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Area of the quadrilateral ABCD = Area of △ABC + Area of △ADC ….(1)
△ABC is a right-angled triangle, which is right-angled at B.
Area of △ABC = 1/2 x Base x Height
= 1/2×AB×BC
= 1/2×3×4
= 6
Area of △ABC = 6 cm
2
……(2)
Now, In △CAD,
Sides are given, apply Heron’s Formula.
Perimeter = 2s = AC + CD + DA
2s = 5 cm + 4 cm + 5 cm
2s = 14 cm
s = 7 cm
Area of the △CAD = 9.16 cm
2
…(3)
Using equations (2) and (3) in (1), we get
Area of quadrilateral ABCD = (6 + 9.16) cm
2
= 15.16 cm
2
.
Here,
AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m
AC is the diagonal joined at A to C point.
Now, in △ADC,
From Pythagoras theorem,
AC
2
= AD
2
+ CD
2
AC
2
= 14
2
+ 7
2
AC = 25
Now, area of △ABC
All the sides are known, Apply Heron’s Formula.
Perimeter of △ABC= 2s = AB + BC + CA
2s = 26 m + 27 m + 25 m
s = 39 m
= 291.84
Area of a triangle ABC = 291.84 m
2
Now, for the area of △ADC, (Right angle triangle)
Area = 1/2 x Base X Height
= 1/2 x 7 x 24
= 84
Thus, the area of a △ADC is 84 m
2
Therefore, the area of rectangular field ABCD = Area of △ABC + Area of △ADC
= 291.84 m
2
+ 84 m
2
= 375.8 m
2
Perimeter of △ADC = 2s = AD + DC + AC
2s = 15 m +14 m +13 m
s = 21 m
= 84
Area of △ADC = 84 m
2
Area of quadrilateral ABCD = Area of △ABC + Area of △ADC
= (30 + 84) m
2
= 114 m
2
Here, AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m.
And BD is a diagonal of ABCD.
Perimeter of △ABD = 2s = 9 m + 8m + 13m
s = 15 m
= 35.49
Area of the △ABD = 35.49 m
2
Area of quadrilateral ABCD = Area of △ABD + Area of △BCD
= (35.496 + 30) m
2
= 65.5m
2
.
Given: AB = 77 m , CD = 60 m, BC = 26 m and AD = 25m
AE and CF are diagonals.
DE and CF are two perpendiculars on AB.
Therefore, we get, DC = EF = 60 m
Let’s say, AE = x
Then BF = 77 – (60 + x)
BF = 17 – x …(1)
The perimeter of a rhombus = 80 m (given)
We know, Perimeter of a rhombus = 4×side
Let a be the side of a rhombus.
4×a = 80
or a = 20
One of the diagonal, AC = 24 m (given)
Therefore OA = 1/2×AC
OA = 12
In △AOB,
Using Pythagoras theorem:
OB
2
= AB
2
− OA
2
= 20
2
−12
2
= 400 – 144 = 256
or OB = 16
Since the diagonal of the rhombus bisect each other at 90 degrees.
And OB = OD
Therefore, BD = 2 OB = 2 x 16 = 32 m
Area of rhombus = 1/2×BD×AC = 1/2×32×24 = 384
Area of rhombus = 384 m
2
.
Then, OA = 1/2×AC
OA = 1/2×10
OA = 5 m
In right triangle AOB,
From Pythagoras theorem;
OB
2
= AB
2
–OA
2
= 8
2
– 5
2
= 64 – 25 = 39
OB = √39 m
And, BD = 2 x OB
BD = 2√39 m
Area of the sheet = 1/2×BD×AC = 1/2 x (2√39 × 10 ) = 10√39
The area of the sheet is 10√39 m
2
Therefore, the cost of printing on both sides of the sheet, at the rate of Rs. 5 per m
2
= Rs. 2 x (10√39 x 5)
= Rs. 625.
S = (3+4+5)/2 = 6
Semi perimeter is 6 cm
Now,
= 6
The area of the given triangle is 6 cm
2
.
In right triangle APC,
Using Pythagoras theorem,
AC
2
= AP
2
+ PC
2
y
2
= h
2
+ (x/2)
2
or h
2
= y
2
– (x/2)
2
