Question

$ABC$ is an equilateral traingle with side length 9 cm.

P, Q, & R, S trisect respectively the sides AC and AB.

Now $\Delta$ PUQ, $\Delta$ STR are equilateral triangles.

Again $U_1,U_2;Q_1,Q_2:T_1,T_2:S_1,S_2$ triset the respective sides PU, QU, RT, TS.

All the small triangles with verticles $U_3,U,Q_3,T_3,T,S_3$ are equilateral.

The total perimeter of the adjoining figure is

• A. 40 cm
• B. 37 cm
• C. 44 cm
• D. 64 cm

Answer 1

Answer: (B)

37 cm

In $△ABC$

$BC=9CM$

Since PQ and RS trisects the side AB and AC and the triangle is equilateral.Hence,

Since $(AR+SB+QC+AP==12cm)$

$\because$ $PQ$ trisect the side $AC$

$PQ=3cm$ and similarly $RS=3cm$

Now

$△PUQ$ Is equilateral, So,

$PU=PQ=UQ=3cm$

$U_1,U_2$ and $Q_1,Q_2$ trisect the side $PU\&QU$ of the $△PUQ$

Hence,

${PU}_1+U_{2}U+{UQ}_2+Q_{1}Q=4cm$

Similarly

${TT}_1+{RT}_2+{TS}_1+S_{2}S=4cm$

Again

$△U_1{U}_2{U}_3,△Q_1{Q}_2{Q}_3,△T_1{T}_2{T}_3\&△S_1{S}_2{S}_3$ are equilateral.

And in $△U_1{U}_2{U}_3,$

$U_1{U}_2=1cm$(Since $U_1{U}_2$ trisect $PU$)

Hence,

$U_1{U}_3+U_2{U}_3=2cm$

Similarly,$Q_1{Q}_3+Q_2{Q}_3=2cm$

$S_1{S}_3+S_2{S}_3=2cm$

$T_1{T}_3+T_2{T}_3=2cm$

$\therefore$ The total perimeter=

$P=9cm+12cm+4cm+4cm+2cm+2cm+2cm+2cm$

$P=37cm$