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Question
∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio `("A"(Δ"MNT"))/("A"(Δ"QRS"))`.
Solution
∆MNT ~ ∆QRS ...(Given)
∴ ∠M ≅ ∠Q ...(Corresponding angles of similar triangles)(i)
In ∆MLT and ∆QPS,
∠M ≅ ∠Q ...[From (i)]
∠MLT ≅ ∠QPS ...(Each angle is of measure 90°)
∴ ∆MLT ~ ∆QPS ...(AA test of similarity)
∴ `"MT"/"QS" = "TL"/"SP" ...("Corresponding sides of similar triangles")`
∴ `"MT"/"QS" = 5/9`
Now, ∆MNT ~ ∆QRS ...(ii) (Given)
By Theorem of areas of similar triangles,
∴ `("A"(∆"MNT"))/("A"(∆"QRS")) = "MT"^2/"QS"^2`
∴ `("A"(∆"MNT"))/("A"(∆"QRS")) = ("MT"/"QS")^2`
∴ `("A"(∆"MNT"))/("A"(∆"QRS")) = (5/9)^2` ...[From (ii)]
∴ `("A"(∆"MNT"))/("A"(∆"QRS")) = 25/81`
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