In figure 1.74, PM = 10 cm, A(\(\triangle\)PQS) = 100 sq.cm, A(\(\triangle\)QRS) = 110 sq.cm then find NR.
Figure: P and R lie on opposite sides of seg QS. PM ⊥ QS and RN ⊥ QS, with M and N on seg QS.
Answer: NR = 11 cm
\(\triangle PQS\) and \(\triangle QRS\) have the common base QS. Their heights are PM and NR respectively.
Areas of triangles with equal bases are proportional to their heights:
\[\frac{A(\triangle PQS)}{A(\triangle QRS)} = \frac{PM}{NR}\] \[\frac{100}{110} = \frac{10}{NR}\] \[NR = \frac{10 \times 110}{100} = 11\]Log in or create a free account to answer this question.
