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RD Chapter 4 Triangles Ex 4.7 Solutions

Question - 21 : - Determine whether the triangle having sides (a – 1) cm, 2 √a cm and (a + 1) cm is a right angled triangle. [CBSE 2010]

Answer - 21 : -

Sides of a triangle are (a – 1) cm, 2 √a cm and (a + 1) cm
Let AB = (a – 1) cm BC = (a + 1) cm
and AC = 2 √a

Question - 22 : - In an acute-angled triangle, express a median in terms of its sides.

Answer - 22 : -

In acute angled ∆ABC,
AD is median and AL ⊥ BC
Using result of theorem, sum of the squares of any two sides is equal to the twice the square of half of the third side together with twice the square of the median which bisects the third side

Question - 23 : - In right-angled triangle ABC in which ∠C = 90°, if D is the mid point of BC, prove that AB² = 4AD² – 3AC².

Answer - 23 : -

Given : In right ∆ABC, ∠C = 90°
D is mid point of BC

Question - 24 : - In the figure, D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that :

Answer - 24 : -


Question - 25 : -
In ∆ABC, ∠A is obtuse, PB x AC and QC x AB. Prove that:
(i) AB x AQ = AC x AP
(ii) BC² = (AC x CP + AB x BQ)

Answer - 25 : - Given : In ∆ABC, ∠A is an obtuse angle PB x AC and QC x AB on producing them

Question - 26 : - In a right ∆ABC right-angled at C, if D is the mid-point of BC, prove that BC² = 4 (AD² – AC²).

Answer - 26 : -

Given : In ∆ABC, ∠C = 90°
D is mid-point of BC
AD is joined
=> 4AD² = 4AC² + BC²
=> BC² = 4AD² – 4AC²
=> BC² = 4 (AD² – AC²)
Hence proved.

Question - 27 : - In a quadrilateral ABCD, ∠B = 90°, AD² = AB² + BC² + CD², prove that ∠ACD = 90°.

Answer - 27 : -

Given : ABCD is a quadrilateral in which ∠B = 90° and AD² = AB² + BC² + CD²
To prove : ∠ACD = 90°
Construction : Join AC
proof : In right ∆ABC, ∠B = 90°
AC² = AB² + BC² ….(i) (Pythagoras Theorem)
But AD² = AB² + BC² + CD² (given)
=> AD² = AC² + CD² {From (i)}
∆ACD is a right angle with right angle ACD
Hence proved.

Question - 28 : - An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/ hr. How far apart will be the two planes after 1(1/2) hours?

Answer - 28 : -


Speed of the first plane = 1000 km/hr
Distance travelled in 1(1/2) hour due north = 1000 x (3/2) = 1500 km
Speed of the second plane = 1200 km/hr
Distance travelled in 1(1/2) hours due west

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