Advertisements
Advertisements
प्रश्न
The perimeter of a triangle ABC is 37 cm and the ratio between the lengths of its altitudes be 6: 5: 4. Find the lengths of its sides.
Let the sides be x cm, y cm, and (37 - x - y) cm. Also, let the lengths of altitudes be 6a cm, 5a cm, and 4a cm.
Advertisements
उत्तर
Consider that the sides be x cm, y cm, and (37 - x - y) cm. also, consider that the lengths of altitudes be 6a cm, 5a cm, and 4a cm.
∴ Area of a triangle = `1/2` x base x altitude
∴ `1/2 xx x xx 6a = 1/2 xx y xx 5a = 1/2 xx ( 37 - x - y) xx 4a`
6x = 5y = 148 - 4x - 4y
6x = 5y and 6x = 148 - 4x - 4y
6x - 5y = 0 and 10x + 4y = 148
Solving both the equations, we have
X = 10 cm, y = 12 cm and ( 37 - x - y ) cm = 15cm.
APPEARS IN
संबंधित प्रश्न
Compute the area of trapezium PQRS is Fig. below.

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects
CD at F.
(i) Prove that ar (ΔADF) = ar (ΔECF)
(ii) If the area of ΔDFB = 3 cm2, find the area of ||gm ABCD.
In square ABCD, P and Q are mid-point of AB and CD respectively. If AB = 8cm and PQand BD intersect at O, then find area of ΔOPB.
The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is
Medians of ΔABC intersect at G. If ar (ΔABC) = 27 cm2, then ar (ΔBGC) =
The medians of a triangle ABC intersect each other at point G. If one of its medians is AD,
prove that:
(i) Area ( ΔABD ) = 3 x Area ( ΔBGD )
(ii) Area ( ΔACD ) = 3 x Area ( ΔCGD )
(iii) Area ( ΔBGC ) = `1/3` x Area ( ΔABC ).
The sides of a rectangular park are in the ratio 4 : 3. If its area is 1728 m2, find
(i) its perimeter
(ii) cost of fencing it at the rate of ₹40 per meter.
Look at a 10 rupee note. Is its area more than hundred square cm?
If each square on this page is equal to 1 square meter of land, how much land will each of her children get? ________ square m
Find all the possible dimensions (in natural numbers) of a rectangle with a perimeter 36 cm and find their areas.
