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प्रश्न
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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उत्तर
(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"`
संबंधित प्रश्न
In the following figure seg AB ⊥ seg BC, seg DC ⊥ seg BC. If AB = 2 and DC = 3, find `(A(triangleABC))/(A(triangleDCB))`

In the following figure RP : PK= 3 : 2, then find the value of A(ΔTRP) : A(ΔTPK).

In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`

In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ

In ∆ABC, B - D - C and BD = 7, BC = 20 then find following ratio.

`"A(∆ ABD)"/"A(∆ ADC)"`
In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then `(A(Δ ABC))/(A(Δ DCB))` = ?

In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.

A roller of diameter 0.9 m and the length 1.8 m is used to press the ground. Find the area of the ground pressed by it in 500 revolutions.
`(pi=3.14)`
If ΔXYZ ~ ΔPQR then `(XY)/(PQ) = (YZ)/(QR)` = ?
In fig., TP = 10 cm, PS = 6 cm. `(A(ΔRTP))/(A(ΔRPS))` = ?
In fig. BD = 8, BC = 12, B-D-C, then `(A(ΔABC))/(A(ΔABD))` = ?

In fig., PM = 10 cm, A(ΔPQS) = 100 sq.cm, A(ΔQRS) = 110 sq.cm, then NR?

ΔPQS and ΔQRS having seg QS common base.
Areas of two triangles whose base is common are in proportion of their corresponding `square`.
`(A(ΔPQS))/(A(ΔQRS)) = (square)/(NR)`,
`100/110 = (square)/(NR)`,
NR = `square` cm
In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `(A(ΔABC))/(A(ΔBCD))` = ?

If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.
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`
