A piece of ice with a stone frozen into it, is floating in a glass vessel filled with water. If the ice melts completely, then level of water in the vessel will

Solution :
Volume of melted ice=water  is less than ice ,then stone volume would remain unchanged ,therefore ,the level of water would fall

In ΔABC, tan(∠CAB) = , and the altitude from A divides CB into segments of length 3 and 17, then the area of ΔABC is

In ΔABC, tan(∠CAB) = , and the altitude from A divides CB into segments of length 3 and 17, then the area of ΔABC is
Solution :

Mnemonic for macro nutrients

See see head master not observing possible passing student(CC HM Not Observing Possible Passing Students)

C-CARBON
C -CALCIUM
H-HYDROGEN
M -MAGNESIUM
N-NITROGEN
O-OXYGEN
P-PHOSPHORUS
P-POTASSIUM
S -SULPHUR 

ABC is a triangle, the incircle touches the sides BC, CA and AB at D, E and F respectively. BD, CE and AF are consecutive natural numbers. I is the incentre of the triangles. The radius of the incircle is 4 units.

ABC is a triangle, the incircle touches the sides BC, CA and AB at D, E and F respectively. BD, CE and AF are consecutive natural numbers. I is the incentre of the triangles. The radius of the incircle is 4 units. The sides of the triangle are  13,14 and 15 unit ,prove

Solution :
Step1:Follow this figure

Step2:You can easily find that BC=2X+1,AC=2X+3 and AB=2X+2.....1
Step3:Find the area of ABC triangle by Heron's formula =[(3x+3)(x)(x+1)(x+2)]^(1/2).....2
Step4:Add area of three triangles namely BDC,CDA and ABD by using their base and perpendicular =12(x+1) square unit......3
Step5:(3x+3)(x)(x+1)(x+2)=144(X+1)^2........4
Step6:Solving 4 ,would give x=6,the sides of triangles are proved to be 13,14 and 15(Answ)

The base of a triangle is axis of x and its other 2 sides are given by the equations :y=[(l+α)/α]x + (1+α) and y=[(l+ß)/ß]x+(1+ß).Prove that the locus of its orthocenter is the line x+y=0

The base of a triangle is axis of x and its other 2 sides are given by the equations :y=[(l+α)/α]x + (1+α) and y=[(l+ß)/ß]x+(1+ß).Prove that the locus of its orthocenter is the line x+y=0

Solution :Steps 

Step 1:Follow this figure 

Step 2:

Step 3:

The eqaution of line which is perpendicular to a given line ax+by+c=0 is bx-ay+k=0 .This deduction would clarify the value of (x,y)=(αβ,-αβ) is the point where perpendicular from C to AB intersect line II.