Question

seg PR ⊥ seg BC, seg AS ⊥ seg BC and seg QT ⊥ seg BC
Find the following ratios:
(i) $\frac{\mathrm{A}\left(∆\mathrm{ABC}\right)}{\mathrm{A}\left(∆\mathrm{PBC}\right)}$
(ii) $\frac{\mathrm{A}\left(∆\mathrm{ABS}\right)}{\mathrm{A}\left(∆\mathrm{ASC}\right)}$
(iii) $\frac{\mathrm{A}\left(∆\mathrm{PRC}\right)}{\mathrm{A}\left(∆\mathrm{BQT}\right)}=\frac{RC}{QT}$
(iv) $\frac{\mathrm{A}\left(∆\mathrm{BPR}\right)}{\mathrm{A}\left(∆\mathrm{CQT}\right)}$

Answer 1

(i)
From the figure, we observe that ΔABC and ΔPBC have the same base BC.
We know that the ratio of the areas of two triangles with the same base is equal to the ratio of their corresponding heights.
Thus, we get:
$\frac{\mathrm{A}\left(∆\mathrm{ABC}\right)}{\mathrm{A}\left(∆\mathrm{PBC}\right)}=\frac{\mathrm{AS}}{\mathrm{PR}}$

(ii)
From the figure, we observe that ΔABS and ΔASC have the common height AS.
We know that the ratio of the areas of two triangles with a common height is equal to the ratio of their corresponding bases.
Thus, we get:
$\frac{\mathrm{A}\left(∆\mathrm{ABS}\right)}{\mathrm{A}\left(∆\mathrm{ASC}\right)}=\frac{\mathrm{BS}}{\mathrm{SC}}$

(iii)
From the figure, we observe that ΔPRC has base RC and height PR and ΔBQT has base BT and height QT.
We know that the ratio of the areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.
Thus, we get:
$\frac{\mathrm{A}\left(∆\mathrm{PRC}\right)}{\mathrm{A}\left(∆\mathrm{BQT}\right)}=\frac{RC}{QT}$

(iv)
From the figure, we observe that ΔBPR has base BR and height PR and ΔCQT has base TC and height QT.
We know that the ratio of the areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.
Thus, we get:

$\frac{\mathrm{A}\left(∆\mathrm{BPR}\right)}{\mathrm{A}\left(∆\mathrm{CQT}\right)}=\frac{\mathrm{BR}\times \mathrm{PR}}{\mathrm{TC}\times \mathrm{QT}}$