PQR is a triangle right angled at P and M is a point on QR such

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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.

In triangle PQR,∵ ∠P = 90° [Given]⇒ ∠2 + ∠3 =

In triangle PQR,
∵ ∠P = 90°    [Given]
⇒    ∠2 + ∠3 = 90°    ...(i)
∵    ∠PM ⊥ QR
Therefore, ∠1 + ∠2 = 90°    ...(ii)
Comparing (i) and (ii), we get
∠1 = ∠3
Similarly,    ∠4 = ∠2.
Now, in ∆PRM and ∆QPM,
∠1 = ∠3
and    ∠2 = ∠4
Therefore, by using AA similar condition

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rightwards double arrow

space PM over QM equals RM over PM

rightwards double arrow

PM squared space equals space QM cross times RM space space space Proved.

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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

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Short Answer Type

217. ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

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218. ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ∆ABC is a right triangle.

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Long Answer Type

219. ABC is an equilateral triangle of side 2a. Find each of its altitudes.

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220. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

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