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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