Q2: Sides of a Triangle are in the ratio of 14 : 20: 25 and its perimeter is 590cm. Find its area.
Answers
Answer:
Draw a line BC parallel to AD
Draw a perpendicular line BE on DF
ABCD is a parallelogram.
BC=AD=25cm
CD=AB=60cm
CF=77-CD=17cm
For Δ BCF
perimeter of triangle=(25+26+17)/2
=68/2=34
By Heron's Formula of triangle=√s(s-a)(s-b)(s-c)
=√34(34-25)(34-26)(34-17)
=204cm²
Now area of ΔBCF=1/2xbase x height
=1/2 BExCF204cm²
=1/2x BEx17BE
=408/17
=24cm
Area of Trapezium =1/2(AB+DF)xBE
=1/2(60+77)x24
=1644cm²
Answer:
let us consider the common ratio between the sides of the triangle be "a"
therefore the sides are 14a , 20a and 25a
perimeter = 590 cm
14a + 20a + 25a = 590cm14a+20a+25a=590cm
59a = 59059a=590
\: a \: = 10a=10
now the sides of the triangles are 140cm ,200cm ,250cm
so,the semi perimeter of the triangle (s)
= \frac{590}{2}=2590
= 295cm=295cm
Using Heron's formula for area of the triangle
\sqrt{s(s - a)(s - b)(s - c)}s(s−a)(s−b)(s−c)
= √295(295-140)(295-200)(295-250)
= \sqrt{295 \times 155 \times 95 \times 45}=295×155×95×45
= \sqrt{195474375}=195474375
= 13981.21 {cm}^{2}=13981.21cm2