In a right-angled, triangle the area of the square drawn on the Hypotenuse is $169 {c m}^{2}$ and the area of the square drawn on the Base is $25 {c m}^{2}$ . What is the length of the Height of the triangle in $c m$ ?

Open in App

Solution

Area of Square $= s i d e^{2}$

Area of Square drawn on Hypotenuse $= 169 c m^{2}$

$S i d e_{H} =$ Hypotenuse of $△ = \sqrt{169} = 13 c m$

Area of Square drawn on Base $= 25 c m^{2}$

$S i d e_{B} = B a s e o f △ = \sqrt{25} = 5 c m$

We know that in a right angled triangle, by Pythagoras theorem,

$H y p o t e n u s e^{2} = b a s e^{2} + h e i g h t^{2}$

$13^{2} = 5^{2} + h e i g h t^{2}$

$h e i g h t^{2} = 144 = 12^{2}$

$h e i g h t = 12 c m$

Suggest Corrections

Similar questions

Q. In a right-angled, triangle the area of the square drawn on the Hypotenuse is

289 {c m}^{2}

and the area of the square drawn on the Base is

225 {c m}^{2}

. What will be the area of the square drawn on the Height of the triangle in

{c m}^{2}

Q. If the Largest area of square among all the squares drawn on the sides of a right-angled triangle is

169 {c m}^{2}

, what is the length of the hypotenuse of the triangle in cm?

Q. What is the sum of areas of squares drawn on base and height of a right-angled triangle, if the length of the hypotenuse is

5 c m

? (Answer in terms of

{c m}^{2}

)

Q. Through a point on the hypotenuse of a right triangle lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is m times the area of the square. The ratio of the area of the other small right triangle to the area of the square is:

In a right angled triangle , if a perpendicular is drawn from right angle ti hypotenuse then prove that the area of the square of the perpendicular is equal to the area of the rectangle formed by the two segments of the hypotenuse .

Join BYJU'S Learning Program

In a right-angled, triangle the area of the square drawn on the Hypotenuse is 169cm2 and the area of the square drawn on the Base is 25cm2. What is the length of the Height of the triangle in cm?

Area of Square = side2Area of Square drawn on Hypotenuse =169 cm2SideH=Hypotenuse of △=√169=13 cmArea of Square drawn on Base =25 cm2SideB=Base of △=√25=5 cmWe know that in a right angled triangle, by Pythagoras theorem, Hypotenuse2=base2+height2132=52+height2height2=144=122height=12 cm

In a right-angled, triangle the area of the square drawn on the Hypotenuse is $169 {c m}^{2}$ and the area of the square drawn on the Base is $25 {c m}^{2}$ . What is the length of the Height of the triangle in $c m$ ?