Question

seg PS is perpendicular to seg RQ, seg QT is perpendicular to seg PR. If RQ=6,PS=6 and PR=12,then find QT. this question is from 10th stf geometry practise set 1.1, 3rd sum. solve it. it is from ‘similarity’ chp

Answer 1

Answer:

Step-by-step explanation:

RQ=6, PS=6 & PR=12..... (GIVEN)

AREA OF TRIANGLE =1/2x BASE x HEIGHT

AREA OF TRIANGLE =1/2 x 6 x 6

THERFORE, AREA OF TRIANGLE PQR =18 SQ UNITS.....(1)

ALSO, AREA OF TRIANGLE PQR =1/2 x PR x QT

THERFORE, 18 =1/2 x 12 x QT.... (FROM 1)

THEREFORE QT =18 DIVIDE BY 6

THEREFORE QT = 3

DONE.....

Answer 2

keoeisjdidjdjdijdjdidjjdidjdksosis