Question

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides

Answer 1

Required answer :

Question:

❥ The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:

Given,

❥ Hypotenuse of right angled triangle = 13 cm

To find:

❥ Other two sides of triangle

Assumptions:

❥ Let base of triangle = x

( As altitude is 7 cm less than its base )

So,

❥ Altitude of triangle = (x - 7)

Step by step explaination:

Using Pythagoras theorem

As, we know

(Perpendicular)² + (Base)² =(Hypotenuse)²

AB² + BC² = AC²

x² + (x - 7)² = (13)²

2x² - 14x + 49 = 169

2x² - 14x + 49 - 169 = 0

x² - 7x - 60 = 0

Now, factorising it.

So, it will be

x - 12x + 5x - 60 = 0

(x - 12) (x + 5) = 0

Thus, x = 12 , x = -5

Taking value of x as 12 because a side of a triangle can't be negative.

Altitude of triangle = (x - 12) → (12 -7)

= 5 cm

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Answer:

Other two sides of triangle are:-

Base = 12 cm

Altitude or perpendicular = 5 cm

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Note:

Diagram is in the attachment

Answer 2

As we are given with the altitude of a right triangle is 7cm less than its base and the hypotenuse of a right triangle is 13cm.

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And, we need to find out the other two sides of right angle triangle.

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⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

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⚘ Let us consider that, the base of right triangle be x cm.

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Given that,

• The altitude/perpendicular of a right triangle is 7cm less than its base.

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Therefore,

$\sf{ \leadsto \: \: Perpendicular = \textsf{\textbf{(x - 7)cm.}}}$

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◗ Let's head to the Question Now:

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We know that, if we are given with the hypotenuse, altitude/perpendicular & base of a right triangle, we have the required formula, that is,

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$\sf{:\implies (Hypotenuse)^2 = (Perpendicular)^2 + (Base)^2}$

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By using the required formula and plugging all the given values in the formula, we get:

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$\sf{:\implies (13)^{2} = (x - 7)^{2} + {(x)}^{2} } \\ \\ \\ \sf{:\implies 169 = {(x - 7)}^{2} + {(x)}^{2} } \\ \\ \\ \sf{:\implies 169 = {x}^{2} + {7}^{2} - 2 \times 7x +{x}^{2} } \\ \\ \\ \sf{:\implies 169 = {x}^{2} + 49 - 14x + {x}^{2} } \\ \\ \\ \sf{:\implies 169 = {x}^{2} + {x}^{2} + 49 - 14x} \\ \\ \\ \sf{:\implies 169 = {2x}^{2} + 49 - 14x} \\ \\ \\ \sf{:\implies 169 - 49 = 2x^{2} - 14x} \\ \\ \\ \sf{:\implies 120 = 2x^{2} - 14x} \\ \\ \\ \sf{:\implies 2x^{2} - 14x - 120 = 0} \\ \\ \\ \sf{:\implies {x}^{2} - 14x - 60 = 0} \\ \\ \\ \sf{:\implies x - 12x + 5x - 60 = 0} \\ \\ \\ \sf{:\implies (x - 12)(x + 5) = 0} \\ \\ \\ \sf{:\implies \frak{ \red{x = 12} \quad | \quad \red{x = - 5}}}$

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We know that, length can not be negative. So, we take x = 12.

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Therefore,

• Base, x = 12cm.
• Altitude, x - 7 = 12 - 7 = 5cm.

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$\sf{\therefore \underline{The \; other\; two \;sides\; of \;right \;triangle \;is \;\textsf{\textbf{12cm}}\; and\; \textsf{\textbf{5cm}}\; respectively.}}$