What is the ratio of the areas of two similar triangles if their corresponding altitudes are in the ratio 1 4?

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.

We have,

ΔABC ~ ΔPQR

AD = 6 cm

And, PS = 9 cm

By area of similar triangle theorem

`("Area"(triangleABC))/("Area"(trianglePQR))="AB"^2/"PQ"^2` ........(i)

In ΔABD and ΔPQS

∠B = ∠Q [ΔABC ~ ΔPQR]

∠ADB = ∠PSQ [Each 90°]

Then, ΔABD ~ ΔPQS [By AA similarity]

`therefore"AB"/"PQ"="AD"/"PS"` [Corresponding parts of similar Δ are proportional]

`rArr"AB"/"PQ"=6/9`

`rArr"AB"/"PQ"=2/3` ......(ii)

Compare equations (i) and (ii)

`("Area"(triangleABC))/("Area"(trianglePQR))=(2/3)^2=4/9`

Concept: Areas of Similar Triangles

Is there an error in this question or solution?

E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.(ii) PE = 4 cm, QE = 4.5 cm. PF = 8 cm and RF = 9 cm.(iii) PQ = 1.28 cm. PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

(i) We have,

From (i) and (ii), we have

Therefore, EF is not parallel to QR [By using converse of Basic proportionality theorem]
(ii) We have,

From (i) and (ii), we have

Therefore,

[Using converse of Basic proportionality theorem]

(iii) We have,

From (i) and (ii), we have

Therefore,

[Using converse of Basic proportionality theorem]

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