## In adjoining figure PQ ⊥ BC, AD ⊥ BC then find following ratios.

(i) PQB and PBC have same height PQ.Ratio of areas of triangles with equal heights are proportional to their corresponding bases.A(PQB )/A(PBC) = BQ/BC (ii) PBC and ABC have same base BC.Ratio of areas of triangles with equal bases are proportional to their corresponding heights.A(PBC) / A( ABC) = PQ/AD (iii) ABC) and ADC have […]

## In adjoining figure, AP ⊥ BC, AD || BC, then find A( ABC):A( BCD).

Given , AP ⊥ BC, and AD || BC. ABC and BCD has same base BC.Areas of triangles with equal bases are proportional to their corresponding heights.Since AP is the perpendicular distance between parallel lines AD and BC, height of ABC and height of BCD are same.A( ABC) /A( BCD) = AP/AP = 1Hence A( […]

## In adjoining figure 1.14 seg PS ⊥ seg RQ, seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.

Given PS ⊥ RQ and QT ⊥ PR.RQ = 6PS = 6PR = 12Area of PQR with base PR and height QT = (1/2)×PR×QTArea of PQR with base QR and height PS = (1/2)×QR×PSA(PQR)/ A(PQR) = (1/2)×PR×QT /(1/2)×QR×PS1 = PR×QT/ QR×PS1 = 12×QT/6×66×6 = QT×12QT = 36/12QT = 3Hence measure of side QT is 3 […]

## In figure 1.13 BC⊥ AB, AD⊥ AB, BC = 4, AD = 8, then find A(ABC) /A( ADB) .

Here ABC and ADB have the same base AB.Areas of triangles with equal bases are proportional to their corresponding heights.Since bases are equal, areas are proportional to heights.Given BC = 4 and AD = 8So, A(ABC) /A( ADB) = BC/AD= 4/8= 1/2Hence ratio of areas of ABC and ADB is 1:2.

## Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.

Let base of the first triangle is b1 and height is h1. Let base of second triangle is b2 and height is h2.Ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.Here b1 = 9h1 = 5b2 = 10h1 = 6Then ratio of their areas = b1×h1/b2×h2= 9×5/10×6= 3/4Hence the ratio […]