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Solution for In the Given Figure, ∠B < 90° and Segment Ad ⊥ Bc, Show that - CBSE Class 10 - Mathematics

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Question

In the given figure, ∠B < 90° and segment AD ⊥ BC, show that

(i) b= h+ a+ x- 2ax

(ii) b2 = a2 + c2 - 2ax

Solution

(i) Since AD perpendicular to BC we obtained two right angled triangles, triangle ADB and triangle ADC.

We will use Pythagoras theorem in the right angled triangle ADC

AC2 = AD2 + DC2              ......(1)

Let us substitute AD = h, AC = b and DC = (a − x) in equation (1) we get,

b2 = h2 + (a − x)2

b2 = h2 + a2 − 2ax + x2

b2 = h2 + a2 + 𝑥2 − 2ax                ......(2)

(ii) Let us use Pythagoras theorem in the right angled triangle ADB as shown below,

AB2 = AD2 + BD2                       ......(3)

Let us substitute AB = c, AD = h and BD = x in equation (3) we get,

c2 = h2 + x2

Let us rewrite the equation (2) as below,

b2 = h2 + x2 + a2 - 2ax                ......(4)

Now we will substitute h2 + x2 = c2 in equation (4) we get,

b2 = c2 + a2 - 2ax

Therefore, b2 = c2 + a2 - 2ax.

  Is there an error in this question or solution?

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Solution for question: In the Given Figure, ∠B < 90° and Segment Ad ⊥ Bc, Show that concept: Application of Pythagoras Theorem in Acute Angle and Obtuse Angle. For the course CBSE
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