1.Find the area of a triangle whose sides are respectively 9 cm, 12 cm and 15 cm.
2.A triangle has sides 35 cm, 54 cm, 61 cm long. Find its area.
Answers
Answer:
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Answer:
1.
We know that medians divide the triangle in
six equal areas.
Let ABC is a triangle ,in which medians AM=9
cm. BN=12 cm. and CP=15 cm.
Let medians meet at O.
OM=1/3 of AM =1/3×9cm= 3 cm.
ON=1/3 ×12 cm. = 4cm.
OP=1/3 ×15 cm.= 5 cm.
Produce AM upto S such MS = OM=3 cm.
join S to C and B.
In triangle COS , OS=OM+MS=3+3= 6 cm.
OC=2/3 ofCP=2/3 ×15 cm.= 10cm.
CS=OB= 2/3 ofBN =2/3 ×12 cm.= 8 cm.
s =(6+8+10)/2=12 cm.
Area of triangle COS =[s(s-a)(s-b)(s-c)]^1/2
=[12(12–6)(12–8)(12–10)]^1/2
=[12×6×4×2]^1/2
=24 cm^2.
Area of triangle COM = 1/2 ×Area of triangle COS.
Area of triangle COM =1/2 ×24 cm^2 =12 cm^2.
Area of triangle ABC = 6 ×Area of triangle COM
= 6 × 12 cm ^2
= 72 cm^2. Answer.
2. Use Heron’s formula
Let a,b,c be the lengths of the sides of a triangle. The area is given by:
Area=√[p(p−a)(p−b)(p−c)]where p is half the perimeter, or (a+b+c)/2
To find smallest altitude [h] use Area [from Heron’s formula]= 1/2* h* [longest side]