Math, asked by anika4, 1 year ago

please solve que 1 and 2. plzzz it's urgent

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Answers

Answered by NabasishGogoi
3
1. Given,
∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m.
In diagram,
BD is joined.
Now,
In ΔBCD,
By applying Pythagoras theorem,
BD^2 = BC^ 2 + CD^2
⇒ BD^2 = 12^2 + 5^2
⇒ BD^2 = 169
⇒ BD = 13 m
Area of ΔBCD = 1/2 × 12 × 5 = 30 m2
Now,
Semi perimeter of ΔABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m
Using heron's formula,
Area of ΔABD = √s (s-a) (s-b) (s-c)
= √15(15 - 13) ( 15 - 9) (15 - 8) m 2
= √15 × 2 × 6 × 7 m2
= 6√35 m 2 = 35.5 m2 (approx)
Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30 m 2 + 35.5m 2 = 65.5m 2

2. Given,
AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm
In ΔABC,
By applying Pythagoras theorem,
AC 2 = AB2 + BC 2
⇒ 5 2 = 3 2 + 4 2
⇒ 25 = 25
Thus, ΔABC is a right angled at B.
Area of ΔBCD = 1/2 × 3 × 4 = 6 cm 2
Now,
Semi perimeter of ΔACD(s) = (5 + 5 + 4)/2 cm = 14/2 cm = 7 m
Using heron's formula,
Area of ΔABD = √s (s-a) (s-b) (s-c)
= √7(7 - 5) (7 - 5 ) (7 - 4 ) cm 2
= √7 × 2 × 2 × 3 cm 2
= 2√21 cm 2 = 9.17 cm2 (approx)
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔABD = 6 cm 2 + 9.17 cm 2 = 15.17 cm 2

NabasishGogoi: you mean Pythagoras theorem?
NabasishGogoi: sorry i went offline solving your problem.. now again online
NabasishGogoi: although you can do in that way too that you'd mentioned
NabasishGogoi: but its NCERT math book as far I remember.. and if you're from CBSE board then you should do in the method I've showed
NabasishGogoi: cause I'm a KVian hence have experience.. so better you do it by using heron's formula
NabasishGogoi: cause the chapter is based on heron's formula and if you use the simple triangle formula to find area they'll simply deduct your marks in exam if you're from CBSE board
NabasishGogoi: so your choice! but I suggest do it using heron's formula since the chapter is based on heron's formula.
anika4: thanks
anika4: now I understand
NabasishGogoi: you're welcome!
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