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Triangles

Short Answer Type

211. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

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212. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4.

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213.

Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(a) 2 : 3 (b) 4 : 9
(c) 81 : 16 (d) 16 : 81.

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Long Answer Type

214. Sides of some triangles are given below. Determine which of them are right triangles. In case of a right angle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm.

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215. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR show that PM² = QM × MR.

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216.
In the given fig, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB² = BC .BD
(ii) AC² = BC. DC
(iii) AD² = BD . CD

Proof : (i) In ∆BAC and ∆BDA,
∠BAC = ∠BDA    ...(i)
In right triangle ABC, we have
∠BAC + ∠CBA = 90°    ...(ii)
In right triangle ABD, we have
∠BDA + ∠CBA = 90°    ...(iii)
Comparing (ii) and (iii), we get
∠BAC = ∠BDA
And,    ∠ACB = ∠DAB
[Each equal to 90°]
∴    ∆BAC ~ ∆BDA
[Using AA similar condition]
∴ $BA over BD equals BC over BA$
[∵ Corresponding sides of two similar triangles are proportional]
⇒    BA² = BC . BD
⇒    AB² = BC . BD.

(ii) In right triangle ACB and DCA,
∠ACB = ∠DCA = 90°
∠BAC = ∠ADC
∴    ∆ACB ~ ∆DCA
[Using AA similar condition]
∴    $AC over DC equals BC over AC$
[∵ Corresponding sides of two similar triangles are proportional]
⇒    AC² = BC × DC.

(iii) In right triangle ADB and ∆CDA,
∠DAB = ∠DCA = 90°
∠BDA = ∠ADC (common)
∴ ∆ADB ~ ∆CDA
[Using AA similar condition]
∴ $AD over CD equals BD over AD$