = 2.5 cm with right angle at A
Area of triangle ΔMAN = ½ × b × h = ½ × 2 × 2.5 = 2.5 cm2
Step 2:- Find the area of ANOD
Consider ANOD,
As NA and DO are perpendicular to AD , NA is parallel to DO ; AD is perpendicular to
both. Here ANOD is a Trapezium .parallel sides are DO and AD
Here AD is height ; AD = MD - MA = 7-2 = 5 cm
Area of trapezium ANOD
Step 3 :- Find the area of ΔDOP
Consider ΔDOP ,
We have OD = 3 cm and DP = MP-MD =9-7 = 2 cm with right angle at D
Area of triangle ΔDOP = ½ × b × h = ½ × 2 × 3 = 3 cm2
Step 4 :- Find the area of ΔMBR
Consider ΔMBR ,
we have MB = 4 cm and RB = 2.5 cm with right angle at B
Area of triangle ΔMAN = ½ × b × h = ½ × 4 × 2.5 = 5 cm2
Step 5 :- Find the area of RBCQ
Consider RBCQ,
As RB and CQ are perpendicular to BC, RB is parallel to CQ ; BC is perpendicular to
both. Here RBCQ is a Trapezium .parallel sides are RB and CQ
Here BC is height ; BC = MC - MB = 6 -4 = 2 cm
Area of trapezium ANOD
Step 6 :- Find the area of ΔQCP
Consider ΔQCP ,
We have QC = 2 cm and CP = MP-MC =9-6 = 3 cm with right angle at C
Area of triangle ΔDOP = ½ × b × h = ½ × 2 × 3 = 3 cm2
Step 7:-Find the area of the polygon.
Area of polygon = area of ΔMAN + area of ANOD+ area of ΔDOP +
area of ΔMBR + area of RBCQ + area of ΔQCP
Area of polygon = 2.5 +13.75 + 3 +5 + 4.5 + 3 = 31.75 cm2
Therefore, area of polygon is 31.75 cm2
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