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

Advertisements
Solution
Ratio of the areas of two triangles with common or equal heights is equal to the ratio of their corresponding bases.
`(A(triangleTRP))/(A(triangleTPK))`
`="RP"/"PK"`
`=3/2`
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.
The ratio of the areas of two triangles with common base is 6:5. Height of the larger triangle of 9 cm, then find the corresponding height of the smaller triangle.
In the given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that:

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

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.

Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle ?
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.

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

`(A(triangleABD))/(A(triangleABC))`
If ΔXYZ ~ ΔPQR then `(XY)/(PQ) = (YZ)/(QR)` = ?
Areas of two similar triangles are in the ratio 144 : 49. Find the ratio of their corresponding sides.
In fig., TP = 10 cm, PS = 6 cm. `(A(ΔRTP))/(A(ΔRPS))` = ?
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))` = ?

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))`
