in the triangle ABC p 3,1 q 5, 6 and are -3, 2 are the midpoints of the ab BC and CA respectively find the vertices of the triangle ABC and show the area of triangle ABC is equals to 4 ×area of triangle pqr
Answers
Step-by-step explanation:
this is answer. use the area formula to prove the area is four times
Given: P (3,1), Q (5, 6) and R (-3, 2 ) are the midpoints of the AB, BC and CA
To find: the vertices of the triangle ABC, show the area of triangle ABC is equals to 4 ×area of triangle PQR
Solution:
- So now, It is given that the P,Q,R are the mid points of the sides of the triangle AB, BC,CA respectively.
- So let the co-ordinates of A be (X,Y), B be (L,M) and C be (O,P)
- So, to find X co-ordinates
( X + L ) / 2 = 3
X + L = 6 ................(i)
- Similarly,
( Y + M ) / 2 = 1
Y + M = 2 ........ ...(ii)
- Similarly,
( L + O ) = 10 . ...(iii)
( M + P ) = 12 .......(iv)
( O + X ) = -6 .........(v)
( P + Y ) = 4 ..........(vi)
- Now subtract (iii) from (i)
(X+L=6)-(O+L=10)
X-O=-4...................... ...(vii)
- We already know that
X+O=-6
- By adding (vii) and (v), we get:
2X=-10
X = -5
- By putting this value in (i)and (iii)
- We get
L=6+5=11
O=10-11=-1
- Solving for the Y co-ordinates
Y + M = 2
P + M = 12
- By subtracting (iv) from(ii)
Y - P = -10. ...(viii)
- Now by adding (vi) and(viii)
2Y=-6
Y=-3
- By putting this value in (ii) and (iv)
- M=2+3=5
P=12-5=7
- Therefore, the co-ordinates of the vertices are
A(X,Y)=A(-5,-3)
B(L,M,)=B(11,5)
C(O,P)=C(-1,7)
Area= (1/2) x { X1 (Y2-Y3) +X2 (Y3-Y1) + X3 (Y1-Y2) }
- Area of triangle formula
- Area OfΔABC=(1/2) {-5(5-7)+11(7+3)-1(-3-5)}
=(1/2)(10+110+8)
=1/2*128
=64 sq.units
- Area of ΔPQR=(1/2){ 3(6-2)+5(2-1)-3(1-6)
=(1/2) (12+5+15)
=(1/2)*32
=16sq. units
- By dividing arΔABC by arΔPQR we get
=AR(ABC)/AR(PQR)
=64/16
=4
Answer:
Hence, it is proved that the area ΔABC Is 4*area ΔPQR