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

- `("A"(∆"PQB"))/("A"(∆"PBC"))`
- `("A"(∆"PBC"))/("A"(∆"ABC"))`
- `("A"(∆"ABC"))/("A"(∆"ADC"))`
- `("A"(∆"ADC"))/("A"(∆"PQC"))`
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Solution
(i) In ∆PQB and ∆PBC,
ΔPQB and ΔPBC have same height PQ. ...(Given)
Areas of triangles with equal heights are proportional to their corresponding bases.
∴ `("A"(∆"PQB"))/("A"(∆"PBC")) = "BQ"/"BC"`
(ii) In ∆PBC and ∆ABC,
∆PBC and ∆ABC have same base BC. ...(Given)
Areas of triangles equal bases are proportional to their corresponding heights.
∴ `("A"(∆"PBC"))/("A"(∆"ABC")) = "PQ"/"AD"`
(iii) In ∆ABC and ∆ADC,
∆ABC and ∆ADC have same height AD. ...(Given)
Areas of triangles with equal heights are proportional to their corresponding bases.
∴ `("A"(∆"ABC"))/("A"(∆"ADC")) = "BC"/"DC"`
(iv) In ∆ADC and ∆PQC,
Ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.
∴ `("A"(∆"ADC"))/("A"(∆"PQC")) = "DC × AD"/"QC × PQ"`
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In the figure, PQ ⊥ BC, AD ⊥ BC. To find the ratio of A(ΔPQB) and A(ΔPBC), complete the following activity.

Given: PQ ⊥ BC, AD ⊥ BC
Now, A(ΔPQB) = `1/2 xx square xx square`
A(ΔPBC) = `1/2 xx square xx square`
Therefore,
`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`
= `square/square`
