Right Triangle Altitude Theorem: Proof & Applications

Ans: It is given that the hypotenuse of a right triangle is 12 cm and the length of a part of it is 7 cm. So the length of the other part of the hypotenuse is (12-7)= 5cm. Now as, we know the altitude of the right triangle is the geometric mean of the line segments that it creates on the hypotenuse, i.e h = \(\sqrt{pq}\), where h is the altitude and p,q are the parts of the hypotenuse. Then putting the values of these we get,h=\(\sqrt{7 \times 5} = \sqrt{35} cm\).

Ans: We note that Q is the projection of P onto RS and that △PRS is a right triangle at P. We can then recall that the corollary of the Euclidean theorem tells us that PQ = RQ × SQ.

Substituting in the given lengths gives PQ = 4 × 9 = 36.

Taking the square root of both sides of the equation, noting that PR is a length and so is nonnegative, we get PQ = √36 =6

Ans: We start by recalling that the area of a triangle is half the length of the triangle’s base multiplied by the triangle’s perpendicular height. Since △HBK is a right triangle at K, we can choose GK as the base and then HK is the perpendicular height, giving us area △HBK = ½(GK×HK)

We can find the length of HK using the Euclidean theorem and then the length of GK using the Pythagorean theorem. First, we note that K is the projection of G onto HL and that △HGL is a right triangle at G . The Euclidean theorem then tells us that HG²=HL× HK.

We substitute HG=22cm and HL=33cm to get 22²=33×HK.

We then divide the equation by 33, giving us HK=\(\frac{22^2}{33}\)=14.6cm

We now have the lengths of two sides of the right triangle HGK.We can find the third length by applying the Pythagorean theorem, which gives HG² =GK²+HK².

Substituting HG=22 cm and HK =14.6cm gives 22²=14.6²+GK².

We can then rearrange and simplify the equation:GK²= 22²-(14.6)²=270.4 

We then take the square root of both sides of the equation, noting that BD is a length and so is nonnegative, which gives GK=√270.4=16.4cm

We now substitute HK=14.6cm and GK=16.4cm into the formula for the area of △HGK to get the area △HGK=½((16.4)×(14.6))=120.04.

Ans: Let us say, the base of the right triangle is y cm.

 Given, the altitude of the right triangle = (y – 7) cmGiven 

 From Pythagoras theorem, we know,

 base² + altitude² = hypotenusee²

∴y² + (y-7)² =169

⇒ y² + y² + 49 – 14y = 169

⇒ 2y² – 14y – 120 = 0

⇒ y² – 7y – 60 = 0

⇒ y² – 12y + 5y – 60 = 0

⇒ y(y – 12) + 5(y – 12) = 0

⇒ (y – 12)(y + 5) = 0

Thus, either y – 12 = 0 or y + 5 = 0,

⇒ y = 12 or y = – 5

Since sides cannot be negative, the only possible value of y is 12

Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be,

(12 - 7) = 5 cm.

Ans: Let's call the length of each leg "x".

According to the Pythagorean theorem, which applies to any right triangle, the length, 

Of the hypotenuse( c),

c² = a² + b²

where a and b are the lengths of the legs.

Since the legs are equal in this problem, we can rewrite this as:

c² = x² + x² = 2x²

We also know that c² = 98 cm², so we can substitute that in:

98 cm² = 2x²

To solve for x, we can divide both sides by 2:

98 cm² = x²

Taking the square root of both sides, we get:

x = √49 cm = 7 cm

So each leg of the right triangle is 5 cm long. 

Ans: In an ABC right triangle with a right angle at C, we have:

a is the hypotenuse (the side opposite the right angle)

b is the length of the side opposite angle B

c is the length of the side opposite angle A

We can use the Pythagorean theorem to find b:

b² = a² – c²

Since the triangle is an ABC right triangle, we know that:

c = h = 8

b² = a² – c² = 17² - 8² = 225

Taking the square root of both sides, we get:

b = 15

So the sides of the triangle are:

a = 17

b = 15

c = 8

The perimeter of the triangle is simply the sum of the lengths of its sides:

P = a + b + c = 15 + 8 + 17 = 40

Therefore, the perimeter of the ABC right triangle is 40 cm.

Ans: Given that length of one side is 3.5 cm

Length of the hypotenuse is 6.5 cm

Using Pythagoras' theorem the length of the third side will be 

\(\sqrt{(6.5)^2 – (3.5)^2} = \sqrt{30} cm\)

Ans: Identify the hypotenuse formed when the altitude

is drawn from the right angle to the hypotenuse.

The lengths of the segments given are 16 and 4.

 The geometric mean of the lengths of the segments is given by the formula x = \(\sqrt{ab}\)

Here a = 16 and b = 4

Thus, x = \(\sqrt{16(4)} =\sqrt{ 64}\) = 8 cm

Ans: In PQR, PQ = 6\(\sqrt{3}\) cm, PR = 12 cm and QR = 6 cm

PQ²= (6\(\sqrt{3}\))²=108

PR² = 12²= 144

QR²= 6² = 36

From this we obtain , 108 + 36 = 144 

In case of a right angled triangle the square of one side is equal to the sum of the squares of the

remaining two sides.

In a right angled triangle if one side is half of the hypotenuse then the angle opposite to that is 30^o

Here , QR is half of PR.

Hence, ∠P = 30^o, .

Ans: The Euclidean and the right triangle altitude theorem will be used to solve the problem.

Using these, we get b = \(\sqrt{x(AB)}\) , h = \(\sqrt{xy}\) and a = \(\sqrt{y(AB)}\)

Starting with , h = \(\sqrt{x(AB)}\), we get , b = \(\sqrt{x(x+7)}\)

Squaring both sides we get,

144 =x²+ 7x 

⇒ x²+ 7x -144 = 0

⇒ (x-9)(x+16) = 0

⇒ x=9, x =- 16 but x = -16 is not possible as length cannot be negative.

⇒ x= 9

AD = 9 cm

So, h = CD= \(\sqrt{xy} = \sqrt{9x7} = \sqrt{37}\) cm 

Now, a = BC = \(\sqrt{y(AB)} = \sqrt{7(16)}\)= 4\(\sqrt{7}\) cm

Ans: According to the Right Triangle Altitude theorem, h = \(\sqrt{xy}\)

Putting the values we get, h = \(\sqrt{xy}\) = \(​​\sqrt{ 15 \times 1}\) = \(\sqrt{15}\) cm 

Now using the Eucleadin theorem, b = \(\sqrt{15(16)}\) = b = 4\(\sqrt{5}\) cm