Similiarity

since PQ is 10 and PV is 6, VQ is 4 so TV is 2. since TVU = 1x1 and PVS is (6x6)/2, the shaded area will be 18-1 =17

Hello,

for the area of shaded region, A = (6 x 5 ) - ( 0.5 x 5 x 5 ) - ( 0.5 x 1 x 1) = 30 - 12.5 - 0.5 = 17 unit^(2)

thanks...

we have PV=PS=6.so PSV=(6 6)/2=18.now we to subtract the area of TUV.Here PV=QT=6 &PQ=10,so TV=2.so TUV=(2 1)/2=1.so the shaded area is PTUS=PSV-TUV =18-1 =17(answer)

Solve first the big triangle that can be formed by the shaded area. Add the area of the rectangle above it including a part that is not shaded so that a rectangle can be formed. Subtract the area of the small triangle that was included in the solution of the rectangle above the big triangle.

i using vector. when i knew big triagle whom shader is 0.5x6x6. then i am looking litle triange. and get the areas is 1. so 18-1 is 17

I using vector. when i knew big triagle whom shader is 0.5x6x6. then i am looking litle triange. and get the areas is 1. so 18-1 is 17

We calculate area of triangle PVS that is 1/2 base X heigft + 3 X6 =18 since triangle TUV is issocellous trianle with base 2 and height 1 so area = 1/2 X 2 X1 =1 Subtracting 1 from 18 we get area of shaded portion 18 -1 =17 Ans

K.K.GARG,India

PV=PS=6 Thus ar(PSV) = 18 Also TUV is right angled at U. Also PQ = 10 give VQ=4 anf TV=2 Thus by applying Pythagoras's theorem to triangle TUV, we get, TU=UV=sqrt(2) Thus ar(TUV)=sqrt(2)*sqrt(2)/2 = 1 Thus area of shaded region = 18-1=17

The area is required for the shared region PTUS. First find the area of the rectangle PQRS=10×6=60 Then find the area of triangle QTR=\frac{1}{2}×6×6=18 Next is the area of triangle SUR=\frac{1}{2}×10×5=25 Now the area will be=60-18-25=17