Advertisements
Advertisements
प्रश्न
The area of the isosceles triangle is 60 cm2, and the length of each one of its equal side is 13cm. Find its base.
Advertisements
उत्तर
Area of a isosceles triangle = `("b" xx sqrt (4"a"^2 - "b"^2))/4`
Here, area = 60, a = 13. Putting this in equation and squaring both sides, we get
(60 x 4)2 = b2 (4 x 13 x 13 - b2 )
⇒ 57600= b2 x (676 - b2) = 676 b2 - b4
⇒ b4 - 676b2 +57600 = 0
⇒ b4 -100b2 - 576b2 + 57600 = 0
⇒ b2 (b2 - 100) - 576 (b2 - 100) = 0
⇒ (b2 - 100 )(b2 - 576) = 0
⇒ b2 = 576 , b2 = 100
⇒ b = 24 , b = 10
⇒ Hence base can be either 24 or 10 cm.
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equations by factorization:
`(3x-2)/(2x-3)=(3x-8)/(x+4)`
Solve the following quadratic equation for x:
x2 − 4ax − b2 + 4a2 = 0
Solve the following quadratic equations by factorization:
\[\frac{16}{x} - 1 = \frac{15}{x + 1}; x \neq 0, - 1\]
If the roots of the equations \[\left( a^2 + b^2 \right) x^2 - 2b\left( a + c \right)x + \left( b^2 + c^2 \right) = 0\] are equal, then
If a and b are roots of the equation x2 + ax + b = 0, then a + b =
The values of k for which the quadratic equation \[16 x^2 + 4kx + 9 = 0\] has real and equal roots are
An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey the speed was increased by 40 km/hr. Write down the expression for the time taken for
The outward journey
Car A travels x km for every litre of petrol, while car B travels (x + 5) km for every litre of petrol.
Write down the number of litres of petrol used by car A and car B in covering a distance of 400 km.
A shopkeeper purchases a certain number of books for Rs. 960. If the cost per book was Rs. 8 less, the number of books that could be purchased for Rs. 960 would be 4 more. Write an equation, taking the original cost of each book to be Rs. x, and Solve it to find the original cost of the books.
Find the roots of the quadratic equation x2 – x – 2 = 0.
