Advertisements
Advertisements
Question
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`
Advertisements
Solution
Given: PQ ⊥ BC, AD ⊥ BC
Now, A(ΔPQB) = `1/2` × BQ × PQ
A(ΔPBC) = `1/2` × BC × PQ
Therefore,
`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx bb(BQ) xx bb(PQ))/(1/2 xx bb(BC) xx bb(PQ))`
= `bb(BQ)/bb(BC)`
APPEARS IN
RELATED QUESTIONS
In the following figure seg AB ⊥ seg BC, seg DC ⊥ seg BC. If AB = 2 and DC = 3, find `(A(triangleABC))/(A(triangleDCB))`

The ratio of the areas of two triangles with the common base is 14 : 9. Height of the larger triangle is 7 cm, then find the corresponding height of the smaller triangle.
In the following figure RP : PK= 3 : 2, then find the value of A(ΔTRP) : A(ΔTPK).

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.
In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD.

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

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

`(A(triangleABD))/(A(triangleABC))`
In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratio.

\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\]
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)` = ?
Ratio of corresponding sides of two similar triangles is 4 : 7, then find the ratio of their areas = ?
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
From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `(A(ΔABC))/(A(ΔBCD))` = ?

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

(i) `(A(ΔABD))/(A(ΔADC))`
(ii) `(A(ΔABD))/(A(ΔABC))`
(iii) `(A(ΔADC))/(A(ΔABC))`
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.
