Question

the length of two adjacent sides of a parallelogram are 17 cm and 12 CM one of its diagonal is 25 cm long find the area of parallelogram also find the length of altitude from vertex on the side of length 12 CM ............. can u give me the answer of second point

Answer 1

Area - A×B=17×12=204

Answer 2

A typical triangle (i.e. not rectangle) can be divided into either a pair of acute triangle or obtuse triangle by its diagonals, depending on which you choose. The relationship between the length of either triangle, i.e. the length of the two sides or the parallelogram and the diagonal that completes the triangle is given by the Cosine rule or Law: c^2 = a^2 + b^2 - 2abCos(C); Where a, b, and c are sides of a triangle and C represents the angle opposite side c.

If we insert the given sides of the triangle into the Cosine rule we get 26^2 = 11^2 +17^2 -2*11*17*Cos(C)

Since 26^2 = 626 and is greater than 11^2+17^2= 121+289=410; 626>410 it is clear the term with Cos(C) must be positive and hence Cos(C) must be negative. Since the domain of our angles are from 0° to 180° this implies C is an obtuse angle (90°< C <180°) and C is the longer of the two diagonals and separates the parallelogram into obtuse triangles. The diagonal we seek will then be the shorter one.

We note on the domain 0° to 180° Cos(180-C) = - Cos(C).

Lets call the shorter diagonal c’ and the angle opposite it C’.

C and C’ are supplementary from the basic properties of parallelograms. Hence C’ = 180 - C.

Using cosine rule again for the acute triangle.

c’^2 = a^2 + b^2 - 2abCos(C’)

c’^2 = a^2+b^2–2abCos(180-C) = a^2+b^2 -2ab[-Cos(C)] = a^2+b^2+2abCos(C)

referring to the earlier equation:26^2 = 11^2 +17^2 -2*11*17*Cos(C)

we can rearrange it to get 2abCos(C) = 2*11*17Cos(C) = 11^2+17^2- 26^2 = 121+289–676 = 410–676 = -266

now c’^2 = a^2+b^2+2abCos(C) = 11^2+17^2 -266 = 410–266 = 144

taking the square root of both sides we get

c’ = √c’^2 = ±√144 = ±12

Taking the positive value we get c’ the shorter diagonal is 12 cm.