the sides of quadrilateral, taken in order are 5,12,14 and 15 meters respectively and the angle contained by the first two sides is a right angle. find it's area
Answers
Given:-
- Sides of quadrilateral are 5, 12, 14 and 15 m.
- The angle contained by the first two sides is a right angle.
════◄••❀••►════
To Find:-
- Area of the quadrilateral
════◄••❀••►════
Formula Used:-
- Area of right angle triangle = ½ × Base × Height
- Area of triangle = By Heron's Formula
- Pythagorean theorem = h² = p² + b²
════◄••❀••►════
Solution:-
➥ Let the quadrilateral be ABCD.
Here;
AB = 5m
BC = 12m
CD = 14m
DA = 15m
════◄••❀••►════
➥ Area of ∆ABC
= ½ × Base × Height
= ½ × BC × AB
= ½ × 12 × 5
= 60 ÷ 2
= 30m²
════◄••❀••►════
Now,
➥ By Pythagorean Theorem
(AC)² = (BC)² + (AB)²
(AC)² = (12)² + (5)²
(AC)² = 144 + 25
AC = 169
AC = 13m.
════◄••❀••►════
➥ For Area of ∆ADC
Value of s
½(AD + AC + CD)
= ½(15 + 13 + 14)
= ½ × 42
= 42 ÷ 2
= 21m
➥ Area of ∆ADC by Heron's Formula
= s(s - AD)(s - AC)(s - CD)
= 21(21 - 15)(21 - 13)(21 - 14)
= 21 × 6 × 8 × 7
= 7 × 3 × 3 × 2 × 2 × 2 × 2 × 7
= 7 × 3 × 2 × 2
= 21 × 4
= 82m²
════◄••❀••►════
➥ Area of quadrilateral ABCD
= Area of ∆ABC + Area of ∆ADC
= 30 + 82
= 112m²
════◄••❀••►════
Answer:-
Area of quadrilateral is 112m²
════◄••❀••►════
Answer:
Given that sides of quadrilateral are AB = 5 m, BC = 12 m, CD = 14 m and DA = 15 m AB = 5m, BC = 12m, CD = 14 m and DA = 15 m Join AC In ∆ABC By applying Pythagoras theorem. By using Heron’s formula Area of quadrilateral ABCD = area of ( ∆ABC) + Area of (∆ADC) = 30+84 = 114 m2Read more on Sarthaks.com - https://www.sarthaks.com/110477/the-sides-of-a-quadrilateral-taken-in-order-are-5-12-14-and-15-meters-respectively