Advertisements
Advertisements
प्रश्न
A wire ; 112 cm long is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.
Advertisements
उत्तर
Perimeter of a right angled triangle = 112 cm
Hypotenuse = 50 cm
∴ Sum of other two sides = 112 – 50 = 62 cm
Let the length of first side = x
and length of other side = 62 – x
According to the condition
(x)2 + (62 – x)2 = (50)2 ...(By Pythagorus Theorem)
⇒ x2 + 3844 – 124x + x2 = 2500
⇒ 2x2 – 124x + 3844 – 2500 = 0
⇒ 2x2 – 124 + 1344 = 0
⇒ x2 – 62x + 672 = 0 ...(Dividing by 2)
⇒ x2 – 48x – 14x + 672 = 0
⇒ x(x – 48) –14(x - 48) = 0
⇒ (x – 48)(x – 14) = 0
Either x – 48 = 0,
then x = 48
or
x – 14 = 0,
then x = 14
(i) If x = 48,
then one side = 48cm
and other side = 62 – 48 = 14cm
(ii) If x = 14,
then one side = 14cm
and other side = 62 – 14 = 48
Hence sides are 14cm, 48cm.
APPEARS IN
संबंधित प्रश्न
Two number differ by 4 and their product is 192. Find the numbers?
The sum of two numbers is 18. The sum of their reciprocals is 1/4. Find the numbers.
Solve:
`1/(x + 1) - 2/(x + 2) = 3/(x + 3) - 4/(x + 4)`
The sum of the squares of two consecutive positive integers is 365. Find the integers.
Find the value of p for which the quadratic equation
\[\left( p + 1 \right) x^2 - 6(p + 1)x + 3(p + 9) = 0, p \neq - 1\] has equal roots. Hence, find the roots of the equation.
Disclaimer: There is a misprinting in the given question. In the question 'q' is printed instead of 9.
If a and b are roots of the equation x2 + ax + b = 0, then a + b =
In each of the following determine whether the given values are solutions of the equation or not.
x2 + x + 1 = 0; x = 1, x = -1.
Solve the following equation by factorization
`(x + 2)/(x + 3) = (2x - 3)/(3x - 7)`
Solve the following equation by factorization
`sqrt(3x + 4) = x`
Find two consecutive integers such that the sum of their squares is 61
