Question on Triangles Congruency

Topic : Congruency of Triangle

Question to Find Length of the Graph of a Function

Topic : Functions and Graphical Representation

Question : Find the arc length of the graph of the function over the indicated interval.
y = 2x^(3/2) + 3 Length of the arc on [0,8]

Solution :
We know that the arc length of curve formed the curve of f(x) in the interval [a, b] is given by

Convex and Non-Convex Polygons

Topic : Polygons

Question : Draw the shape of convex and non-convex polygons.

Solution :
Polygons are as shown below,

Dimensions of Rectangle when Value of Diagonal is Given

Topic : Dimension of Rectangle

Question : Find the dimensions of rectangle RECT if the diagonals intersect at a 60º angle and RC = 18√5 inches. Leave your answer in simplest radical form.

Solution :

Given RC = 18√5 inches

Since RECT is a rectangle both diagonals are equal and bisect eachother

we have OE = OC = OT = OR = 1/2(18√5) = 9√5inches

Consider triangle OEC as
∟EOC = 60º and OE = OC

It is an equilateral triangle. Hence EC = OE = OC = 9√5

To find TC :

Consider triangle OCT; using sine rule

9√5 / Sin 30º = TC / Sin 120º

TC = (Sin 120º * 9√5 ) / Sin 30º

= [(√3/2) * 9√5] /1/2

= 9√5 * √3

= 9√15 inches

So the sides TC = RE = 9√15 inches

Hence the dimensions of RECT are 9√15 inches , 9√5 inches

Question to Find the Unknown Value of a Angles

Topic : Similar Triangles

Question :

If two triangles with vertex A and ECD are similar to each other. If ∟A + ∟ECD = 80º , then find ∟B + ∟D ?

Solution :

∟∆BAC ~ ∆ECD
So , ∟A = ∟EDC
∟B = ∟DEC
∟ACB = ∟D

AB = AC (as ∟BAC is an isosceles triangle)
DE = DC(as ∟ECD is an isosceles triangle)
So, AB = ∟B = ACB
And ∟DEC = ∟ECD

Now ∟A + ∟ECD = 80º (given)
So, ∟A = ∟ECD = 80º/2 = 40º (as ∟A = ∟ECD)

In ∆ABC, ∟A + ∟B + ∟ACB = 180º
40º + ∟B + ∟B = 180º (as ∟A = 40º, ∟B = ∟ACB)
2 ∟B = 180º - 40º
∟B = 140º / 2
∟B =70º

Similarly, in ∆ECD, ∟D =70º
So, ∟B + ∟D = 70º + 70º =140º

Question on Profit and Loss

Topic : Profit and Loss

Question : Vimmi buys a sewing machine for Rs 3725 and spends Rs 275 on repairs. She then sells it for Rs 5000. Find her profit percent or loss percent.

Solution :
C.P = Rs 3725
Expenses = Rs 275
So C.P = Rs 3725 + Rs 275 = Rs 4000
S.P = Rs 5000
Profit = S.P – C.P
Profit = Rs 5000 – Rs 4000
Profit = Rs 1000
Profit Percent = (Profit Amt x 100 ) / C.P
Profit Percent = (1000 x 100) / 4000
Profit Percent = 100 / 4
Profit Percent = 25 %

Question on Factorisation using Identities

Topic : Factorisation using identities

Problem : Factorise 49p^2 – 36

Solution :
We know that (a + b)^ 2 = a^2 + 2ab + b^2 …………… (I)
(a – b) 2= a^2 – 2ab + b62………….. (II)
(a + b) (a – b) = a^2 – b^2………… (III)
Let us use these identities for factorisation.

There are two terms; both are squares and the second is negative. The expression is of the form (a^2 – b^2). Identity III is applicable here;
49 p^2– 36 = (7p)^ 2 – (6)^2
= (7p – 6) (7p + 6) (required factorisation)